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UNIT-1 LESSON-1 : PREPARATION OF FREQUENCY DISTRIBUTION AND THEIR GRAPHICAL PRESENTATION 1. STRUCTURE 1.0 1.1 1.2 1.3 1.4 Objective What is Frequency Distribution Types of Frequency Distribution Principles of Constructing Frequency Distribution Graphs of Frequency Distributions 1.4.1 Histogram 1.4.2 Frequency Polygon 1.4.3 Smoothed Frequency Curves 1.4.4 Cumulative Frequency Curves or Ogives 1.5 Summary 1.6 Self Assessment Questions 1.0 OBJECTIVE After reading this lesson, you should be able to : (a) Learn a frequency distribution and types of distributions (b) Learn the principles and procedure of preparing a frequency distribution (c) Learn the graphical presentation of distribution with the help of histogram, frequency polygon, smoothed frequency curves and ogives. 1.1 WHAT IS FREQUENCY DISTRIBUTION Collected and classified data are presented in a form of frequency distribution. Frequency distribution is simply a table in which the data are grouped into classes on the basis of common characteristics and the number of cases which fall in each class are recorded. It shows the frequency of occurrence of different values of a single variable. A frequency distribution is constructed for satisfying three objectives : (i) to facilitate the analysis of data (ii) to estimate frequencies of the unknown population distribution from the distribution of sample data and (iii) to facilitate the computation of various statistical measures. 1.2 TYPES OF FREQUENCY DISTRIBUTION 1. Univariate Frequency Distribution 2. Bivariate Frequency Distribution 1 In this lesson, we shall discuss the Univariate frequency distribution. Univariate distribution incorporates different values of one variable only whereas the Bivariate frequency distribution incorporates the values of two variables only. The Univariate frequency distribution is classified further into three categories : (i) Series of Individual observations (ii) Discrete frequency distribution, and (iii) Continuous frequency distribution. Series of individual observations, is a simple listing of items of each observation. If marks of 20 students in statistics of a class are given individually, it will form a series of Individual observations. Marks obtained in Statistics: Roll Nos. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Marks : 60 71 80 41 94 33 81 41 78 66 85 35 61 55 98 52 50 91 30 88 Marks in Ascending Order Marks in Descending Order 30 98 33 94 35 91 41 88 41 85 50 81 52 80 55 78 60 71 61 66 66 61 71 60 78 55 80 52 81 50 85 41 88 41 91 35 94 33 98 30 2 Discrete Frequency Distribution : In a discrete series, the data are presented in such a way that exact measurements of units are indicated. In a discrete frequency distribution, we count the number of times each value of the variable in data given to you. This is facilitated through the technique of tally bars. In the first column, we write all values of the variable. In the second column, a vertical bar called tally bar against the variable, we write a particular value has occurred four times, for the fifth occurrence, we put a cross tally mark (/) on the four tally bars to make a block of 5. The technique of putting cross tally bars at every fifth repetition facilitates the counting of the number of occurrences of the value. After putting tally bars for all the values in the data; we count the number of times each value is repeated and write it against the corresponding value of the variable in the third column entitled frequency. This type of representation of the data is called discrete frequency distribution. We are given marks of 50 students : 70 55 51 42 57 40 26 43 46 41 46 48 33 40 26 40 40 41 43 53 45 60 47 63 53 33 50 40 33 40 26 53 59 33 65 78 39 55 48 15 26 43 59 51 39 15 45 26 61 15 We can construct a discrete frequency distribution from the above given marks. Marks of 50 Students Marks Tally Bars Frequency 15 ||| 3 26 |||| 5 33 |||| 4 39 40 || |||| 2 5 41 || 2 42 | 1 43 ||| 3 45 | 2 46 || 2 47 | 1 48 || 2 50 | 1 51 || 2 53 ||| 3 55 ||| 3 57 | 1 3 59 || 2 60 | 1 61 | 1 63 | 1 65 | 1 70 | 1 78 | 1 Total 50 The presentation of the data in the form of a discrete frequency distribution is better than arranging but it does not condense the data as needed and is quite difficult to grasp and comprehend. This distribution is quite simple in case the values of the variable are repeated otherwise there will be hardly any condensation. Continuous Frequency Distribution : If the identity of the units about a particular information is collected, is not relevant nor is the order in which the observations occur, then the first step of condensation is to classify the data into different classes by dividing the entire group of values of the variable into a suitable number of groups and then recording the number of observations in each group. Thus, if we divide the total range of values of the variable (marks of 50 students) i.e. 78 – 15 = 63 into groups of 10 each, then we shall get (63/10) 6 groups and the distribution of marks is displayed by the following frequency distribution : Marks of 50 students Marks (×) Tally Bars Number of Students (f) 15–25 25–34 35–45 ||| |||| |||| |||| |||| ||| 3 9 13 45–55 55–65 65–75 75–85 |||| |||| ||| 13 9 2 1 |||| |||| || | Total 50 The various groups into which the values of the variable are classified are known as classes, the length of the class interval (10) is called the width of magnitude of the class. Two values, specifying the class, are called the class limits. The presentation of the data into continuous classes with the corresponding frequencies is known as continuous frequency distribution. There are two methods of classifying the data according to class intervals : (i) exclusive method (ii) inclusive method 4 In an exclusive method, the class intervals are fixed in such a manner that upper limit of one class becomes the lower limit of the following class. Moreover, an item equal to the upper limit of a class would be excluded from that class and included subsequently in the next class. The following data are classified on this basis. Income (Rs.) No. of Persons 200–250 250–300 300–350 350–400 400–450 450–500 50 100 70 130 50 100 Total 500 It is clear from the example that the exclusive method ensures continuity of the data in as much as the upper limit of one class is the lower limit of the next class. Therefore, 50 persons have their incomes between 200 to 249.99 and a person whose income is 250 shall be included in the next class of 250–300. According to the inclusive method, an item equal to upper limit of a class is included in that class itself. The following table demonstrates this method. Income(Rs.) No. of Persons 200–249 250–299 300–349 350–399 400–49 450–99 50 100 70 130 50 100 Total 500 Hence in the class 200–249, we include persons whose income is between Rs. 200 and Rs. 249. 1.3 PRINCIPLES OF CONSTRUCTING FREQUENCY DISTRIBUTIONS Inspite of the great importance of classification in statistical analysis, no hard and fast rules be laid down for it. A statistician uses his discretion for classifying a frequency distribution and sound experience, wisdom, skill and aptness for an appropriate classification of the data. However, the following guidelines must be considered to construct a frequency distribution: 1. Types of classes : The classes should be clearly defined and should not lead to any ambiguity. They should be exhaustive and mutually exclusive so that any value of variable corresponds to only class. 5 2. Number of classes : The choice about the number of classes into which a given frequency distribution should be divided depends upon (i) The total frequency which means the total number of observations in the distribution. (ii) The nature of the data which means the size or magnitude of the values of the variable. (iii) The desired accuracy. (iv) The convenience regarding computation of the various descriptive measures of the frequency distribution such as means, variance etc. The number of classes should neither be too small nor too large. In case the classes are few, the classification becomes very broad and rough which might obscure some important features and characteristics of the data. The accuracy of the results decreases as the number of classes becomes smaller. On the other hand, too many classes will result in very few frequencies in each class. This will give an irregular pattern of frequencies in different classes thus makes the frequency distribution irregular. Moreover a large number of classes will render the distribution too unwieldy to handle. The computational work for further processing of the data will become quite tedious and time consuming without any proportionate gain in the accuracy of the results. Hence a balance should be maintained between the loss of information in the first case and irregularity of frequency distribution in the second case, to arrive at a pleasing compromise giving the optimum number of classes. Normally, the number of classes should not be less than 5 and more than 20. Prof. Sturges has given a formula : k = 1+3.322 log n where k refers to the number of classes and n is the total frequency or number of observations. The value of k is rounded to the next higher integer : If n = 100 k = 1 + 3.322 log 100 = 1 + 6.6448 = 8 If n = 10,000 k = 1 + 3.322 log 10,000 = 1 + 13.288 = 14 However, this rule should be applied only when the number of observations are not very small. Moreover, the number or class intervals should be such that they give uniform and unimodal distribution which means that the frequencies in the given classes increase and decrease steadily and there are no sudden jumps. The number of classes should be an integer preferably 5 or some multiples of 5, 10, 15, 20, 25 etc. which are quite convenient for numerical computations. 3. Size of class intervals : Because the size of the class interval is inversely proportional to the number of classes in a given distribution, the choice about the size of the class interval will also depend upon the sound subjective judgment of the statistician. An approximate value of the magnitude of the class interval say i can be calculated with the help of Sturge’s Rule : i= Range 1 + 3.322 log n where i stands for class magnitude or interval, Range is calculated by taking the difference between largest and smallest value of the distribution, and n refers to total number of observations. 6 If we are given the following information; n = 400, Largest item = 1300 and Smallest item = 340. then, i = 1300 − 340 960 960 = = = 99.54(100 approx.) 1 + 3.322 log 400 1 + 3.322 × 2.6021 9.644 Another rule of thumb for determining the size of the class interval is that the length of the class interval should not be greater than 1 4 th of the estimated population standard deviation. If 6 is the estimate of population standard deviation then the length of class interval is given by: i ≥ 6/4 The size of class intervals should be taken as 5 or multiples of 5,10,15 or 20 for easy computations of various statistical measures of the frequency distribution, class intervals should be so fixed that each class has a convenient mid-point around which all the observations in that class cluster. It means that the entire frequency of the class is concentrated at the mid value of the class. This assumption will be true only if the frequencies of the different classes are uniformly distributed in the respective class intervals. It is always desirable to take the class intervals of equal or uniform magnitude throughout the frequency distribution. 4. Class boundaries : If in a grouped frequency distribution there are gaps between the upper limit of any class and lower limit of the succeeding class (as in case of inclusive type of classification), there is a need to convert the data into a continuous distribution by applying a correction factor for continuity for determining new classes of exclusive type. The lower and upper class limits of new exclusive type classes are called class boundaries. If d is the gap between the upper limit of any class and lower limit of succeeding class, the class boundaries for any class are given by : 1 d 2 1 Lower class boundary = Lower class limit − d 2 Upper class boundary = Upper class limit + d/2 is called the correction factor. Let us consider the following example to understand: Marks Class Boundaries 20 – 24 (20 – 0.5,24+ 0.5) i.e., 19.5 – 24.5 25 – 29 (25 – 0.5,29 + 0.5) i.e., 24.5 – 29.5 30 – 34 (30 – 0.5,34 + 0.5) i.e., 29.5 – 34.5 35 – 39 (35 – 0.5,39 + 0.5) i.e., 34.5 – 39.5 40 – 44 (40 – 0.5,44 + 0.5) i.e., 39.5 – 44.5 Correction factor = d 35 − 34 1 = = = 0.5 2 2 2 7 5. Mid-value or class mark: Mid value or class mark is the value of a variable which lies exactly at the middle of a class. Mid-value of any class is obtained on dividing the sum of the upper and lower class limits by 2. Mid value of a class = 1 [Lower class limit + Upper class limit] 2 The class limits should be selected in such a manner that the observations in any class are evenly distributed throughout the class interval so that the actual average of the observations in any class is very close to the mid-value of the class. 6. Open end classes : The classification is termed as open end classification if the lower limit of the first class or the upper limit of the last class or both are not specified and such classes in which one of the limits is missing are called open end classes. For example, the classes like the marks less than 20 or age below 60 years. As far as possible open end classes should be avoided because in such classes the mid-value cannot be accurately obtained. But if the open end classes are inevitable then it is customary to estimate the class mark or mid-value for the first class with reference to the succeeding class. In other words, we assume that the magnitude of the first class is same as that of the second class. Example 1 : Construct a frequency distribution from the following data by inclusive method taking 4 as the class interval : 10 17 15 22 11 16 19 24 29 18 25 26 32 14 17 20 23 27 30 12 15 18 24 36 18 15 21 28 33 38 34 13 10 16 20 22 29 19 23 31 Solution : Because the minimum value of the variable is 10 which is a very convenient figure for taking the lower limit of the first class and the magnitude of the class interval is given to be 4, the classes for preparing frequency distribution by the Inclusive Method will be 10–13, 14–17, 18–21, 22–25,............ 38–41. Frequency Distribution Class Interval Tally Bars Frequency (f) 10 – 13 |||| 5 14 – 17 |||| ||| 8 18 – 21 |||| ||| 8 22 – 25 |||| || 7 26 – 29 |||| 5 30 – 33 |||| 4 34 – 37 || 2 38 – 41 | 1 8 Example 2 : Prepare a statistical table from the following : Weekly wages (Rs.) of 100 workers of Factory A 88 23 27 28 86 96 94 93 86 99 82 24 24 55 88 99 55 86 82 36 96 39 26 54 87 100 56 84 83 46 102 48 27 26 29 100 59 83 84 48 104 46 30 29 40 101 60 89 46 49 106 33 36 30 40 103 70 90 49 50 104 36 37 40 40 106 72 94 50 60 24 39 49 46 66 107 76 96 46 67 26 78 50 44 43 46 79 99 36 68 29 67 56 99 93 48 80 102 32 51 Solution : The lowest value is 23 and the highest 106. The difference in the lowest and highest value is 83. If we take a class interval of 10, nine classes would be made. The first class should be taken as 20–30 instead of 23–33 as per the guidelines of classification. Frequency Distribution of the Wages of 100 Workers Wages (Rs.) Tally Bars Frequency (f) 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 100 – 110 |||| |||| ||| |||| |||| | 13 11 18 10 6 5 14 12 11 |||| |||| |||| ||| |||| |||| |||| | |||| |||| |||| |||| |||| |||| || |||| |||| | Total 100 1.4 GRAPHS OF FREQUENCY DISTRIBUTIONS The guiding principles for the graphic representation of the frequency distributions are precisely the same as for the diagrammatic and graphic representation of other types of data. The information contained in a frequency distribution can be shown in graphs which reveals the important characteristics and relationships that are not easily discernible on a simple examination of the frequency tables. The most commonly used graphs for charting a frequency distribution for the general understanding of the details of the data are : 9 1. Histogram 2. Frequency polygon 3. Smoothed frequency curves 4. Ogives or cumulative frequency curves. 1.4.1 Histogram The term ‘histogram’ must not be confused with the term ‘historigram’ which relates to time charts. Histogram is the best way of presenting graphically a simple frequency distribution. The statistical meaning of histogram is that it is a graph that represents the class frequencies in a frequency distribution by vertical adjacent rectangles. While constructing histogram the variable is always taken on the X-axis and the corresponding frequencies on the Y-axis. Each class is then represented by a distance on the scale that is proportional to its class-interval. The distance for each rectangle on the X-axis shall remain the same in case the class-intervals are uniform throughout; if they are different the width of the rectangles shall also change proportionately. The Y-axis represents the frequencies of each class which constitute the height of its rectangle. We get a series of rectangles each having a class interval distance as its width and the frequency distance as its height. The area of the histogram represents the total frequency. The histogram should be clearly distinguished from a bar diagram. A bar diagram is onedimensional i.e., only the length of the bar is important and not the width, a histogram is twodimensional, that is, in a histogram both the length and the width are important. However, a histogram can be misleading if the distribution has unequal class-intervals and suitable adjustments in frequencies are not made. The technique of constructing histogram is explained for : (i) distributions having equal class-intervals and (ii) distributions having unequal class-intervals. When class-intervals are equal, take frequency on the Y-axis, the variable on the X-axis and construct rectangles. In such a case the heights of the rectangles will be proportional to the frequencies. Example 3 : Draw a histogram from the following data : Classes Frequency 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 5 11 19 21 16 10 8 6 3 1 10 Solution: Y HISTOGRAM 25 FREQUENCY 20 15 10 5 0 10 20 30 40 50 60 70 80 90 X 100 CLASSES When class-intervals are unequal the frequencies must be adjusted before constructing a histogram. We take that class which has the lowest class-interval and adjust the frequencies of other classes accordingly. If one class-interval is twice as wide as the one having the lowest class-interval we divide the height of its rectangle by two, if it is three times more we divide it by three etc., the heights will be proportional to the ratios of the frequencies to the width of the classes. Example 4 : Represent the following data on a histogram. Average monthly income of 1035 employees in a construction industry is given below: Monthly Income (Rs.) No. of Workers 600 – 700 700 – 800 800 – 900 900 – 1000 1000 – 1200 1200 – 1400 1400 – 1500 1500 – 1800 1800 or more 25 100 150 200 240 160 50 90 20 Solution : Histogram showing monthly incomes of workers NUMBER OF WORKERS 200 Y 150 100 50 600 700 800 900 1000 1100 1200 1300 1400 1500 MONTHLY INCOME 11 1800 X When mid point are given, first we ascertain the upper and lower limits of each class and then construct the histogram in the same manner. Example 5 : Draw a histogram of the following distribution : Life of Electric Lamps Firm A Firm B 1010 10 287 1030 130 105 1050 482 26 1070 360 230 1090 18 352 in hours Solution : Since we are given the mid points, we should ascertain the class limits. To calculate the class limits of various classes, take difference of two consecutive mid-points and divide the difference by 2, then add and subtract the value obtained from each mid-point to calculate lower and higher class-limits. 500 Life of Electric Frequency Frequency Lamps Firm A Firm B 1000–1020 10 287 1020–1040 130 105 1040–1060 482 76 1060–1080 360 230 1080–1100 18 352 HISTOGRAM (FIRM A) 400 FREQUENCY FREQUENCY 400 300 200 1.4.2 300 200 100 100 1000 HISTOGRAM (FIRM A) 500 1020 1040 1060 LIFE OF LAMPS 1080 1000 1100 1020 1040 1060 1080 1100 LIFE OF LAMPS Frequency Polygon This is a graph of frequency distribution which has more than four sides. It is particularly effective in comparing two or more frequency distributions. There are two ways of constructing a frequency polygon. 12 (i) We may draw a histogram of the given data and then join by straight line the mid-points of the upper horizontal side of each rectangle with the adjacent ones. The figure so formed shall be frequency polygon. Both the ends of the polygon should be extended to the base line in order to make the area under frequency polygons equal to the area under Histogram. NUMBER OF STUDENTS (FREQUENCY) 400 Y 300 200 100 0 X CLASS MARK (ii) Another method of constructing frequency polygon is to take the mid-points of the various class-intervals and then plot the frequency corresonding to each point and join all these points by straight lines. The figure obtained by both the methods would be equal. 400 Y 2 5 4 300 5 3 5 NUMBER OF STUDENTS (FREQUENCY) 1 200 7 3 2 100 r 8 2 9 1 X 0 CLASS MARK 13 Frequency polygon has an advantage over the histogram. The frequency polygons of several distributions can be drawn on the same axis, which makes comparisons possible whereas histogram can not be usefully employed in the same way. To compare histograms we draw them on separate graphs. 1.4.3 Smoothed Frequency Curve A smoothed frequency curve can be drawn through the various points of the polygon. The curve is drawn by free hand in such a manner that the area included under the curve is approximately the same as that of the polygon. The object of drawing a smoothed curve is to eliminate as far as possible all accidental variations which exists in the original data, while smoothening, the top of the curve would overtop the highest point of polygon particularly when the magnitude of the class interval is large. The curve should look as regular as possible and all sudden turns should be avoided. The extent of smoothening would depend upon the nature of the data. For drawing smoothed frequency curve it is necessary to first draw the polygon and then smoothen it. We must keep in mind the following points to smoothen a frequency graph : (i) Only frequency distribution based on samples should be smoothened. (ii) Only continuous series should be smoothened. (iii) The total area under the curve should be equal to the area under the histogram or polygon. The diagram given below will illustrate the point: HISTOGRAM FREQUENCY POLYGON AND CURVE 50 HISTOGRAM 40 FREQUENCY CURVE 20 14.5 11.5 12.5 9.5 10.5 7.5 8.5 13.5 FREQUENCY POLYGON 10 6.5 NO. OF LEAVES 30 LENGTH OF LEAVES (cm) 1.4.4 Cumulative Frequency Curves or Ogives We have discussed the charting of simple distributions where each frequency refers to the measurement of the class-interval against which it is placed. Sometimes it becomes necessary to know the number of items whose values are greater or less than a certain amount. We may, for example, be interested in knowing the number of students whose weight is less than 65 lbs. or more than say 15.5 lbs. To get this information, it is necessary to change the form of frequency distribution from a simple to a cumulative distribution. In a cumulative frequency distribution of the frequency of each class is made to include the frequencies of all the lower or all the upper classes depending 14 upon the manner in which cumulation is done. The graph of such a distribution is called a cumulative frequency curve or an Ogive. There are two method of constructing ogives, namely : (i) less than method and (ii) more than method. In the less than method, we start with the upper limit of each class and go on adding the frequencies. When these frequencies are plotted we get a rising curve. In the more than method, we start with the lower limit of each class and we subtract the frequency of each class from total frequencies. When these frequencies are plotted, we get a declining curve. This example would illustrate both types of ogives. Example 6 : Draw ogives by both the methods from the following data. Distribution of weight of the students of a college (lbs.) Weights No. of Students 90.5–100.5 5 100.5–110.5 34 110.5–120.5 139 120.5–130.5 300 130.5–140.5 367 140.5–150.5 319 150.5–160.5 205 160.5–170.5 76 170.5–180.5 43 180.5–190.5 16 190.5–200.5 3 200.5–210.5 4 210.5–220.5 3 220.5–230.5 1 Solution : First of all we shall find out the cumulative frequencies of the given data by less than method. Less than (Weights) Cumulative frequency 100.5 5 110.5 39 120.5 178 130.5 478 140.5 845 15 150.5 1164 160.5 1369 170.5 1445 180.5 1488 190.5 1504 200.5 1507 210.5 1511 220.5 1514 230.5 1515 Plot these frequencies and weights on a graph paper. The curve formed is called an Ogive. 1500 1250 750 500 220.5 230.5 200.5 210.5 180.5 190.5 150.5 160.5 170.5 130.5 140.5 0 110.5 120.5 250 90.5 100.5 CUMULATIVE FREQUENCY 1000 X SIZES Now we calculate the cumulative frequencies of the given data by more than method. More than (Weights) Cumulative frequencies 90.5 1515 100.5 1510 110.5 1476 120.5 1337 130.5 1037 140.5 670 16 150.5 351 160.5 146 170.5 70 180.5 27 190.5 11 200.5 8 210.5 4 220.5 1 By plotting these frequencies on a graph paper, we will get a declining curve which will be our cumulative frequency curve or Ogive by More than method. Y 1500 1250 750 500 230.5 210.5 220.5 190.5 200.5 180.5 160.5 170.5 140.5 150.5 120.5 130.5 0 90.5 250 100.5 110.5 CUMULATIVE FREQUENCY 1000 X SIZES Although the graphs are a powerful and effective media of presenting statistical data, they are not under all circumstances and for all purposes complete substitutes for tabular and other forms of presentation. The specialist in this field is one who recognizes not only the advantages but also the limitations of these techniques. He knows when to use and when not to use these methods and from his experience and expertise is able to select the most appropriate method for every purpose. Example 7 : Draw an ogive by less than method and determine the number of companies getting profits between Rs. 45 crores and Rs. 75 crores : 17 Profits (Rs. crores) No. of Companies 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100 8 12 20 24 15 10 7 3 1 Solution : OGlVE BY LESS THAN METHOD Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 Less than 100 No. of Companies 100 92 8 20 40 64 79 89 96 99 100 NO. OF COMPANIES Profit (Rs. Crores) OGIVE BY LESS THAN METHOD 80 92–51 = 41 60 51 40 20 20 30 40 45 50 60 70 75 80 85 PROFIT RS. IN CRORES It is clear from the graph that the number of companies getting profits less than Rs. 75 crores is 92 and the number of companies getting profits less than Rs. 45 crores is 51. Hence the number of companies getting profits between Rs. 45 crores and Rs. 75 crores is 92 – 51 = 41. Example 8 : The following distribution is with regard to weight in grams of mangoes of a given variety. If mangoes of weight less than 443 grams be considered unsuitable for foreign market, what is the percentage of total yield suitable for it? Assume the given frequency distribution to be typical of the variety: Weight in gms. No. of mangoes 410–419 420–429 430–439 440–449 450–459 460–469 470–479 10 20 42 54 45 18 7 18 Draw an ogive of ‘more than’ type of the above data and deduce how many mangoes will be more than 443 grams. Solution : Mangoes weighing more than 443 gms. are suitable for foreign market. Number of mangoes weighing more than 443 gms lies in the last four classes. Number of mangoes weighing between 444 and 449 grams would be 6 324 × 54 = = 32.4 10 10 Total number of mangoes weighing more than 443 gms. = 32.4 + 45 + 18 + 7 = 102.4 Percentage of mangoes = 102.4 × 100 = 52.25 196 Therefore, the percentage of the total mangoes suitable for foreign market is 52.25. OGIVE BY MORE THAN METHOD Weight more than (gms.) No. of Mangoes OGIVE BY MORE THAN METHOD 410 196 420 186 430 166 440 124 450 70 200 180 No. of mangoes 160 140 120 100 80 60 40 460 25 20 410 470 420 430 440 450 460 470 Weight in grams 7 From the graph it can be seen that there are 103 mangoes whose weight will be more than 443 gms and are suitable for foreign market. 1.5 SUMMARY ● A frequency distribution aims to reduce the size of the given set of data for a better comprehension. ● An array, which is an arrangement of data in an ascending or descending order of magnitude, is a useful step in preparing a frequency distribution. ● To prepare a frequency distribution, we have to decide about the class intervals to be taken. The width of class intervals depends on the number of classes. The number of classes should not be very small or very large. 19 ● Given values are considered one by one and placed in appropriate class intervals. The number of observations in each class is called the class frequency. ● The class intervals may be overlapping Like 10–20, 20–30, etc. or inclusive like 10–19, 20– 29, etc. ● Inclusive class intervals should be transformed into exclusive classes, depending on the way the given data are recorded. ● Class mid-points are the points that lie halfway between the two class limits. ● The frequencies of a distribution can also be cumulated in ascending or descending order. They are known as ACF and DCF. respectively. ● The ACF are ‘less than’ cumulative frequencies while the DCF are ‘more than’ cumulative frequencies. ● Absolute class frequencies may also be expressed as relative frequencies, either as proportions or percentages. ● A frequency distribution may have class intervals with equal or unequal width. ● A frequency distribution may be shown graphically by a histogram and frequency polygon. ● A histogram consists of bars drawn over class limits with heights of bars such that the areas of the bars are proportional to the frequencies of various class intervals. ● A frequency polygon is a line chart and is drawn by joining points given by the class midpoints and class frequencies. ● Cumulative frequencies arc shown graphically by means of ogives. 1.6 SELF ASSESSMENT QUESTIONS Exercise 1 : True or False Statements (i) Before constructing a frequency distribution, it is necessary that the data be arranged as an array. (ii) If the class intervals are given in the exclusive form as 10–20. 20–30. etc.. then a value exactly equal to 20 may be included in either of these classes. (iii) In the case of inclusive class intervals, the class mid-points are determined only after converting them into exclusive form. (iv) The number and width of class intervals are determined independently of each other. (v) A frequency distribution must have all class intervals of equal width. (vi) A distribution can have both ends open. (vii) A bivariate frequency distribution can be prepared only when both the variables involved are discrete or are continuous. (viii) Relative frequencies are obtained by dividing the frequencies of various classes by the width of the respective classes. 20 (ix) Frequency density is another name for relative frequency. (x) The proportionate frequencies facilitate comparison between distributions better than absolute frequencies. (xi) It is never possible to calculate absolute frequencies from the proportionate frequencies for a distribution. (xii) In presenting a distribution graphically, the variable is shown horizontally while the frequencies are shown vertically. (xiii) It is necessary that the widths of bars representing various class intervals of a frequency distribution be always equal. (xiv) The areas covered by a histogram and a frequency polygon are equal. (xv) Strictly speaking, a histogram cannot be drawn for an open-ended distribution. Ans. 1. F 2. F 3. T 4. F 5. F 6. T 7. F 8. F 9. F 10. T 11. F 12. T 13. F 14. T 15. T Exercise 2 : Questions and Answers (i) What is a frequency distribution? Explain the process of preparing a univariate frequency distribution. (ii) Explain the following: (a) Grouping error (b) Cumulative frequencies (c) Relative frequencies (d) Frequency density (iii) What is a bivariate frequency distribution? How is it constructed? Can we prepare a bivariate frequency distribution if one of the variables is discrete and the other is continuous? (iv) Explain the drawing of histogram when class intervals are equal and when they are not equal. (v) What are ogives? How are they constructed and what information do they provide? (vi) From the time cards of a factory, the following information has been obtained about the number of days each one of the 48 workers has reported late for the work during the last month: 3 0 5 0 6 2 1 0 4 6 5 2 1 1 1 3 4 2 2 5 6 3 0 2 2 3 2 5 4 2 4 3 5 2 2 2 4 6 4 0 3 1 1 4 5 2 1 1 Prepare a frequency distribution using this information. Also, indicate percentage frequencies. (vii) XYZ Company collected data regarding the number of interviews required for each of its 40 sales persons to make their most recent sale. Following are those numbers: 102 95 90 90 101 60 80 113 102 110 126 66 121 116 139 72 101 93 114 99 112 105 97 100 99 115 129 111 119 81 91 93 119 113 128 110 75 87 107 108 (a) Construct a frequency distribution with six class intervals. (b) Construct a histogram from the data. 21 (viii) If the class mid-points in a frequency distribution of weights of a group of students are 125, 132, 139, 146, 153, 160, 167, 174 and 181 lbs. find (a) Size of the class interval. (b) Class limits assuming weights have been measured to the nearest pound. (ix) Convert the following class intervals into exclusive form: (a) (b) (c) Diameters (in cm) Age in years Height in inches 0.5–0.9 1.0–1.4 1.5–1.9 2.0–2.4 2.5–2.9 3.0–3.4 5–9 10–14 15–24 25–39 40–59 60–79 60–64 65–69 70–74 75–79 80–84 85–89 (x) The monthly profits earnedby 100 companies during the last financial year are given below. Monthly Profit (Rs. lakhs) No. of Companies Monthly Profit (Rs. lakhs) No. of Companies 20–30 4 60–70 15 30–40 8 70–80 10 40–50 18 80–90 8 50–60 30 90–100 7 (a) Draw an ogive by ‘less than’ method and ‘more than’ method. (b) Obtain the limits of monthly profits of central 50 percent of the companies and check these values against the formula calculated values. (xi) The salary distribution of employees of a company is given below Salary (in ‘000 Rs.) No. of Employees 8–10 18 10–12 32 12–14 70 14–16 88 16 –18 64 18–20 44 20–22 24 22–24 10 22 (a) Show these data by means of a histogram and frequency polygon on the same graph. (b) Draw a more-than ogive and using it estimate (i) the number of employees earning more than Rs. 16,500; and (ii) the number of employees earning less than Rs. 13,000. (xii) The following table gives the distribution of weekly income of 160 families: Weekly Income (Rs.) No. of Families 2,000–4,000 20 4,000–6,000 40 6,000–8,000 50 8,000–12,000 32 12,000–16,000 16 16,000–20,000 2 Draw a ‘less than’ ogive and answer the following from it: (a) What are the limits within which incomes of the middle 50 percent of the families lie? (b) It is decided that 80 percent of the families should pay income tax. What is the minimum taxable income? (c) What is the minimum income of the richest 30 percent of the families? Ans. 10. (b) = 47 and 70, 11. (b) = 126 and 86, 12. (a) = 5000 – 9250 (b) = 4600 (c) = 8250 23 LESSON : 2 MEASURES OF CENTRAL TENDENCY – MATHEMATICAL AND POSITIONAL AVERAGES 2. STRUCTURE 2.0 2.1 2.2 2.3 2.4 Objective What is Central Tendency? What are the Objectives of Central Tendency? Characteristics of a Good Average Types of Averages 2.4.1 Arithmetic Mean 2.4.2 Mathematical Properties of Arithmetic Mean 2.4.3 Weighted Mean 2.5 Geometric Mean 2.5.1 Specific uses of G.M. 2.5.2 Weighted G.M. 2.6 Harmonic Mean 2.6.1 Application of Harmonic Mean 2.7 Median 2.8 Other Positional Averages 2.9 Calculation of Missing Frequencies 2.10 Mode 2.10.1 Determination of Mode by Graph 2.11 Summary 2.12 Self Assessment Questions 2.0 OBJECTIVE After reading this lesson, you should be able to : (a) Learn the meaning of central tendency and other averages (b) Learn the process of computing arithmetic mean, weighted Mean, Harmonic mean, Geometric mean, Median, Deciles, Quartiles, Percentiles and Mode under different situations (c) Comprehend mathematical properties of Arithmetic average (d) Learn specific uses of different averages. 2.1 WHAT IS CENTRAL TENDENCY One of the important objectives of statistical is to find out various numerical values which explains the inherent characteristics of a frequency distribution. The first of such measures are averages. The averages are the measures which condense a huge unwieldy set of numerical data into single numerical values which represent the entire distribution. The inherent inability of the human mind to a large body of numerical data compels us to few constants that will describe the data. Averages provide us the gist and give a bird’s eye view of the huge mass of unwieldy numerical data. Averages are the 24 typical values around which other items of the distribution congregate. This value lie between the two extreme observation of the distribution and give us an idea about the concentration of the values in the central part of the distribution. They are called the measures of central tendency. Averages are also called measures of location since they enable us to locate the position or place of the distribution in question. Averages are statistical constants which enables us to comprehend in a single value the significance of the whole. According to Croxton and Cowden, an average value is a single value within the range of the data that is used to represent all the values in that series. Since an average is somewhere within the range of the data, it is sometimes called a measure of central value. An average, known as the measure of central tendency, is the most typical representative item of the group to which it belongs and which is capable of revealing all important characteristics of that group or distribution. 2.2 WHAT ARE THE OBJECTS OF CENTRAL TENDENCY The most important object of calculating an average or measuring central tendency is to determine a single figure which may be used to represent a whole series involving magnitudes of the same variable. Second object is that an average represents the entire data, it facilitates comparison within one group or between group of data. Thus, the performance of the members of a group can be compared with the average performance of different group. Third object is that an average helps in computing various other statistical measures such as dispersion, skewness. kurtosis etc. 2.3 CHARACTERISTICS OF A GOOD AVERAGE An average represents the statistical data and it is used for purposes of comparison, it must possess the following properties. 1. It must be rigidly defined and not left to the mere estimation of the observer. If the definition is rigid, the computed value of the average obtained by different persons shall be similar. 2. The average must be based upon all values given in the distribution. If the item is not based on all values it might not be representative of the entire group of data. 3. It should be easily understood. The average should possess simple and obvious properties. It should be too abstract for the common people. 4. It should be capable of being calculated with reasonable care and rapidity. 5. It should be stable and unaffected by sampling fluctuations. 6. It should be capable of further algebraic manipulation. 2.4 TYPES OF AVERAGES Different methods of measuring “Central Tendency” provide us with different kinds of averages. The following are the main types of averages that are commonly used : 25 1. Mean (i) Arithmetic mean (ii) Weighted mean (iii) Geometric mean (iv) Harmonic mean 2. Median 3. Mode 2.4.1 Arithmetic Mean The arithmetic mean of a series is the quotient obtained by dividing the sum of the values by the number of items. In algebraic language, if X1, X2, X3,.........Xn are the n values of a variate X, then the Arithmetic Mean (X) is defined by the following formula : X= = 1 (X1 + X2 + X3 + ............. + X n ) n ∑X 1 n Xi = ∑ n i=1 N Example 1 : The following are the monthly salaries (Rs.) of ten employees in an office. Calculate the mean salary of the employees: 250, 275, 265, 280, 400, 490, 670, 890, 1100, 1250. Solution : X = X= ∑X N 250 + 275 + 265 + 280 + 400 + 490 + 670 + 890 + 1100 + 1250 5870 = = Rs. 587 10 10 Short-cut Method : Direct method is suitable where the number of items is moderate and the figures are small sizes and integers. But if the number of items is large and/or the values of the variate are big, then the process of adding together all the values may be a lengthy process. To overcome this difficulty of computations, a short-cut method may be used. Short cut method of computation is based on an important characteristic of the arithmetic mean, that is, the algebraic sum of the deviations of a series of individual observations from their mean is always equal to zero. Thus deviations of the various values of the variate from an assumed mean computed and the sum is divided by the number of items. The quotient obtained is added to the assumed mean to find the arithmetic mean. Symbolically, X = A + Σdx , where A is assumed mean and deviations or dx = (X – A). N We can solve the previous example by short-cut method. 26 Computation of Arithmetic Mean Serial Number Salary (Rupees) X 1. 250 – 150 2. 275 – 125 3. 265 – 135 4. 280 – 120 5. 400 0 6. 490 + 90 7. 670 + 270 8. 890 + 490 9. 1100 + 700 10. 1250 + 850 N = 10 Σdx = 1870 X=A+ Deviations from assumed mean where dx = (X – A), A = 400 Σdx N By substituting the values in the formula, we get X = 400 + 1870 = Rs. 587 10 Computation of Arithmetic Mean in Discrete Series. In discrete series, arithmetic mean may be computed by both direct and short cut methods. The formula according to direct method is : X = 1 Σ ( fX ) ( f1 X 1 + f 2 X 2 + ........... + f n X n ) = n N where the variable values X1, X2, ........Xn have frequencies f1, f2 ,........fn and N = Σf. Example 2 : The following table gives the distribution of 100 accidents during area days of the week in a given month. During a particular month there were 5 Fridays and Saturdays and only four each of other days. Calculate the average number of accidents per day. Days : Sun Mon Tue Wed Thur Fri Sat Total Number of accidents : 20 22 10 9 11 8 20 = 100 27 Solution : Calculation of Number of Accidents per Day Day No. of No. of days Accidents in month X f fX 20 22 10 9 11 8 20 4 4 4 4 4 5 5 80 88 40 36 44 40 100 100 N = 30 Sunday Monday Tuesday Wednesday Thursday Friday Saturday X= Total accidents Σf X = 428 ΣfX 428 = = 14.27 = 14 accidents per day N 30 The formula for computation of arithmetic mean according to short cut method is X=A+ Σfdx N where A is Assumed mean, dx = (X – A) and N = Σf . We can solve the previous example by short-cut method as given below : Calculation of Average Accidents per day Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday X=A+ X 20 22 10 9 11 8 20 dx = X–A (where A = 10) + 10 + 12 +0 –1 +1 –2 + 10 f fdx 4 4 4 4 4 5 5 + 40 + 48 +0 –4 +4 – 10 + 50 30 + 128 Σfdx 128 = 10 + = 14.27 = 14 accidents per day N 30 Calculation of Arithmetic Mean for Continuous Series : The arithmetic mean can be computed both by direct and short-cut method. In addition, a coding method or step deviation method is also applied for simplification of calculations. In any case, it is necessary to find out the mid-values of the various classes in the frequency distribution before arithmetic mean of the frequency 28 distribution can be computed. Once the mid-points of various classes are found out, then the process of the calculation of arithmetic mean is same as in the case of discrete series. In case of direct method, the formula to be used : X= Σfm , N when m = mid points of various classes and N = the total frequency In the short-cut method, the following formula is applied: X=A+ Σfdx N where dx = (m – A) and N = Σf The short-cut method can further be simplified in practice and is named coding method. The deviations from the assumed mean are divided by a common factor to reduce their size. The sum of the products of the deviations and frequencies is multiplied by this common factor and then it is divided by the total frequency and added to the assumed mean. Symbolically X=A+ Σfd ' x ×i, N where d ' x = m− A and i = common factor i Example 3 : Following is the frequency distribution of marks obtained by 50 students in a test of Statistics : Marks Number of Students 0–10 4 10–20 6 20–30 20 30–40 10 40–50 7 50–60 3 Calculate arithmetic mean by; (i) direct method, (ii) short-cut method, and (iii) coding method. Solution : Calculation of Arithmetic Mean X f m fm m–A i (where A = 25) (where i = 10) dx = m – A d'x = fdx fd′ x 0–10 4 5 20 – 20 –2 – 80 –8 10–20 6 15 90 – 10 –1 – 60 –6 20–30 20 25 500 0 0 0 0 29 30–40 10 35 350 + 10 +1 100 + 10 40–50 7 45 315 + 20 +2 140 + 14 50–60 3 55 165 + 30 +3 90 +9 Σfdx = 190 Σfd ' x = + 19 Σfm = 1440 N = 50 Direct Method : X= Σfm 1440 = = 28.8 marks. N 50 Short-cut Method: X=A+ Σfdx 190 = 25 + = 28.8 marks. N 50 X=A+ Σfd ' x 19 × i = 25 + × 10 = 25 + 3.8 = 28.8 marks. N 50 Coding Method: We can observe that answer of average marks i.e. 28.8 is identical by all methods. 2.4.2 Mathematical Properties of Arithmetic Mean (i) The sum of the deviation of a given set of individual observations from the arithmetic mean is always zero. Symbolically, ∑ ( X – X ) = 0. It is due to this property that the arithmetic mean is characterised as the centre of gravity i.e., the sum of positive deviations from the mean is equal to the sum of negative deviations. (ii) The sum of squares of deviations of a set of observations is the minimum when deviations are taken from the arithmetic average. Symbolically, ∑ ( X – X ) 2 = smaller than Σ(X – any other value)2. We can verify the above properties with the help of the following data : Values Deviations from X Deviations from assumed mean X (X – X) ( X – X )2 (X – A) ( X – A) 2 3 –6 36 –7 49 5 –4 16 –5 25 10 1 1 0 0 12 3 9 2 4 15 6 36 5 25 Total = 45 0 98 –5 103 30 X= ∑ X 45 = =9, n 5 where A (assumed mean) = 10 (iii) If each value of a variable X is increased or decreased or multiplied by a constant k, the arithmetic mean also increases or decreases or multiplies by the same constant. (iv) If we are given the arithmetic mean and number of items of two or more groups, we can compute the combined average of these groups by applying the following formula : X12 = N1X1 + N 2 X 2 N1 + N 2 where X12 refers to combined average of two groups, X1 refers to arithmetic mean of first group, X 2 refers to arithmetic mean of second group, N1 refers to number of items of first group, and N2 refers to number of items of second group We can understand the property with the help of the following examples. Example 4 : The average marks of 25 male students in a section is 61 and average marks of 35 female students in the same section is 58. Find combined average marks of 60 students. Solution : We are given the following information, X1 = 61, Apply N1 = 25, X12 = X 2 = 58, N2 = 35 N1X1 + N 2 X 2 (25 × 61) + (35 × 58) = = 59.25 marks. N1 + N 2 25 + 35 Example 5 : The mean wage of 100 workers in a factor, running two shifts of 60 and 40 workers respectively is Rs. 38. The mean wage of 60 workers in morning shift is Rs. 40. Find the mean wage of 40 workers working in the evening shift. Solution : We are given the following information, X1 = 40, N1 = 60, X 2 = ?, N2 = 40, X12 = 38, and N = 100 Apply X12 = N1X1 +N 2 X 2 N1 + N 2 38 = (60 × 40) + (40 × X 2 ) 60 + 40 X2 = 3800 − 2400 = 35. 40 or 31 3800 = 2400 + 40 X 2 Example 6 : The mean age of a combined group of men and women is 30 years. If the mean age of the group of men is 32 and that of women group is 27. find out the percentage of men and women in the group. Solution : Let us take group of men as first group and women as second group. Therefore, X1 = 32 years, X 2 = 27 years, and X12 = 30 years. In the problem, we are not given the number of men and women. We can assume N1 + N2 = 100 and therefore, N1 = 100 – N2 Apply X12 = 30 = N1X1 + N 2 X 2 N1 + N 2 32N1 + 27N 2 (Substitute N1 = 100 – N2) 100 30 × 100 = 32 (100 – N 2 ) + 27 N 2 or 5N 2 = 200 N 2 = 200 / 5 = 40% N1 = (100 – N 2 ) = (100 – 40) = 60% Therefore, the percentage of men in the group is 60 and that of women is 40. (v) Because X = ∑ X N ∴ Σf = N.X If we replace each item in the series by the mean, the sum of these substitutions will be equal to the sum of the individual items. This property is used to find out the aggregate values and corrected averages. We can understand the property with the help of an example. Example 7 : Mean of 100 observations is found to be 44. If at the time of computation two items are wrongly taken as 30 and 27 inplace of 3 and 72. Find the corrected average. Solution : ∴ X= ΣX N ∑ X = N.X = 100×44 = 4400 Corrected ∑ X = ∑ X + correct items – wrong items = 4400 + 3 + 72 – 30 – 27 = 4418 Corrected average = Corrected ∑ X 4418 = = 44.18 N 100 Calculation of Arithmetic mean in Case of Open-End Classes : Open-end classes are those in which lower limit of the first class and the upper limit of the last class are not defined. In these series, we can not calculate mean unless we make an assumption about the unknown limits. The assumption depends upon the class-interval following the first class and preceding the last class, For example : 32 Marks No. of students Below 15 4 15–30 6 30–45 12 45–60 8 Above 60 7 In this example, because all defined class-intervals are same, the assumption would be that the first and last class shall have same class-interval of 15 and hence the lower limit of the first class shall be zero and upper limit of last class shall be 75. Hence first class would be 0–15 and the last class 60–75. What happens in this case ? Marks No. of students Below 10 4 10–30 7 30–60 10 60–100 8 Above 100 4 In this problem because the class interval is 20 in the second class, 30 in the third, 40 in the fourth class and so on. The class interval is increasing by 10. Therefore the appropriate assumption in this case would be that the lower limit of the first class is zero and the upper limit of the last class is 150. In case of other open-end class distributions the first class limit should be fixed on the basis of succeeding class interval and the last class limit should be fixed on the basis of preceding class interval. If the class intervals are of varying width, an effort should be made to avoid calculating mean and mode. It is advisable to calculate median. 2.4.3 Weighted Mean In the computation of arithmetic mean, we give equal importance to each item in the series. Raja Toy Shop sell, Toy Cars at Rs. 3 each, Toy Locomotives at Rs. 5 each, Toy Aeroplane at Rs. 7 each and Toy Double Decker at Rs. 9 each. What shall be the average price of the toys sold ? If the shop sells 4 toys one of each kind. ∑ X 24 X (Mean Price) = N = 4 = Rs. 6. In this case the importance of each toy is equal as one toy of each variety has been sold. While computing the arithmetic mean this fact has been taken care of including the price of each toy once only. 33 But if the shop sells 100 toys, 50 cars, 25 locomotives, 15 aeroplanes and 10 double deckers, the importance of the four toys to the dealer is not equal as a source of earning revenue. In fact their respective importance is equal to the number of units of each toy sold, i.e. the importance of Toy car is 50; the importance of Locomotive is 25; the importance of Aeroplane is 15; and the importance of Double Decker is 10. It may be noted that 50, 25, 15, 10 are the quantities of the various classes of toys sold. These quantities are called as ‘weights’ in statistical language. Weight is represented by symbol W and ΣW represents the sum of weights. While determining the average price of toy sold these weights are of great importance and are taken into account to compute weighted mean. Xw = ∑[(W1X1 ) + (W2 X 2 ) + (W3X 3 ) + (W4 X 4 )] ∑ WX = ∑W W1 + W2 + W3 + W4 where W1, W2, W3, W4 are weights and X1, X2, X3, X4 represents the price of 4 varieties of toy. Hence by substituting the values of W1, W2, W3, W4 and X1, X2, X3, X4, we get (50 × 3) + (25 × 5) + (15 × 7) + (10 × 9) 50 + 25 + 15 + 10 150 + 125 + 105 + 90 470 Xw = = = Rs. 4.70 100 100 The table given below demonstrates the procedure of computing the weighted Mean. Xw = Weighted Arithmetic mean of Toys by the Raja Shop. Toy Car Locomotive Aeroplane Double Decker Price per toy (Rs.) Number sold Price × weight X 3 5 7 9 W 50 25 15 10 WX 150 125 105 90 ∑ W = 100 ∑ WX = 470 ∑ WX 470 = = Rs. 4.70 100 ∑X Example 8 : The table below shows the number of skilled and unskilled workers in two localities along with their average hourly wages. ∴ Xw = Ram Nagar Worker Category Shyam Nagar Number Wages (per hour) Number Wages (per hour) Skilled 150 1.80 350 1.75 Unskilled 850 1.30 650 1.25 Determine the average hourly wage in each locality. Also give reasons why the results show 34 that the average hourly wage in Shyam Nagar exceed the average hourly wage in Ram Nagar, even though in Shyam Nagar the average hourly wages of both categories of workers is lower. It is required to compute weighted mean. Solution : Ram Nagar Shyam Nagar X W WX X W WX Skilled 1.80 150 270 1.75 350 612.50 Unskilled 1.30 850 1105 1.25 650 812.50 1000 1375 1000 1425 Total Xw = 1375 = Rs.1.375 1000 Xw = 1425 = Rs. 1.425 1000 It may be noted that weights are more evenly assigned to the different categories of workers in Shyam Nagar than in Ram Nagar. 2.5 GEOMETRIC MEAN In general, if we have n numbers (none of them being zero), then the G.M. is defined as G.M. = x1 , x2 ,...........xn = ( x1 , x2 ...........xn )1/ n In the case of a discrete series, if x1, x2,...........xn occur f1, f2,.......fn times respectively and N is the total frequency (i.e. N = f1+, f2+,.........fn), then G.M. = n x1 f1 , x2 f 2 ,...........xn f n For convenience, use of logarithms is made extensively to calculate the nth root. In terms of logarithms log x1 + log x2 + ........... + log xn G.M. = AL n ∑ log x , = AL N where AL stands for anti log. In discrete series, G.M. = AL ∑ f log x N ∑ f log m N Example 9 : Calculate G.M. of the following data : and in case of continuous series, G.M. = AL 2, 4, 8 35 Solution : G.M. = 3 2 × 4 × 8 = 3 64 = 4 In terms of logarithms, the question can be solved as follows : log 2 = 0.3010, log 4 = 0.6021, and log 8 = 9.9031 Apply the formula: G.M. = AL ∑ log x 1.8062 = AL = AL 0.60206 = 4 N 3 Example 10. Calculate geometric mean of the following data : x 5 6 7 8 9 10 11 f 2 4 7 10 9 6 2 Solution : Calculation of G.M. x log.x f f log x 5 0.6990 2 1.3980 6 0.7782 4 3.1128 7 0.8451 7 5.9157 8 0.9031 10 9.0310 9 0.9542 9 8.5878 10 1.0000 6 6.0000 11 1.0414 2 2.0828 N = 40 Σf log x = 36.1281 36.1281 Σf log x = AL G.M. = AL = AL (0.9032) = 8.002 40 N Example 11 : Calculate G.M. from the following data : X f 9.5–14.5 10 14.5–19.5 15 19.5–24.5 17 24.5–29.5 25 29.5–34.5 18 34.5–39.5 12 39.5–44.5 8 36 Solution : Calculation of G.M. X m log m f f log m 9.5–14.5 12 1.0792 10 10.7920 14.5–19.5 17 1.2304 15 18.4560 19.5–24.5 22 1.3424 17 22.8208 24.5–29.5 27 1.4314 25 35.7850 29.5–34.5 32 1.5051 18 27.0918 34.5–39.5 37 1.5682 12 18.8184 39.5–44.5 42 1.6232 8 12.9856 N = 105 Σf log m = 146.7496 146.7496 = AL (1.3976) = 24.98 G.M. = AL 105 2.5.1 Specific uses of G.M. The Geometric Mean has certain specific uses, some of them are : (i) It is used in the construction of index numbers, (ii) It is also helpful in finding out the compound rates of change such as the rate of growth of population in a country. (iii) It is suitable where the data are expressed in terms of rates, ratios and percentage. (iv) It is quite useful in computing the average rates of depreciation or appreciation. (v) It is most suitable when large weights are to be assigned to small items and small weights to large items. Example 12. The gross national product of a country was Rs. 1,000 crores 10 years earlier. It is Rs. 2,000 crores now. Calculate the rate of growth in G.N.P. Solution : In this case compound interest formula will be used for computing the average annual per cent increase of growth. Pn = Po(1 + r)n where Pn = prinicipal sum (or any other variate) at the end of the period. Po = prinicipal sum in the beginning of the period. r = rate of increase or decrease. n = number of years. It may he noted that the above formula can also be written in the following form : r = n Pn −1 Po 37 Substituting the values given in the formula, we have r= 10 2000 − 1 = 10 2 − 1 1000 log 2 0.30103 − 1 = AL − 1 = 1.0718 − 1 = 0.0718 = 7.18% = AL 10 10 Hence, the rate of growth in GNP is 7.18%. Example 13 : The price of commodity increased by 5 per cent from 1998 to 1999, 8 per cent from 1999 to 2000 and 77 per cent from 2000 to 2001. The average increase from 1998 to 2001 is quoted at 26 per cent and not 30 per cent. Explain this statement and verify your result. Solution : Taking Pn as the price at the end of the period. Po as the price in the beginning, we can substitute the values of Pn and Po in the compound interest formula. Taking Po = 100; Pn = 200.72 Pn = Po(1 + r)n 200.72 = 100(1 + r)3 (1 + r)3 = or r = 200.72 or 1 + r = 100 3 3 200.72 100 200.72 − 1 = 1.260 − 1 = 0.260 = 26% 100 Thus increase is not average of (5 + 8 + 77)/3 = 30 per cent. It is 26% as found out by G.M. 2.5.2 Weighted G.M. The weighted G.M. is calculated with the help of the following formula : G.M. = = x1w1.x2 w2 .........xn wn log x1 × w1 + log x2 × w2 + ...log xn × wn w1 + w2 ...wn Σ (log x × w) = AL Σw Example 14 : Find out weighted G.M. from the following data : Group Index number Weights Food 352 48 Fuel 220 10 Cloth 230 8 House Rent 160 12 Misc. 190 15 38 Solution : Calculation of Weighted G.M. Group Index Number(x) Weights (w) log x w log x Food 352 48 2.5465 122.2320 Fuel 220 10 2.3424 23.4240 Cloth 230 8 2.3617 18.8936 House Rent 160 12 2.2041 26.4492 Misc. 190 15 2.2788 34.1820 93 225.1808 225.1808 Σw log x = AL = 263.8 G.M. (weighted) = AL 93 Σw Example 15 : A machine depreciates at the rate of 35.5% per annum in the first year, at the rate of 22.5% per annum in the second year, and at the rate of 9.5% per annum in the third year, each percentage being computed on the actual value. What is the average rate of depreciation? Solution : Average rate of depreciation can be calculated by taking G.M. X (values taking 100 as base) log X I 100 – 35.5 = 64.5 1.8096 II 100 – 22.5 = 77.5 1.8893 III 100 – 9.5 = 90.5 1.9566 Year Σlog X = 5.6555 Σ log x 5.6555 = = AL 1.8851 = 76.77 Apply G.M.= AL 3 w ∴ Average rate of depreciation = 100 – 76.77 = 23.33%. Example 16 : The arithmetic mean and geometric mean of two values are 10 and 8 respectively. Find the values. Solution : If two values are taken as a and b, then a+b = 10, 2 Or a + b = 20, then a – b = and ab = 8 ab = 64 (a + b) 2 − 4ab = (20) 2 − 4 × 64 = 400 − 256 = 144 = 12 Now, we have a + b = 20, a – b = 12 Solving for a and b, we get a = 4 and b = 16. 39 2.6 HARMONIC MEAN The harmonic mean is defined as the reciprocals of the average of reciprocals of all items in a series. Symbolically, H.M. = N N = 1 1 1 1 1 x + x + x + ....... x Σ x n 1 2 3 In case of a discrete series, H.M. = N { } 1 Σ f× x and in case of a continuous series, H.M. = N { } Σ f× 1 m It may be noted that none of the values of the variable should be zero. Example 17 : Calculate harmonic mean from the following data : 5, 15, 25, 35 and 45 Solution : X 1 X 5 0.20 15 0.067 25 0.040 35 0.029 45 0.022 N=5 1 Σ = 0.358 X H.M. = 5 N = = 14 approx. 1 0.358 Σ x Example 18 : From the following data compute the value of the harmonic mean : x: 5 15 25 35 45 f: 5 15 10 15 5 40 Solution : Calculation of Harmonic Mean x f 1 x 5 5 0.200 1.000 15 15 0.067 1.005 25 10 0.040 0.400 35 15 0.29 0.435 45 5 0.022 0.110 1 Σ f = 2.950 x Σf = 50 N H.M. = 1 x f 1 Σf × x 50 = 17 approx. 2.95 = Example 19 : Calculate harmonic mean from the following distribution: x f 0–10 5 10–20 15 20–30 10 30–40 15 40–50 5 Solution : First of all, we shall find out mid points of the various classes. They are 5, 15, 25, 35 and 45. Then we will calculate the H.M. by applying the following formula : H.M. = N { } Σ f× 1 m Calculation of Harmonic Mean x (mid points) f 1 x 5 5 0.200 1.000 15 15 0.067 1.005 25 10 0.040 0.400 41 f 1 x 35 15 0.29 0.435 45 5 0.022 0.110 1 Σ f = 2.950 x Σf = 50 H.M. = N 1 Σ f × m = 50 = 17 approximately 2.950 2.6.1. Application of Harmonic Mean Like Geometric means, the harmonic mean is also applicable to certain special types of problems. Some of them are: (i) If, in averaging time rates, distance is constant, then H.M. is to be calculated. Example 20 : A man travels 480 km. a day. On the first day he travels for 12 hours @ 40 km. per hour and second day for 10 hours @ 48 km. per hour. On the third day he travels for 15 hours @ 32 km. per hour. Find his average speed. Solution: We shall use the harmonic mean, H.M. = 3 3 N = = = 39 km. per hour (approx.) 1 1 1 37 480 1 + + Σ X 40 48 32 48 + 40 + 32 = 40 km. per hour The arithmetic mean would be 3 (ii) If, in averaging the price data, the prices are expressed as “quantity per rupee”. Then harmonic mean should be applied. Example 21 : A man purchased one kilo of cabbage form each of four places at the rate of 20 kg., 16 kg., 12 kg. and 10 kg. per rupees respectively. On the average how many kilos of cabbages he has purchased per rupee. Solution: H.M. = 4 4 4 × 240 N = = = = 13.5 kg. per rupee. 1 1 1 1 71 250 71 1 + + + Σ x 20 16 12 10 Example 22 : Find two numbers whose geometric mean is 18 and arithmetic mean is 19.5. Also, calculate their harmonic mean. Solution : Let us assume two numbers are x and y. Now with the help of given information, X = or x+ y = 19.5 2 x + y = 39 ...(i) 42 Also, or Now, G.M. = xy = 18 xy = 324 ...(ii) (x – y)2 = ( x + y ) 2 − 4 xy = 392 − 4 × 324 = 225 ∴ x – y = ± 15 ...(iii) With the help of equations (i) and (iii), we get x = 27 and y = 12 Therefore the two numbers are 12 and 27. Now, G.M. = AM × HM G.M.2 = AM × HM or H.M. = GM2/AM H.M. = 182/19.5 = 16.62 2.7 MEDIAN The median is that value of the variable which divides the group in two equal parts. One part comprising the values greater than and the other all values less than median. Median of a distribution may be defined as that value of the variable which exceeds and is exceeded by the same number of observation. It is the value such that the number of observations above it is equal to the number of observations below it. Thus we know that the arithmetic mean is based on all items of the distribution, the median is positional average, that is, it depends upon the position occupied by a value in the frequency distribution. When the items of a series are arranged in ascending or descending order of magnitude the value of the middle item in the series in known as median in the case of individual observations. Symbolically, N + 1 th item Median = size of 2 If the number of items is even, then there is no value exactly in the middle of the series. In such a situation the median is arbitrarily taken to be halfway between the two middle items. Symbolically, N N + 1 size of th item + size of th item 2 2 Median = 2 Example 23. Find the median of the following series: (i) 8, 4, 8, 3, 4, 8, 6, 5, 10. (ii) 15, 12, 5, 7, 9, 5, 11, 28. 43 Solution : Computation of Median (i) (ii) Serial No. X Serial No. X 1 3 1 5 2 4 2 5 3 4 3 7 4 5 4 9 5 6 5 11 6 8 6 12 7 8 7 15 8 8 8 28 9 10 N=9 For (i) series N=8 9 +1 N + 1 th item = size of the th item = size of 5th item = 6 Median = size of the 2 2 8 +1 N + 1 th item = size of the For (ii) series Median = size of the th item 2 2 = size of 4th + size of 5th item 9 + 11 = = 10 2 2 Location of Median in Discrete series: In a discrete series, medium is computed in the following manner: (i) Arrange the given variable data in ascending or descending order. (ii) Find cumulative frequencies. (iii) Apply Med. = size of ( N + 1) th item 2 (iv) Locate median according to the size i.e., variable corresponding to the size or for next cumulative frequency. Example 24 : Following are the number of rooms in the houses of a particular locality. Find median of the data : No. of rooms : 3 4 5 6 7 8 No. of houses : 38 654 311 42 12 2 44 Solution : Computation of Median No. of rooms No. of houses Cumulative frequency X f Cf 3 38 38 4 654 692 5 311 1003 6 42 1045 7 12 1057 8 2 1059 1059 + 1 N + 1 th item = size of Median = size of th item = 530th item. 2 2 Median lies in the cumulative frequency of 692 and the value corresponding to this is 4. Therefore, Median = 4 rooms In a continuous series, median is computed in the following manner : (i) Arrange the given variable data in ascending or descending order. (ii) If inclusive series is given, it must be converted into exclusive series to find real class intervals. (iii) Find cumulative frequencies. (iv) Apply Median = size of N th item to ascertain median class. 2 (v) Apply formula of interpolation to ascertain the value of median. N − cf 0 Median = l1 + 2 × (l2 − l1 ) f or N − cf 0 Median = l2 − 2 × (l2 − l1 ) f where, l1 refers to lower limit of median class l2 refers to higher limit of median class cf0 refers cumulative frequency of previous class f refers to frequency of median class. Example 25 : The following table gives you the distribution of marks secured by some students in an examination 45 Marks No. of Students 0–20 21–30 31–40 41–50 51–60 61–70 71–80 42 48 120 84 48 36 31 Find the median marks. Solution : Calculation of Median Marks Marks No.of students (x) (f) 0–20 21–30 31–40 41–50 51–60 61–70 71–80 42 38 120 84 48 36 31 Median = size of cf 42 80 200 284 332 368 399 N 399 th item = size of th = 199.5th item 2 2 which lies in (31–40) group, therefore the median class is 30.5–40.5 Applying the formula of interpolation. N − cf 0 Median = l1 + 2 × (l2 − l1 ) f = 30.5 + 199.5 − 80 119.5 × (10) = 30.5 + = 40.46 marks. 120 12 2.8 OTHER POSITIONAL AVERAGES The median divides the series into two equal parts. Similarly there are certain other measures which divide the series into certain equal parts. There are first quartile, third quartile, deciles percentiles etc. If the items are arranged in ascending or descending order of magnitude, Q1 is that value which covers l/4th of the total number of items. Similarly, if the total number of items are divided into ten 46 equal parts, then, there shall be nine deciles. Symbolically, N + 1 th item First quartile (Q1) = size of 4 Third quartile (Q3) = size of 3( N + 1) th item 4 N + 1 th item First decile (D1) = size of 10 Sixth decile (D6) = size of 6( N + 1) th item 10 N + 1 th item First percentile (P1) = size of 100 Once values of the items are found out, then formulae of interpolation are applied for ascertaining the value of Q1, Q3, D1, D4, P40 etc. Example 26. Calculate Q1, Q3, D2 and P5 from the following data : Marks: Below 10 10–20 20–40 40–60 60–80 8 10 22 25 10 No. of Students: Solution: Calculation of Positional values Marks No. of Students (f) c.f. 8 10 22 25 10 5 8 18 40 65 75 80 Below 10 10–20 20–40 40–60 60–80 Above 80 N = 80 Q1 = size of N 80 th item = = 20th item 4 4 Hence Q1 lies in the class 20–40, apply N − Cf 0 N Q1 = l1 + 4 × i where l1 = 20, = 20, Cf0 = 18, f = 22 and i = (l2 − l1 ) = 20 4 f 47 Above 80 5 By substituting the values, we get Q1 = 20 + (20 − 18) × 20 = 20 + 1.8 = 21.8 22 Similarly, we can calculate Q3 = size of 3N 3 × 80 th item = th item = 60th item. 4 4 Hence Q3 lies in the class 40–60 3N − Cf 0 3N Q3 = l1 + 4 × i where l1 = 40, 4 = 60, Cf0 = 40, f = 25, i = 20. f ∴ Q3 = 40 + D2 = size of (60 − 40) × 20 = 40 + 16 = 56 25 2N th item = 16th item. Hence D2 lies in the class 10–20. 10 2N − Cf 0 2N × i where l1 = 10, D2 = l1 + 10 = 16, Cf0 = 8, f = 10, i = 10. 4 f D2 = 10+ (16 − 8) × 10 = 10 + 8 = 18 10 P5 = size of 5N 5 × 80 th item = th item = 4th item. Hence P5 lies in the class 0–10 100 100 5N − Cf 0 5N P5 = l1 + 100 × i where l1 = 0, 100 = 4, Cf0 = 0, f = 8, i = 10 f P5 = 0 + 4−0 × 10 = 0 + 5 = 5. 8 2.9. CALCULATION OF MISSING FREQUENCIES Example 27: In the frequency distribution of 100 families given below; the number of families corresponding to expenditure groups 20–40 and 60–80 are missing from the table. However the median is known to be 50. Find out the missing frequencies. Expenditure: 0–20 No. of families: 14 20–40 40–60 60–80 80–100 ? 27 ? 15 48 Solution: We shall assume the missing frequencies for the classes 20–40 to be x and 60–80 to y Expenditure (Rs.) 0–20 20–40 40–60 60–80 80–100 No. of families 14 x 27 y 15 N = 100 = 56 + x + y C.f 14 14 + x 14 + 27+ x 41 + x + y 41 + 15 + x + y From the table we have N = Σ F = 56 + x + y = 100. ∴ x + y = 100 – 56 + 44 Median is given as 50 which lies in the class 40–60, which becomes the median class. By using the median formula we get: N − Cf 0 Median = l1 + 2 ×i f ∴ 50 = 40 + 50 − (14 + x) × (60 − 40) 27 or 50 = 40 + 50 − (14 + x) × 20 27 36 − x 20 × 20 or 50 − 40 = (36 − x) × 27 27 or 50 − 40 = or 10 × 27 = 720 − 20x or ∴ 20x = 720 – 270 270 = 720 – 20x 450 = 22.5 20 By substituting the value of x in the equation, x + y = 44 we get, 22.5 + y = 44 ∴ y = 44 – 22.5 = 21.5. Hence frequency for the class 20–40 is 22.5 and 60–80 is 21.5. x= 2.10 MODE Mode is that value of the variable which occurs or repeats itself maximum number of times. The mode is the most “fashionable” size in the sense that it is the most common and typical and is defined by Zizek as “the value occurring most frequently in series of items and around which the other items are distributed most densely.” In the words of Croxton and Cowden, the mode of a distribution is the value at the point where the items tend to be most heavily concentrated. According to A.M. Tuttle, Mode is the value which has the greater frequency density in its immediate neighbourhood. In the case of individual observations, the mode is that value which is repeated the maximum number of times in the series. The value of mode can be denoted by the alphabet z also. 49 Example 28 : Calculate mode from the following data: Sr. Number : 1 2 3 4 5 Marks obtained : 10 27 24 12 27 6 27 7 20 8 18 9 15 10 30 Solution : Marks No. of Students 10 12 15 18 20 24 27 30 1 1 1 1 1 1 3 1 Calculation of Mode in Discrete series. In discrete series, it is quite often determined by inspection. We can understand with the help of an example : X 1 2 3 4 5 6 7 f 4 5 13 6 12 8 6 By inspection, the modal size is 3 as it has the maximum frequency. But this test of greatest frequency is not fool proof as it is not the frequency of a single class, but also the frequencies of the neighbour classes that decide the mode. In such cases, we shall be using the method of Grouping and Analysis table. Size of shoe 1 2 3 4 5 6 7 frequency 4 5 13 6 12 8 6 Solution : By inspection, the mode is 3, but the size of mode may be 5. This is so because the neighbouring frequencies of size 5 are greater than the neighbouring frequencies of size 3. This effect of neighbouring frequencies is seen with the help of grouping and analysis table technique. Grouping table Size of Shoe Frequency 1 2 3 50 4 5 6 When there exist two groups of frequencies in equal magnitude, then we should consider either both or omit both while analysing the sizes of items. Analysis Table Column Size of items with maximum frequency 1 3 2 5, 6 3 1, 2, 3, 4, 5 4 4, 5, 6 5 5, 6, 7 6 3, 4, 5 Item 5 occurs maximum number of times, therefore, mode is 5. We can note that by inspection we had determined 3 to be the mode. Determination of mode in continuous series : In the continuous series, the determination of mode requires one additional step. Once the modal class is determined by inspection or with the help of grouping technique, then the following formula of interpolation is applied: f1 − f0 Mode = l1 + 2 f − f − f (l2 − l1 ) 1 0 2 f1 − f0 Mode = l2 − 2 f − f − f (l2 − l1 ) 1 0 2 or l1 = lower limit of the class, where mode lies. l2 = upper limit of the class, where mode lies. f0 = frequency of the class proceeding the modal class. f1 = frequency of the class, where mode lies. f2 = frequency of the class succeeding the modal class. Example 29 : Calculate mode of the following frequency distribution : Variable Frequency 0–10 5 10–20 10 20–30 15 30-40 14 40–50 10 50–60 5 60–70 3 51 Solution : Grouping Table X 1 0–10 5 2 3 4 5 6 15 30 10–20 10 30 25 20–30 15 39 29 30–40 14 40–50 10 39 24 29 15 50–60 5 18 8 60–70 3 Analysis Table Column Size of item with maximum frequency 1 20–30 2 20–30, 30–40 3 10–20, 20–30 4 0–10, 10–20, 20–30 5 10–20, 20–30, 30–40 6 20–30, 30–40, 40–50 Modal group is 20–30 because it has occurred 6 times. Applying the formula of interpolation. f1 − f 0 Mode = l1 + 2 f − f − f (l2 − l1 ) 1 0 2 = 20 + 15 − 10 5 (30 − 20) = 20 + (10) = 28.3 30 − 10 − 14 6 Calculation of mode where it is ill defined. The above formula is not applied where there are many modal values in a series or a distribution. For instance there may be two or more than two items having the maximum frequency. In these cases, the series will be known as bimodal or multimodal series. The mode is said to be ill-defined and in such cases the following formula is applied. Mode = 3 Median – 2 Mean. 52 Example 30. Calculate mode of the following frequency data : Variate value Frequency 10–20 5 20–30 9 30–40 13 40–50 21 50–60 20 60–70 15 70–80 8 80–90 3 Solution : First of all, ascertain the modal group with the help of process of grouping. Grouping Table X 1 10–20 5 2 3 4 5 6 14 20–30 9 27 22 30–40 13 43 34 40–50 21 54 41 50–60 20 56 35 60–70 15 43 23 70–80 8 26 11 80–90 3 Analysis Table Column Size of item with maximum frequency 1 40–50 2 50–60, 60–70 3 40–50, 50–60 4 40–50, 50–60, 60–70 5 20–30, 30–40, 40–50, 50–60, 60–70, 70–80 6 30–40, 40–50, 50–60 53 There are two groups which occur equal number of items. They are 40–50 and 50–60. Therefore, we will apply the following formula: Mode = 3 median – 2 mean and for this purpose the values of mean and median are required to be computed. Calculation of Mean and Median Variate m – 45 10 frequency mid values X f m d′x fd′x cf 10–20 5 15 –3 – 15 5 20–30 9 25 –2 – 18 14 30–40 13 35 –1 – 13 27 40–50 21 45 0 0 50–60 20 55 +1 + 20 68 value of 60–70 15 65 +2 + 30 83 item which lies 70–80 8 75 +3 + 24 91 in (40–50) group 80–90 3 85 +4 + 12 94 48 Median is the N th 2 Σfd′ = +40 N = 94 N − cf 0 Med. = l1 + 2 × i. f 40 47 − 27 200 (10) = 40 + = 49.5 = 45 + (10) = 45 + 4.2 = 49.2 = 40 + 94 21 21 Mode = 3 median – 2 mean Σfd ′x X = A+ × i. N = 3 (49.5) –2 (49.2) = 148.5 – 98.4 = 50.1 2.10.1. Determination of mode by graph Mode can also be computed by curve fitting. The following steps are to be taken; (i) Draw a histogram of the data. (ii) Draw the lines diagonally inside the modal class rectangle, starting from each upper corner of the rectangle to the upper corner of the adjacent rectangle. (iii) Draw a perpendicular line from the intersection of the two diagonal lines to the X-axis. The abscissa of the point at which the perpendicular line meets is the value of the mode. Example 31 : Construct a histogram for the following distribution and, determine the mode graphically : 54 X : 0–10 10–20 20–30 30–40 40–50 f : 5 8 15 12 7 Verify the result with the help of interpolation. Solution : Mode = l1 + f1 − f 0 (l2 − l1 ) 2 f1 − f 0 − f 2 = 20 + 15 − 8 7 (30 − 20) = 20 + (10) = 27 30 − 8 − 12 10 Example 32 : Calculate mode from the following data : Marks No. of Students Below 10 4 ′′ 20 6 ′′ 30 24 ′′ 40 46 ′′ 50 67 ′′ 60 86 ′′ 70 96 ′′ 80 99 ′′ 90 100 Solution : Since we are given the cumulative frequency distribution of marks, first we shall convert it into the normal frequency distribution: 55 Marks Frequencies 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90 4 6–4=2 24 – 6 = 18 46 – 24 = 22 67 – 46 = 21 86 – 67 = 19 96 – 86 = 10 99 – 96 = 3 100 – 99 = 1 It is evident from the table that the distribution is irregular and maximum chances are that the distribution would be having more than one mode. You can verify by applying the grouping and analysing table. The formula calculate the value of mode in cases of bio-modal distributions is : Mode = 3 median – 2 mean. Computation of Mean and Median Marks X – 45 10 Mid-value Frequency (x) (f) cf (dx) 5 15 25 35 45 55 65 75 85 4 2 18 22 21 19 10 3 1 4 6 24 46 67 86 96 99 100 –4 –3 –2 –1 0 1 2 3 4 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90 Σf = 100 Mean = A + – 16 –6 – 36 – 22 0 19 20 9 4 Σfdx = – 28 Σfdx −28 × i = 45 + × 10 = 42.2 N 100 Median = size of fdx N 100 th item = = 50th item. 2 2 Because 50 is smaller to 67 in cf column. Median class is 40–50 N − Cf 0 Median = l1 + 2 ×i f 56 50 − 46 4 × 10 = 40 + × 10 = 41.9 21 21 Apply, Mode = 3 median –2 mean Median = 40 + Mode = 3 × 41.9 –2 × 42.2 = 125.7 – 84.3 = 41.3. Example 33 : Median and mode of the wage distribution are known to be Rs. 33.5 and 34 respectively. Find the missing values. Wages (Rs.) No. of workers 0–10 10–20 20–30 30–40 40–50 50–60 60–70 4 16 ? ? ? 6 4 Total = 230 Solution : We assume the missing frequencies as 20–30 as x, 30–40 as y, and 40–50 as 230 – (4 + 16 + x + y + 6 + 4) = 200 – x – y. We now proceed further to compute missing frequencies : Wages (Rs.) No. of workers Cumulative frequencies X f cf 0–10 10–20 20–30 30–40 40–50 50–60 60–70 4 16 x y 200 – x – y 6 4 4 20 20 + x 20 + x + y 220 226 230 N = 230 Apply, N − cf 0 Median = l1 + 2 × (l2 − l1 ) f = 30 + 115 − (20 + x) × (40 − 30) y y (33.5 – 30) = (115 – 20 – x)10 3.5 y = 1150 – 200 – 10x 57 10x + 3.5y = 950 ...................(i) f1 − f0 Mode = l1 + 2 f − f − f (l2 − l1 ) 1 0 2 Apply, = 20 + 15 − 8 (30 − 20) 30 − 8 − 12 4(3y – 200) = 10 (y – x) 10x + 2y = 800 ....................(ii) Subtract equation (ii) from equation (i), 1.5y = 150, y= 150 = 100 1.5 Substitute the value of y = 100 in equation (i), we get 10x + 3.5 (100) = 950 10x = 950 – 350 x = 600/10 = 60. ∴ Third missing frequency = 200 – x – y = 200 – 60 – 100 = 40. 2.11 SUMMARY ● Central tendency indicates the location of the centre of a set of data. It is the average value. ● An average is a typical value which is used to represent the entire set of values and is used as a benchmark to make comparisons. ● A good average is expected to be based on all values; not affected unduly by the presence of extremely large or small values in the data, amenable to further algebraic treatment and having sampling stability. ● Averages are distinguished as mathematical and positional. ● Arithmetic mean is a mathematical average which is most commonly used and understood and also very extensively used in statistical work. Obtained by dividing the sum of values by their number, it is easy to calculate. It enjoys well defined algebraic properties like zero-sum deviations, least squares, and combined mean. It meets most of the requisites of a good average. ● Geometric mean and harmonic mean are other mathematical averages but they have limited and specific uses. Their calculation is restricted only to positive values. It is possible to calculate combined average for two or more sets of data for each of these. ● Geometric mean, which is nth root of the product of n values, is basically applied to obtain average growth rates, price changes and depreciation rates. ● Harmonic mean is equal to the reciprocal of the arithmetic mean of reciprocals. It is used to average rates. Harmonic mean is used when the weights are in terms of the numerator factor of 58 the given rates. Arithmetic mean is correct to use when weights are in terms of the denominator factor. ● Mathematical averages can be simple or weighted and used accordingly as all values enjoy an equal or unequal weightage. Their values can be calculated only by using well-defined formulae and cannot be obtained graphically. Being based on all values, they are affected in a larger measure by the presence of extreme values. ● The positional averages include median and mode. While median refers to the central value in a set of arrayed values, the mode is that value in a series which appears the maximum number of times. A given set of individual values or a frequency distribution may have one or more modal values. If values in a given set of data are all unique, there is no mode. Mode suffers from these drawbacks. ● The positional averages do not possess any mathematical properties, except that the sum of absolute deviations from median is the least. ● In addition to median, there are a number of partition values that divide given distribution into a certain number of parts. They include quartiles, deciles and percentiles. The partition values are not averages but they are discussed here for the reason that their calculation proceeds in the same manner as that of median. They are used to locate relative position of different values clearly (like use of percentiles in the CAT entrance examinations) and also to calculate measures of variation, skewness etc. 2.12 SELF ASSESSMENT QUESTIONS Exercise 1 : True or False Statements (i) An average serves as a benchmark for comparisons. (ii) Mean, median and mode are called positional averages while geometric mean and harmonic mean are designated as mathematical averages. (iii) In the deviation method of calculating arithmetic mean, the mean is obtained by adding the mean of the deviations to the assumed mean value. (iv) Arithmetic mean is not suitable for open-ended frequency distributions. (v) All averages can be distinguished as being simple and weighted. (vi) In the weighted arithmetic mean calculation, it is immaterial whether the weights are expressed as, say, 20% and 80% or as 4 and 16. (vii) The sum of squares of deviations as well as the sum of deviations from mean is equal to zero. (viii) For calculating different measures of central tendency, it is necessary that all class intervals have equal width. (ix) Median cannot be calculated in open-ended class frequency distributions. (x) In an array of 41 items, median is equal to (41 + 1 )/2 = 21. (xi) Two sets of values. A and B. are identical except that their respective largest values are 80 and 8,000. The median of both the distributions shall be same. 59 (xii) The sum of absolute deviations from median is equal to zero. (xiii) The quartiles divide a distribution into four equal parts. (xiv) A distribution has 10 deciles and 100 percentiles. (xv) In a distribution of wages of the workers of a factory, the 95th percentile indicates the maximum wage earned by the top 95 percent of the workers. (xvi) The lower quartile in a distribution with a total frequency of 800 is equal to n/4 = 200. (xvii) The median, quartiles and percentiles can be determined graphically only by means of a “less than’ ogive. (xviii) The lower and the upper quartiles mark off the limits within which the middle 50 percent of the cases fall. (xix) It is possible to have more than one median in a given distribution. (xx) For every frequency distribution, the upper and lower quartiles are located at equal distance from median. (xxi) A distribution can have more than two modes. (xxii) In an array of marks (out of 100) scored by students of a class, the mark 47 appears 48 times (total number of students is 90). The mode of the set of values is 48. (xxiii) It is possible to estimate mode from median and mean of a distribution. (xxiv) For every distribution, X > Median > Mode. (xxv) Geometric mean is the nth root of the sum of n values. (xxvi) Geometric mean is the appropriate measure for averaging ratios and percentages. (xxvii) The geometric mean of two unequal values, x and v. is equal to the geometric mean of their arithmetic mean and harmonic mean values. (xxviii) The positional averages are not amenable to algebraic manipulations. (xxix) Only mathematical averages are distinguishable as being simple or weighted. (xxx) Mathematical averages cannot be determined graphically. Ans. 1. T, 2. F, 3. T, 4. T, 5. F, 6. T, 7. F, 8. F, 9. F, 10. F, 11. T, 12. F, 13. T, 14. F, 15. F, 16. F, 17. F, 18. T, 19. F, 20. F, 21. T, 22. F, 23. T, 24. F, 25. F, 26. T, 27. T, 28. T, 29. T, 30. T Exercise 2 : Questions and Answers (i) What do you understand by an average? Discuss the desirable properties of a good measure of central tendency. (ii) “An average is a number indicating the central value of a group of observations.” How far is it true for mean, median and mode? Give illustrations. (iii) State and explain the properties of arithmetic mean, geometric mean and harmonic mean. (iv) Write a note on how you would decide whether arithmetic mean or harmonic mean should be used to calculate average in a given case. 60 (v) Define median, quartiles, deciles and percentiles. State the property of median. Does it have any practical application? (vi) Write a detailed note on the choice of an average. Which average would be more suitable in the following cases: (a) Average size of ready-made garments sold by a store. (b) Average intelligence level of students of a class. (c) Average rate of growth of population per decade. (vii) A taxi ride in New Delhi costs Rs. 20 for the first kilometer and Rs. 11 per kilometer thereafter. Assume that the cost of each kilometer is incurred at the beginning of the kilometer. The waiting charges are Rs. 30 per hour or a part thereof, subject to a minimum of 15 minutes stay. Calculate the effective average cost per kilometer to a customer who rides a taxi from the Railway Station for her home 21.7 kilometers away and chooses to stay for a coffee for 25 minutes on the way. (viii) Find the arithmetic mean of the first 100 natural numbers. (ix) The arithmetic mean of a distribution is known to be 55.45. It is written below with the variate given in codified values. You are required to determine the class intervals. d′ –3 –2 –1 0 1 2 3 f 10 28 30 42 65 15 10 It is known that various d′ values have been calculated as (X – A)/10. (x) In a hotel, a total of 500 bulbs were installed simultaneously and their failure over time was observed as detailed below. End of week : 1 No. of failures : 12 2 3 4 5 6 7 40 108 242 346 428 500 You are required to calculate the mean life of the bulbs. (xi) A factory employs 100 workers. The mean daily wages of 99 of these workers is Rs. 85 while the wages of the 100th worker are Rs. 99 more than the mean wages of all the workers. Obtain mean wages of the workers of the factory. (xii) The following data gives the distribution of accidents in a large city over weekdays of the last month : Day: Average number of accidents: Sun. Mon. Tue. Wed. Thu. Fri. Sat. 26 16 12 10 8 10 8 Over the particular month, there were 5 Mondays and 5 Tuesdays. Calculate the mean number of accidents per day. (xiii) The table below shows the number of skilled and unskilled workers in two factories and their average daily wages. 61 Worker Factory A Factory B category Number Wages per day (Rs.) Number Wages per day (Rs.) Skilled 150 180 350 175 Unskilled 850 130 650 125 Determine the average daily wages for each factory. Also, give reasons why the results show that the average daily wages in Factory B are higher than the average daily wages in Factory A, even though in Factory B the average wages for both categories of workers are lower. (xiv) The average sales made by 18 salesmen in a company were reported to be Rs. 73,560 during the last month. It was discovered later that the sales made by one of the salesmen was recorded as Rs. 92,280 instead of the actual Rs. 22, 280 and that of another salesman at Rs. 16,630 instead of the actual Rs. 76,630. Compute the actual average sales made. (xv) The arithmetic mean of daily wages of 300 weavers and 250 spinning machine workers are Rs. 198 and Rs. 179, respectively. It is given that the arithmetic mean of all the workers in the factory is Rs. 187. Find the total number of workers in the factory if it is also given that the arithmetic mean of the daily wages of the remaining workers is Rs. 180.5. (xvi) Of a batch of 20 students, four students failed while 7 students passed with distinction. The remaining students, who passed ordinarily, scored the following marks: 47, 49, 52, 52, 57, 58, 59, 64, 66 The students who failed secured 23 marks on an average while the students who passed with distinction fetched 76 marks on an average. With this information, you are required to obtain the mean and median marks. (xvii) A survey of 350 families in a town yielded the following information : No. of children: 0 1 2 3 4 or more No. of families: 13 94 146 67 30 Find the median number of children in the families. (xviii) The following table gives the distribution of monthly income of 600 families in a certain city: Monthly income (in ‘000 Rs.) No. of families 10 – 20 60 20 – 30 170 30 – 40 200 40 – 50 60 50 – 60 50 60 – 70 40 70 – 80 20 62 Draw a ‘less than’ ogive and a ‘more than’ ogive for these data on a graph. Read the median income and the limits within which the middle 50 percent of families have their income. (xix) Find the missing frequencies in the following distribution if N = 100 and median = 30. Marks No. of students 0 – 10 10 10 – 20 ? 20 – 30 25 30 – 40 30 40 – 50 ? 50 – 60 10 (xx) The number of sales made by a shoe store in the City Mall during the past 20 days is as follows: 7 6 13 16 8 5 9 9 10 19 16 8 11 13 7 24 22 15 21 21 Find the 50th, 75th and 88th percentiles. Ans. 7. Rs. 12.14/km 8. 50.5 9. 20–30, 30–40, etc. 10. 4.15 Weeks 11. Rs.86 12. 14.27 13. A Rs. 138 B Rs. 143 14. Rs. 73.004 15. 750 16. Mean = 56.4, Median = 58.5 17. 2 18. Med = 33.5, 25–43 app. 19. 15 and 10 20. 12, 18.25 and 21.48 63 LESSON-3 MEASURES OF VARIATION – ABSOLUTE AND RELATIVE 3. STRUCTURE 3.0 3.1 3.2 3.3 3.4 3.5 Objective Need of Variation What is Variation? Requisites of a Good Measure of Variation Types of Variation Methods of Computing Variation 3.5.1 Mathematical Methods – Range 3.5.2 Quartile Deviation 3.5.3 Average Deviation 3.5.4 Standard Deviation 3.5.5 Mathematical Properties of Standard Deviation 3.5.6 Graphic Method of Variation 3.6 Revisionary Problems 3.7 Summary 3.8 Self Assessment Questions 3.0 OBJECTIVE After reading this lesson, you should be able to : (a) Understand the meaning, need and requisites of a good measure of variation (b) Differentiate between absolute and relative measure of variation (c) Identify and compute different types of variation such as range, quartile deviation, average deviation, standard deviation and variance (d) Comprehend merits and demerits of different measures and properties of standard deviation (e) Comment upon the variability of problems with the help of coefficient of variation. 3.1 NEED OF VARIATION Measures of central tendency, Mean, Median, Mode, etc., indicate the central position of a series. They indicate the general magnitude of the data but fail to reveal all the peculiarities and characteristics of the series. In other words, they fail to reveal the degree of the spread out or the extent of the variability in individual items of the distribution. This can be known by certain other measures, known as ‘Measures of Dispersion’ or Variation. We can understand variation with the help of the following example : 64 Series I Series II Series III 10 2 10 10 8 12 10 20 8 ΣX = 30 30 30 X= ΣX 30 = = 10 N 3 X= 30 = 10 3 X= 30 = 10 3 In all three series, the value of arithmetic mean is 10. On the basis of this average, we can say that the series are alike. If we carefully examine the composition of three series, we find the following differences : (i) In case of 1st series, the value are equal; but in 2nd and 3rd series, the values are unequal and do not follow any specific order. (ii) The magnitude of deviation, item-wise, is specific different for the 1st, 2nd and 3rd series. But all these deviations cannot be ascertained if the value of ‘simple mean’ is taken into consideration. (iii) In these three series, it is quite possible that the value of arithmetic mean is 10; but the value of median may differ from each other. This can be understood as follows : I II III 10 2 8 10 Median 8 Median 10 Median 10 20 12 The value of ‘Median’ in 1st series is 10, in 2nd series = 8 and in 3rd series = 10. Therefore, the value of Mean and Median are not identical. (iv) Even though the average remains the same, the nature and extent of the distribution of the size of the items may vary. In other words, the structure of the frequency distributions may differ even though their means are identical. 3.2 WHAT IS VARIATION Simplest meaning that can be attached to the word ‘dispersion’ is a lack of uniformity in the sizes or quantities of the items of a group or series. According to Reiglemen, “Dispersion is the extent to which the magnitudes or qualities of the items differ, the degree of diversity.” The word dispersion may also be used to indicate the spread of the data. In all these definitions, we can find the basic property of dispersion as a value that indicates the extent to which all other values are dispersed about the central value in a particular distribution. 65 3.3 REQUISITES OF A GOOD MEASURE OF VARIATION There are certain pre-requisites for a good measure of dispersion : 1. It should be simple to understand. 2. It should be easy to compute. 3. It should be rigidly defined. 4. It should be based on each individual item of the distribution. 5. It should be capable of further algebraic treatment. 6. It should have sampling stability. 7. It should not be unduly affected by the extreme items. 3.4 TYPES OF VARIATION The measures of dispersion can be either ‘absolute’ or ‘relative’. Absolute measures of dispersion are expressed in the same units in which the original data are expressed. For example, if the series is expressed as Marks of the students in a particular subject; the absolute dispersion, will provide the value in Marks. The only difficulty is that if two or more series are expressed in different units, the series cannot be compared on the basis of dispersion. ‘Relative’ or ‘Coefficient’ of dispersion is the ratio or the percentage of a measure of absolute dispersion to an appropriate average. The basic advantage of this measure is that two or more series can be compared with each other, despite the fact they are expressed in different units. Theoretically, ‘Absolute measure’ of dispersion is better. But from a practical point of view, relative or coefficient of dispersion is considered better as it is used to make comparison between series. 3.5 METHODS OF COMPUTING VARIATION Methods of studying dispersion are divided into two types : (i) Mathematical Methods : We can study the ‘degree’ and ‘extent’ of variation by these methods. In this category, commonly used measures of dispersion are : (a) Range (b) Quartile Deviation (c) Average Deviation (d) Standard deviation and coefficient of variation. (ii) Graphic Methods : Where we want to study only the extent of variation, whether it is higher or lesser a Lorenz-curve is used. 3.5.1 Mathematical Methods – Range: It is the simpest method of studying dispersion. Range is the difference between the smallest 66 value and the largest value of a series. While computing range, we do not take into account frequencies of different groups. Formula: Absolute Range = L – S Coefficient of Range = L −S L+S where, L represents largest value in a distribution S represents smallest value in a distribution We can understand the computation of range with the help of examples of different series. (i) Raw Data Example 1 : Marks out of 50 in a subject of 12 students, in a class are given as follows : 12, 18, 20, 12, 16, 14, 30, 32, 28, 12, 12 and 35. In the example, the maximum or the highest marks obtained by a candidate is ‘35’ and the lowest marks obtained by a candidate is ‘12’. Therefore, we can calculate range; L = 35 and S = 12 Absolute Range = L – S = 35 – 12 = 23 marks Coefficient of Range = L − S 35 − 12 23 = = = 0.49 approx. L + S 35 + 12 47 (ii) Discrete Series Example 2 : Marks of the Students in Accounts (out of 50) Smallest Largest No. of students (X) (f) 10 4 12 10 18 16 20 15 Total 45 Absolute Range = 20 – 10 = 10 marks Coefficient of Range = 20 − 10 10 = = 0.34 approx. 20 + 10 30 67 (iii) Continuous Series Example 3 : X frequencies 10 – 15 4 S = 10 15 – 20 10 L = 30 20 – 25 26 25 – 30 8 Absolute Range = L – S = 30 –10 = 20 marks Coefficient of Range = L − S 35 − 12 20 = = = 0.5 approx. L + S 35 + 12 40 Range is a simplest method of studying dispersion. It takes lesser time to compute the ‘absolute’ and ‘relative’ range. Range does not take into account all the values of a series, i.e. it considers only the extreme items and middle items are not given any importance. Therefore, Range cannot tell us anything about the character of the distribution. Range cannot be computed in the case of ‘open ends’ distribution i.e., a distribution where the lower limit of the first group and upper limit of the higher group is not given. The concept of range is useful in the field of quality control and, to study the variations in the prices of the shares etc. 3.5.2 Quartile Deviation The concept of ‘Quartile Deviation’ does take into account only the values of the ‘Upper quartile’ (Q3) and the ‘Lower quartile’ (Q1). Quartile Deviation is also called ‘inter-quartile range’. It is a comparatively better method when we are interested in knowing the range within which certain proportion of the items fall. ‘Quartile Deviation’ can be obtained as : (i) Inter-quartile range = Q3 – Q1 (ii) Semi-quartile range = Q3 − Q1 2 Q3 − Q1 Q3 + Q1 Calculation of Inter-quartile Range, semi-quartile Range and Coefficient of Quartile Deviation in case of Raw Data (iii) Coefficient of Quartile Deviation = Example 4 : Suppose the values of X are : 20, 12, 18, 25, 32, 10 In case of quartile-deviation, it is necessary to calculate the values of Q1 and Q3 by arranging the given data in ascending or descending order. Therefore, the arranged data are : (in ascending order) X = 10, 12, 18, 20, 25, 32 No. of items = 6 68 N + 1 th item = Q1 = the value of 4 6 + 1 = 1.75th item 4 = the value of 1st item + 0.75 (value of 2nd item – value of 1st item) = 10 + 0.75 (12 – 10) = 10 + .75 (2) 10 + 1.50 = 11.50 N + 1 6 + 1 th item = 3 Q3 = the value of 3 4 4 = the value of 3(7/4)th item = the value of 5.25th item = the value of 5th item + 0.25 (the value of 6th item minus the value of 5th item) = 25 + 0.25 (32 – 25) = 25 + 0.25 (7) = 26.75. Therefore, (i) Inter-quartile range = Q3 – Q1 = 26.75 – 11.50 = 15.25 (ii) Semi-quartile range = Q3 − Q1 15.25 = = 7.625 2 2 (iii) Coefficient of Quartile Deviation = Q3 − Q1 26.75 − 11.50 15.25 = = = 0.39 approx. Q3 + Q1 26.75 + 11.50 38.25 Calculation of Inter-quartile Range, semi-quartile Range and Coefficient of Quartile Deviation in discrete series Example 5 : Suppose a series consists of the salaries (Rs.) and number of the Workers in a factory : Salaries (Rs.) No. of workers 60 4 100 20 120 21 140 16 160 9 Solution : In the problem, we will first, compute the values of Q3 and Q1. Salaries (Rs.) No. of workers Cumulative frequencies (x) (f) (c.f.) 60 4 4 100 20 24 – Q1 lies in this cumulative 120 21 45 140 16 61 160 9 70 N = Σf = 70 69 frequency Calculation of Q1 : Calculation of Q3 : N + 1 th item Q1 = size of 4 N + 1 th item Q3 = size of 3 4 70 + 1 th item = 17.75th item = size of 4 70 + 1 th item = 53.25th item = size of 3 4 17.75 lies in the cumulative frequency 24, 53.25 lies in the cumulative frequency 61 which is corresponding to the value Rs. 100 which is corresponding to Rs. 140 ∴ Q1 = Rs. 100 ∴ Q3 = Rs. 140 (i) Inter-quartile range = Q3 – Q1 = Rs. 140 – Rs. 100 = Rs. 40 (ii) Semi-quartile range = Q3 − Q1 140 − 100 = = Rs. 20 2 2 Q3 − Q1 140 − 100 40 (iii) Coefficient of Quartile Deviation = Q + Q = 140 + 100 = 240 = 0.17 approx. 3 1 Calculation of Inter-quartile range, semi-quartile range and Coefficient of Quartile Deviation in the case of continuous series Example 6 : We are given the following data : Salaries (Rs.) No. of workers 10–20 4 20–30 6 30–40 10 40–50 5 Total 25 In this example, the values of Q3 and Q1 are obtained as follows : Salaries (Rs.) (x) No. of workers (f) Cumulative frequencies (c.f.) 10–20 4 4 20–30 6 10 30–40 10 20 40–50 5 25 N = 25 70 N − cf 0 N Q1 = l1 + 4 is used to find out Q1 group ×i 4 f where l1 = lower limit of Q1 group f = frequency of Q1 group i = magnitude of Q1 group (l2 – l1) cf0 = cumulative frequency of the group preceeding Q1 group. N 25 or or 6.25. It lies in the cumulative frequency 10, which is corresponding to 4 4 class 20–30. Therefore, Therefore, Q1 group is 20–30. Q1 = 20 + where, l1= 20, f = 6, i = 10, 6.25 − 4 × 10 = 20 + 3.75 = 23.75 6 N = 6.25, cf0 = 4 4 3N − Cf 0 Q3 = l1 + 4 ×i f 3 N 3 × 25 75 = = 18.75 which lies in the cumulative frequency 20, which is corresponding = 4 4 4 to class 30–40. Therefore Q3 group is 30–40. where, l1= 30, i = 10, 3N = 18.75, cf0 = 10, f = 10 4 Q3 = 30 + 18.75 − 10 × 10 = Rs. 38.75 10 Therefore : (i) Inter-quartile range = Q3 – Q1 = Rs. 38.75 – Rs. 23.75 = Rs. 15.00 (ii) Semi-quartile range = Q3 − Q1 15.00 = = 7.50 2 2 (iii) Coefficient of Quartile Deviation = Q3 − Q1 Rs. 38.75 – Rs. 23.75 15 = = = 0.24. Q3 + Q1 Rs. 38.75 + Rs. 23.75 62.50 71 Advantages of Quartile Deviation Some of the important advantages of this measure of dispersion are : (i) It is easy to calculate. We are required simply to find the values of Q1 and Q3 and then apply the formula of absolute and coefficient of quartile deviation. (ii) It has better results than range method. While calculating range, we take only the extreme values that make dispersion erratic. In the case of quartile deviation, we take into account middle 50% items. (iii) The quartile deviation is not affected by the extreme items. Disadvantages (i) It is completely dependent on the central items. If these values are irregular and abnormal the result is bound to be affected. (ii) All the items of the frequency distribution are not given equal importance in finding the values of Q1 and Q3. (iii) Because it does not take into account all the items of the series, considered to be inaccurate of dispersion. Similarly, somtimes we calculate percentile range, say, 90th and 10th percentile as it gives slightly better measure of dispersion, in certain cases. If we consider the calculations, then (i) Absolute percentile range = P90 – P10 (ii) Coefficient of percentile range = P90 − P10 P90 + P10 This method of calculating dispersion can be applied generally in the case of open end series where the importance of extreme values are not considered. 3.5.3 Average Deviation Average deviation’ is defined as a value, which is obtained by taking the average of the deviations of various items, from a measure of central tendency, Mean or Median or Mode, after ignoring negative signs. Generally, the measure of central-tendency, from which the deviations are taken, is specified in the problem. If nothing is mentioned regarding the measure of central tendency specified then deviations are taken from median because the sum of the deviations (after ignoring negative signs) is minimum. Computation in case of raw data (i) Absolute Average Deviation about Mean or Median or Mode = Σ|d | N where: N = Number of observations, |d| = deviations taken from Mean or Median or Mode ignoring signs. (ii) Coefficient of A.D. = Average Deviation about Mean or Median or Mode Mean or Median or Mode 72 Steps to Compute Average Deviation : (i) Calculate the value of Mean or Median or Mode (ii) Take deviations from the given measure of central-tendency and they are shown as d. (iii) Ignore the negative signs of the deviation that can be shown as |d| and add them to find Σ |d|. (iv) Apply the formula to get Average Deviation about Mean or Median or Mode. Example 7 : Suppose the values are 5, 5, 10, 15, 20. We want to calculate Average Deviation and Coefficient of Average Deviation about Mean or Median or Mode. Solution : Average Deviation about mean (Absolute and Coefficient). Deviation from mean Deviations after ignoring signs d |d| 5 –6 6 5 –6 6 10 +1 1 15 +4 4 20 +9 9 (X) ΣX = 55 X = ΣX N where N = 5, ΣX = 55 X = 55 = 11 5 Σ |d| = 26 Average Deviation about Mean = Σ | d | 26 = = 5.2. N 5 Coefficient of Average Deviation about Mean = Mean Deviation about Mean 5.2 = = 0.47. Mean 11 Average Deviation (Absolute and Coefficient) about Median X Deviation from median d Deviations after ignoring negative signs |d| 5 –5 5 5 –5 5 Median 10 0 0 15 +5 5 20 + 10 10 Σ|d| = 25 N=5 73 Average deviation about Median = Σ | d | 25 = = 5.2 . N 5 Coefficient of Average Deviation about median = A.D. about Median 5 = = 0.5 Median 10 Average Deviation (Absolute and Coefficient) about Mode X Deviation from mode d |d| 5 0 0 Mode 5 0 0 10 +5 5 15 + 10 10 20 + 15 15 Σ | d | = 30 N=5 Average deviation about Mode = Σ | d | 30 = = 6. N 5 Coefficient of Average Deviation about mode = A.D. about Mode 6 = = 1.2. Mode 5 Average deviation in case of discrete and continuous series Average Deviation about Mean or Median or Mode = Σf | d | N where N = No. of items | d | = deviations from Mean or Median or Mode, after ignoring negative signs. Coefficient of A.D. about Mean or Median or Mode = A.D. about Mean or Median or Mode Value of Mean or Median or Mode Example 8 : Suppose we want to calculate coefficient of Average Deviation about Mean from the following discrete series: X Frequency 10 5 15 10 20 15 25 10 30 5 74 Solution : First of all, we shall calculate the value of arithmetic Mean, Calculation of Simple Mean X f fX 10 5 50 15 10 150 20 25 30 15 10 5 N = 45 300 250 150 ΣfX = 900 X= ΣfX 900 = = 20 N 45 Calculation of Coefficient of Average Deviation about Mean Deviation from mean Deviations after ignoring X f d negative signs | d | 10 15 20 25 30 5 10 15 10 5 – 10 –5 0 +5 + 10 10 5 0 5 10 Σ f|d| 50 50 0 50 50 Σf |d| = 200 N = 45 Coefficient of Average Deviation about Mean = Average Deviation about Mean = A.D. about Mean 4.4 = = 0.22 Mean 20 Σ | d | 200 = = 4.44 approx. N 45 In case we want to calculate coefficient of Average Deviation about Median from the following data : Class Interval Frequency 10–14 15–19 20–24 25–29 30–34 5 10 15 10 5 N = 45 First of all we shall calculate the value of Median but it is necessary to find the ‘real limits’ of the given class-intervals. This is possible by subtracting 0.5 from the lower-limits and added to the 75 upper limits of the given classes. Hence, the real limits shall be : 9.5–14.5, 14.5–19.5, 19.5–24.5, 24.5–29.5 and 29.5–34.5. Calculation of Median f c.f. 5 10 15 10 5 5 15 30 40 45 Class Intervals 9.5–14.5 14.5–19.5 19.5–24.5 24.5–29.5 29.5–34.5 N = 45 N − Cf0 Median = l1 + 2 ×i f where l1 = lower limit of median group i = magnitude of median group f = frequency of median group Cf0 = cumulative frequency of the group preceeding median group n 2 = size of median group N 45 th item i.e. = 22.5 2 2 It lies in the cumulative frequency 30, which is corresponding to class interval 19.5–24.5. ∴ Median size = Median group is 19.5–24.5 Median = 19.5 + 22.5 − 15 7.5 × 5 = 19.5 + × 5 = 19.5 + 2.5 = 19.5 + 2.5 = 22 15 15 Calculation of Coefficient of Average Deviation about Median Class Intervals Frequency Mid points Deviation from Deviations after ignoring f x median (22) negative signs |d| f|d| 9.5–14.5 5 12 – 10 10 50 14.5–19.5 10 17 –5 5 50 19.5–24.5 15 22 0 0 0 24.5–29.5 10 27 +5 5 50 29.5–34.5 5 32 + 10 10 50 Σf |d| = 200 N = 45 76 Coefficient of Average Deviation about Median = Average Deviation about Mean = A.D. about Median Median Σ | d | 200 = = 4.44 approx. N 45 Coefficient of A.D. about Median = 4.4 = 0.2. 22 Advantages of Average Deviations 1. Average deviation takes into account all the items of a series and hence, it provides sufficiently representative results. 2. It simplifies calculations since all signs of the deviations are taken as positive. 3. Average Deviation may be calculated either by taking deviations from Mean or Median or Mode. 4. Average Deviation is not affected by extreme items. 5. It is easy to calculate and understand. 6. Average deviation is used to make healthy comparisons. Disadvantages of Average Deviations 1. It is illogical and mathematically unsound to assume all negative signs as positive signs. 2. Because the method is not mathematically sound, the results obtained by this method are not reliable. 3. This method is unsuitable for making comparisons either of the series or structure of the series. This method is more effective during the reports presented to the general public or to groups who are not familiar with statistical methods. 3.5.4 Standard Deviation The standard deviation, which is shown by greek letter σ (read as sigma) is extremely useful in judging the representativeness of the mean. The concept of standard deviation which was introduced by Karl Pearson, has a practical significance because it is free from all defects which exists in case of range, quartile deviation or average deviation. Standard deviation is calculated as the square root of average of squared deviations taken from actual mean. It is also called root mean square deviation. The square of standard deviation i.e. σ2 is called ‘variance’ in statistics. Calculation of standard deviation in case of raw data There are four ways of calculating standard deviation for raw data : (i) When actual values are considered; (ii) When deviations are taken from actual mean; (iii) When deviations are taken from assumed mean; and (iv) When ‘step deviations’ are taken from assumed mean. 77 (i) When the actual values are considered : σ= ΣX 2 − ( X )2 N where N = Number of the items, ΣX 2 − ( X )2 or σ = N 2 X = Given values in the series X = Arithmetic mean of the values. We can also write the formula as follows : σ= ΣX 2 ΣX − N N 2 ΣX where X = N Steps to calculate σ (i) Compute simple mean of the given values. (ii) Square the given values and aggregate them (iii) Apply the formula to find the value of standard deviation. Example 9 : Suppose the values are given 2, 4, 6, 8, 10. We want to apply the formula σ= ΣX 2 − ( X )2 N Solution : We are required to calculate the values of N , X , ΣX 2 . They are calculated as follows : 220 − (6)2 5 X X2 σ= 2 4 = 4 16 Variance (σ2) = ( 8) 2 = 8 6 36 X = 8 64 10 100 N=5 ΣX2 = 220 44 − 36 = 8 = 2.828 ΣX 30 = =6 N 5 There are certain specific problems, where the method can be applied. It is different type of problem which is given as follows : 78 (ii) When the deviations are taken from actual mean σ= Σx 2 N where N = no. of items and x = ( X − X ) Steps to Calculate σ (i) Compute the deviations of given values from actual mean i.e., ( X − X ) and represent them by x. (ii) Square these deviations and aggregate them (iii) Use the formula, σ = Σx 2 N Example 10 : We are given values as 2, 4, 6, 8, 10. We want to find out standard deviation. X x2 x = (X – X ) 2 2 – 6 = –4 (– 4 )2 = 16 4 4 – 6 = –2 (–2)2 = 4 6 6–6 =0 8 8–6 =+2 (2)2 = 4 10 10 – 6 = + 4 (4)2 = 16 =0 Σx2 = 40 N=5 ΣX 30 X = 6 N = 5 Σx 2 40 = = 8 = 2.828 N 5 (iii) When the deviations are taken from assumed mean σ= σ= where, Σdx 2 Σdx − N N 2 N = no. of items. dx = deviations from assumed mean i.e., (X – A). A = assumed mean Steps to Calculate : (i) We consider any value as assumed mean. The value may be given in the series or may not be given in the series. (ii) We take deviations from the assumed value i.e., (X – A), to obtain dx for the series and aggregate them to find Σdx. 79 (iii) We square these deviations to obtain dx2 and aggregate them to find Σdx2. (iv) Apply the formula given above to get standard deviation : Example 11 : Suppose the values are given as 2, 4, 6, 8 and 10. We can obtain the standard deviation as: assumed mean (A) X dx = (X – A) dx2 2 – 2 = (2 – 4) 4 4 0 = (4 – 4) 0 6 + 2 = (6 – 4) 4 8 + 4 = (8 – 4) 16 10 + 6 = (10 – 4) 36 N=5 Σ dx = 10 Σ dx2 = 60 2 σ= Σdx 2 Σdx − = N N 2 60 10 − = 12 − 4 = 8 = 2.828. 5 5 (iv) When step deviations are taken from assumed mean 2 σ= Σdx 2 Σdx − ×i N N X − A where i = Common factor, N = Number of items, dx = Step-deviations = i Steps to Calculate σ : (i) We consider any value as assumed mean from the given values or from outside. (ii) We take deviation from the assumed mean i.e., (X – A), (iii) We divide the deviations obtained in step (ii) with a common factor to find step deviations and represent them as dx and aggregate them to obtain Σ dx. (iv) We square the step deviations to obtain dx2 and aggregate them to find Σdx2. Example 12 : We continue with the same example to understand the computation of Standard Deviation. d dx = ,i = 2 i dx2 X d = (X – A) 2 –2 –1 1 A=4 0 0 0 6 +2 1 1 8 +4 2 4 10 +6 3 9 Σdx = 5 Σdx2 = 15 N=5 80 2 σ = Σdx 2 Σdx − ×i N N σ = 15 5 − × 2 = 3 − 1 × 2 = 2 × 2 = 1.414 × 2 = 2.828. 5 5 where N = 5, i = 2, dx = 5 Σdx2 = 15 2 Note : We can notice an important point that the standard deviation value is identical by four methods. Therefore, any of the four formulae can be applied to find the value of standard deviation. But the suitability of a formula depends on the magnitude of items in a question. σ X In the above given example, σ = 2.828 and X = 6 Coefficient of Standard-deviation = Therefore, coefficient of standard deviation = σ 2.828 = = 0.471 X 6 Coefficient of Variation or C.V. = σ 2.828 × 100 = × 100 = 47.1% X 6 Generally, coefficient of variaton is used to compare two or more series. If coefficient of variation (C.V.) is more in one series as compared to the other, there will be more variations in that series, lesser stability or consistency in its composition. If coefficient of variation is lesser as compared to other series, it will be more stable, or consistent. Moreover, that series is always better where coefficient of variation is lesser or coefficient of standard deviation is lesser. Example 13 : Suppose we want to compare two firms where the salaries of the employees are given as follows : Firm A Firm B No. of workers 100 100 Mean salary (Rs.) 100 80 40 45 Standard-deviation (Rs.) Solution : We can compare these firms either with the help of coefficient of standard deviation or coefficient of variation. If we use coefficient of variation, then we shall apply the formula : σ C.V. = × 100 X Firm A C.V. = Firm B 40 × 100 = 40% 100 C.V. = 45 × 100 = 56.25% 80 X = 100, σ = 40 X = 80, σ = 45. Because the coefficient of variation is lesser for firm A as compared to firm B, therefore, firm A is better. 81 Calculation of standard-deviation in discrete and continuous series We use the same formula for calculating standard deviation for a continuous series and a discrete series. The only difference that in discrete series, values and frequencies are given whereas in a continuous series, class-intervals and frequencies are given. When the mid-points of these classintervals are obtained, a continuous series takes the shape of a discrete series. Alphabet X denotes values in a discrete series and mid points in a continuous series. When the deviations are taken from actual mean We use the same formula for calculating standard deviation for a continuous series σ = where Σfx 2 N N = Number of the items (Σf) f = Frequencies corresponding to different values or class-intervals. x = Deviations from actual mean ( X − X ) X = Values in a discrete series and mid-points in a continuous series. Step to calculate σ (i) Compute the arithmetic mean by applying the required formula. (ii) Take deviations from the arithmetic mean and represent these deviations by x. (iii) Square the deviations to obtain values of x2. (iv) Multiply the frequencies of the different class-intervals with x2 to find fx2. Aggregate fx2 column to obtain Σfx2. (v) Apply the formula to obtain the value of standard deviation. 2 If we want to calculate variance then we can take σ = Σfx 2 N Example 14 : We can understand the procedure by taking an example : Class Intervals Frequency (f) Midpoints (m) fm 10 – 14 5 12 60 15 – 19 10 17 170 20 – 24 15 22 330 25 – 29 10 27 270 30 – 34 5 32 160 Σ fm = 990 N= 45 Therefore, X = Σfm 990 = = 22 N 45 where, N = 45, Σ fm = 990 82 Calculation of Standard Deviation Class Intervals Mid points Deviations from actual median = 22 f X x x2 f x2 10 – 14 5 12 –10 100 500 15 – 19 10 17 –5 25 250 20 – 24 15 22 0 0 0 25 – 29 10 27 +5 25 250 30 – 34 5 32 + 10 100 500 Σfx2 = 1500 N = 45 σ = Σfx 2 where, N = 45, Σfx 2 = 1500 N σ = 1500 = 33.33 = 5.77 approx. 45 When the deviations are taken from assumed mean In some cases, the value of simple mean may be in fractions, then it becomes time consuming to take deviations and square them. Alternatively, we can take deviations from the assumed mean. σ = where Σfdx 2 Σfdx − N N 2 N = number of items, dx = deviations from assumed mean (X – A), f = frequency of the different groups, A = assumed mean and X = values or mid points. Step to calculate σ (i) Take the assumed mean from the given values or mid points. (ii) Take deviations from the assumed mean and represent them by dx. (iii) Square the deviations to get dx2 . (iv) Multiply f with dx of different groups to obtain fdx and add them up to get Σ fdx. (v) Multiply f with dx2 of different groups to obtain fdx2 and add them up to get Σ fdx2. (vi) Apply the formula to get the value of standard deviation. 83 Example 15 : We can understand the procedure with the help of an example. Class Intervals Frequency Mid Deviations from point assumed Mean = (17) f x dx dx2 fdx fdx2 10 – 14 5 12 –5 25 – 25 125 15 – 19 10 17 0 0 0 0 20 – 24 15 22 +5 25 75 375 25 – 29 10 27 + 10 100 100 1000 30 – 34 5 32 + 15 225 75 1125 Σ fdx =225 Σfdx2 = 2625 N = 45 σ= Σfdx 2 Σfdx − N N 2 where, N = 45, Σfdx 2 = 2625, Σfdx = 225 2 ∴σ = 2625 225 − = 58.33 − 25 = 33.33 = 5.77 approx. 45 45 When the step deviations are taken from the assumed mean 2 σ= where Σfdx 2 Σfdx − ×i N N N = Number of the items (Σf), i = common factor f = frequencies correspondig to the different groups, X − A dx = step-deviations i Steps to calculate σ (i) Take deviations from the assumed mean of the calculated mid-points and divide all deviations by a common factor (i) and represent these values by dx. (ii) Square these step deviations dx to obtain dx2 for different groups. (iii) Multiply f with dx of different groups to find fdx and add them to obtain Σ fdx. (iv) Multiply f with dx2 of different groups to find fdx2 for different groups and add them to obtain Σ fdx2. (v) Apply the formula to get standard deviation. Example 16 : Suppose we are given the series and we want to calculate standard deviation with the 84 help of step deviation method. According to the given formula, we are required to calculate the value of i, N, Σfdx and Σfdx2. Class Frequency Mid Deviations from point assumed mean (22) i=5 f x X X – A i dx 10 – 14 5 12 – 10 –2 4 – 10 20 15 – 19 10 17 –5 –1 1 – 10 10 20 – 24 15 22 +0 0 0 0 0 25 – 29 10 27 +5 +1 1 10 10 30 – 34 5 32 + 10 +2 4 10 20 Intervals dx2 fdx fdx2 Σfdx = 0 Σfdx2 = 60 N =45 2 σ = Σfdx 2 Σfdx − × i, N N 2 ∴σ = 60 0 − ×5= 45 45 where, N = 45, i = 5, Σfdx = 0, Σfdx2 = 60 4 × 5 = 1.33 × 5 = 1.154 × 5 = 5.77 approx. 3 Advantages of Standard Deviation (i) Standard deviation is the best measure of dispersion because it takes into account all the items and is capable of future algebraic treatment and statistical analysis. (ii) It is possible to calculate standard deviation for two or more series. (iii) The measure is most suitable for making comparisons among two or more series about variability. Disadvantages (i) It is difficult to compute. (ii) It assign more weights to extreme items and less weights to items that are nearer to mean. It is because of this fact that the squares of the deviations which are large in size would be proportionately greater than the squares of those deviations which are comparatively small. 3.5.5 Mathematical Properties of Standard Deviation (i) If deviations of given items are taken from arithmetic mean and squared then the sum of squared deviation should be minimum, i.e., ( X − X ) 2 = Minimum. (ii) If different values are increased or decreased by a constant, the standard deviation will remain the same. Whereas if different values are multiplied or divided by a constant than the standard deviation will be multiplied or divided by that constant. 85 (iii) Combined standard deviation can be obtained for two or more series with formula given below : σ12 = N1σ12 + N2σ 22 + N1d12 + N 2 d 22 N1 + N 2 where : N1 represents number of items in first series, N2 represents number of items in second series, σ12 represents variance of first series, σ22 represents variance of second series, d1 represents the difference between X 12 − X 1 , d2 represents the difference between X 12 − X 2 , X 1 represents arithmetic mean of first series, X 2 represents arithmetic mean of second series, X12 represents combined arithmetic mean of both the series. Example 17 : Find the combined standard deviation of two series, from the below given information : First Series Second Series No. of items 10 15 Arithmetic means 15 20 Standard deviation 4 5 Solution : Since we are considering two series, therefore combined standard deviation is computed by the following formula : σ12 = N1σ12 + N2σ 22 + N1d12 + N 2 d 22 N1 + N 2 where : N1= 10, N2 = 15, X1 = 15, X 2 = 20, σ1 = 4, σ 2 = 5 X 12 = or X12 = X1 N1 + X 2 N 2 N1 + N 2 (15 × 10) + (20 × 15) 150 + 300 450 = = = 18 10 + 15 25 25 d1 = ( X 12 − X 1 ) = 18 − 15 = 3 d 2 = ( X 12 − X 2 ) = 18 − 20 = −2. and By applying the formula of combined standard deviations, we get: 86 10(4) 2 + 15(5) 2 + 10(18 − 15)2 + 15(18 − 20) 2 10 + 15 σ12 = = (10 × 16) + (15 × 25) + (10 × 9) + (15 × 4) 25 = 160 + 375 + 90 + 60 685 = = 27.4 = 5.2 approx. 25 25 (iv) Standard deviation of n natural numbers can be computed as : σ = 1 ( N 2 − 1) where, N represents number of items 12 (v) For a symmetrical distribution, X ± σ covers 68.27% of items, X ± 2σ covers 95.45% of items, X ± 3σ covers 99.73% of items. Example 18 : You are heading a rationing department in a State affected by food shortage. Local investigators, submit the following report: Daily calorie value of food available per adult during current period : Area Mean Standard Deviation A 2,500 400 B 2,000 200 The estimated requirement of an adult is taken as 2,800 calories daily and the absolute minimum is 1,350. Comment on the reported figures, and determine which area, in your opinion, need more urgent attention. Solution: We know that X ± σ covers 68.27% of items, X ± 2σ covers 95.45% of items, X ± 3σ covers 99.73% cases. In the given problem if we take into consideration 99.73%, i.e., almost the whole population, the limits; would be X ± 3σ. For Area A these limits are : X + 3σ = 2,500 + (3 × 400) = 3,700 X – 3σ = 2,500 – (3 × 400) = 1,300 87 For Area B these limits are : X + 3σ = 2,000 + (3 × 200) = 2,600 X – 3σ = 2,000 – (3 × 200) = 1,400 It is clear from above limits that in Area A there are some persons who are getting 1300 calories, i.e. below the minimum which is 1,350. But in case of area B there is no one who is getting less than the minimum. Hence area A needs more urgent attention. (vi) Relationship between quartile deviation, average deviation and standard deviation is given as: Quartile deviation = 2/3 Standard deviation Average deviation = 4/5 Standard deviation (vii) We can also compute corrected standard deviation by using the following formula : Corrected σ = (a) Compute corrected X = where Corrected ΣX 2 − (corrected X )2 N Corrected ΣX N corrected ΣX = ΣX + correct items – Wrong items where ΣX = N.X (b) Compute corrected ΣX2 = ΣX2 + (Each correct item)2 – (Each wrong item)2 where ΣX2 = Nσ 2 + NX 2 Example 19 : (a) Find out the coefficient of variation of a series for which the following results are given: N = 50, ΣX′ = 25, ΣX′2 = 500 where : X′ = deviation from the assumed average 5. (b) For a frequency distribution of marks, in statistics of 100 candidates (grouped in class intervals of 0–10, 10–20) the mean and, standard deviation, were found to be 45 and 20. Later it was discovered that the score 54 was misread as 64 in obtaining frequency distribution. Find out the correct mean and correct standard deviation of the frequency distribution. (c) Can, coefficient of variation be greater than 100%? If so, when? Solution : (a) We want to calculate, coefficient of variation, which is = σ × 100. X Therefore, we are required to calculate mean and standard deviation. Calculation of simple mean X = A+ 88 ΣX ′ where, A = 5, N = 50, ΣX′ = 25 N X = 5+ ∴ 25 = 5.5 50 Calculation of standard deviation 2 σ= 2 500 25 ΣX ′ 2 ΣX ′ − = − = 5 − 0.25 = 4.75 = 2.179 N N 50 50 Calculation of Coefficient of variation C.V. = σ 2.179 217.9 × 100 = × 100 = = 39.6% X 5.5 5.5 (b) Given X = 45, σ = 20, N = 100, wrong value = 64, correct value = 54. Since this is a case of continuous series, therefore, we will apply the formulae for mean and standard deviation that are applicable in continuous series. Calculation of correct Mean X = Σfx or NX = ΣfX N By substituting the values, we get 100 × 45 = 4500 Correct ΣfX = 4500 – 64 + 54 = 4490 ∴ Correct X = Correct Σfx 4490 = = 44.9 N 100 Calculation of correct σ σ= ΣfX 2 ΣfX 2 2 2 − ( X )2 − ( X ) or σ = N N where, σ = 20, N = 100, X = 45. ΣfX 2 − (45) 2 (20) = 100 2 ΣfX 2 − 2025 100 or 400 = or ΣfX 2 400 + 2025 = 100 or 2425 × 100 = ΣfX 2 = 242500 89 ∴ Correct ΣfX 2 = 242500 – (64)2 + (54)2 = 242500 – 4096 + 2916 = 242500 – 1180 = 241320 Correct σ = = Correct ΣfX 2 − (Correct ( X ))2 N 241320 − (44.9) 2 = 2413.20 − 2016.01 = 397.19 = 39.9 approx. 100 (c) The formulae for the computation of coefficient of variation is = { } σ × 100 . X Hence, coefficient of variation can be greater than 100% only when the value of standard deviation is greater than the value of mean. This will happen when data contains a large number of small items and few items are quite large. In such a case the value of simple mean will be pulled down and the value of standard deviation will go up. Similarly, if there are negative items in a series, the value of mean will come down and the value of standard deviation shall not be affected because of squaring the deviations. Example 20 : In a distribution of 10 observations, the value of mean and standard deviation are given as 20 and 8. By mistake, two values are taken as 2 and 6 instead of 4 and 8. Find out the value of correct mean and variance. Solution : We are given; N = 10, X = 20, σ = 3 Wrong values = 2 and 6 and Correct values = 4 and 8 Calculation of correct Mean ΣX X = N or X = ΣX ∴ ΣX = 10 × 20 = 200 But ΣX is incorrect. Therefore we shall find correct ΣX. Correct ΣX = 200 – 2 – 6 + 4 + 8 = 204 Correct Mean = Correct ΣX 204 = = 20.4 N 10 Calculation of correct variance σ2 = or σ2 = ΣX 2 − ( X )2 N ΣX 2 − ( X )2 N 90 ΣX 2 − (20) 2 (8) = 10 2 or ΣX 2 − 400 10 or 64 = or ΣX 2 64 + 400 = 10 or ΣX2 = 4640 But this is wrong and hence we shall compute correct ΣX2 Correct ΣX2 = 4640 –22 – 62 + 42 + 82 = 4640 – 4 – 36 + 16 + 64 = 4680 Correct ΣX 2 − Correct ( X )2 Correct σ = N 4680 − (20.4) 2 = 468 − 416.16 = 51.84 = 10 2 3.5.6 Graphic Method of Variation The concept of Lorenz-curve was devised by Max-o-Lorenz. It is also called a cumulative percentage curve, in which the percentage of the items is combined with the percentage of other items as wealth, profits, etc. The Lorenz-curve can be drawn in the following manner: (i) Size of the items and the frequencies are converted into percentages. (ii) Cumulative percentages are obtained both for the items and frequencies. (iii) We take cumulative frequency on OX-axis. The values on OX-axis starts from 100% and decreases to 0%. (iv) We take into account cummulative item percentages and divide the axis in equal parts in such a manner that different parts on both the axis are equal. On OY-axis, we start from 0% and increases to 100%. (v) Join 0% of OX-axis and 100% of OY-axis by a straight line which is called line of Equal Distribution. (vi) We plot points of different series, on the basis of cummulative item percentages and cumulative frequency percentages and join these points to draw a ‘Lorenz curve’ for different series. (vii) The ‘Lorenz-curves’ are compared with the line of Equal Distribution and the distance between them will determine variability. The Lorenz-curve located far away from the line of equal distribution is more variable. 91 Example 21 : Draw Lorenz curves and interpret the results of the below given profits of companies located in two different sites. No. of companies Area B Profit earned (Rs. lakhs) Area A 10 4 24 20 10 18 30 6 5 40 5 3 Solution : Profits earned (Rs. lakhs) X No. of Companies Area A Area B fA fB Percentage X fA fB Cumulative Percentage CX CfA CfB 10 4 24 10 16 48 10 16 48 20 10 18 20 40 36 30 56 84 30 6 5 30 24 10 60 80 94 40 5 3 40 20 06 100 100 100 Total: 100 25 50 100 100 100 Solution : Y 80 NE LI OF EQ L UA 60 LORENZ CURVE OF AREA A COMPANIES ST DI T BU RI 40 N IO CUMULATIVE ITEM PERCENTAGES 100 20 0 100 80 60 40 LORENZ CURVE OF AREA B COMPANIES X 0 20 CUMULATIVE FREQUENCY PERCENTAGES Figure We notice in the above diagram that the Lorenz curve for Area B companies is away from the line of equal distribution in comparison with Lorenz curve for Area A. Therefore, we can conclude that there is more variability in Area B companies as compared to Area A companies. 92 3.6 REVISIONARY PROBLEMS Example 22 : Compute (a) Inter-quartile range, (b) Semi-quartile range, and (c) Coefficient of quartile deviation from the following data : Farm Size (acres) No. of firms Farm Size (acres) No. of firms 0–40 394 161–200 169 41–80 461 201–240 113 81–120 391 241 and over 148 121–160 334 Solution : In this case, the real limits of the class intervals can be obtained by subtracting 0.5 from the lower limits of the class intervals and adding 0.5 to the upper limits of the different class intervals. This adjustment is necessary to calculate median and quartiles of the series. Farm Size (acres) No. of firms Cumulative frequency (c.f.) 0–40 394 394 41–80 461 855 81–120 391 1246 121–160 334 1580 161–200 169 1749 201–240 113 1862 241 and over 148 2010 N = 2010 Q1 = l1 + N 4 − c. f 0 ×i f 2010 n = 502 th item = 4 4 Q Q1 lies in the cumulative frequency of the group 41–80, where the real limits of class intervals are 40.5–80.5 and l1 = 40.5, f = 461, i = 40, c.f0 = 394, ∴ Similarly, n = 502.5 4 502.5 − 394 × 40 = 40.5 + 9.4 = 49.9 acres 461 3n − c. f 0 Q3 = l1 + 4 ×i f Q1 = 40.5 + 93 3n 3 × 2010 = × 1507.5 th item 4 4 Q3 lies in the cumulative frequency of the group 121–160, where the real limits of the class interval are 120.5–160.5 and l1 = 120.5, i = 40, f = 334, 1507.5 − 1246 × 40 = 120.5 + 31.3 = 151.8 acres 334 Q3 = 120.5 + ∴ 3n = 1507.5, c.f. = 1246 4 Inter-quartile range = Q3 – Q1 = 151.8 – 49.9 = 101.9 acres Semi-quartile range = Q3 − Q1 151.8 − 49.9 = = 50.95 approx. 2 2 Coefficient of quartile deviation = Q3 − Q1 151.8 − 49.9 101.9 = = = 0.5 approx. Q3 + Q1 151.8 + 49.9 201.7 Example 23 : Calculate mean and coefficient of mean deviation about mean from the following data : Marks less than No. of students 10 4 20 10 30 20 40 40 50 50 60 56 70 60 Solution : In this question, we are given less than type series alongwith the cumulative frequencies. Therefore, we are required first of all to find out class intervals and frequencies for calculating mean and coefficient of mean deviation about mean. Marks No. of Mid Deviations from Step Deviation Deviations from mean (35) (A = 35) i = 10 (ignoring signs) students points assumed Mean f X X′ X – A dx = i |dx| fdx f|dx| 0–10 4 5 – 30 –3 3 – 12 12 10–20 6 15 – 20 –2 2 – 12 12 20–30 10 25 – 10 –1 1 – 10 10 94 30–40 20 35 0 0 0 0 0 40–50 10 45 + 10 +1 1 + 10 10 50–60 6 55 + 20 +2 2 + 12 12 60–70 4 65 + 30 +3 3 + 12 12 Σfdx = 0 Σf |dx| = 68 N = 60 Σfdx X = A+ N ×i where, N = 60, A = 35, i = 10, Σ.fdx = 0 0 ∴ X = 35 + × 10 = 35 60 Σf | dx | 68 ×i = × 10 = 11.33 M.D. about mean = N 60 M.D. about mean 11.33 = = 0.324 approx. mean 35 Example 24 : Calculate standard deviation from the following data: Coefficient of M.D. about mean = Class interval Frequency –30 to –20 5 –20 to –10 10 –10 to 0 15 0 to 10 10 10 to 20 5 N = 45 Solution : Calculation of Standard Deviation Class Intervals Frequency Mid points Deviations from assumed Mean (A = –5) Step Deviations when i = 10 f X X′ –30 to –20 5 – 25 – 20 ( X – A) i –2 –20 to –10 10 – 15 – 10 10 to 0 15 –5 0 to 10 10 10 to 20 5 dx2 fdx fdx2 4 – 10 20 –1 1 – 10 10 +0 0 0 0 0 5 + 10 1 1 10 10 15 + 20 2 4 10 20 dx = Σfdx = 0 Σfdx2 = 60 N = 45 95 2 Σfdx 2 Σfdx − ×i N N σ= where N = 45, i = 10, Σfdx = 0, Σfdx2 = 60 2 ∴ 60 0 − × 10 = 45 45 σ= 60 × 10 = 1.33 × 10 = 1.153 45 Example 25 : For two firms A and B belonging to same industry, the following details are available : Number of Employees: Average wage per month : Standard deviation of the wage per month : Firm A Firm B 100 200 Rs. 240 Rs. 170 Rs. 6 Rs. 8 Find (i) Which firm pays out larger amount as monthly wages? (ii) Which firm shows greater variability in the distribution of wages? (iii) Find average monthly wage and the standard deviation of the wages of all employees firms. Solution : (i) For finding out which firm pays larger amount, we have to find out ΣX. ΣX N or Firm A : N = 100, X = 240 ∴ Σ X = 100 × 240 = 24000 Firm B : N = 200. X = 170 ∴ Σ X = 200 × 170 = 34000 X= Σ X = NX Hence firm B pays larger amount as monthly wages. (ii) For finding out which firm shows greater variability in the distribution of wages, we have to calculate coefficient of variation Firm A : C.V. = σ 6 × 100 = × 100 = 2.50 X 240 Firm B : C.V. = σ 8 × 100 = × 100 = 4.71 X 170 Since coefficient of variation is greater for firm B, hence it shows greater variability in the distribution of wages. (iii) Combined wage : X12 = N1 X1 + N 2 X 2 N1 + N 2 where, N1 = 100, X1 = 240, N2 = 200, X 2 = 170 96 (100 × 240) + (200 × 170) 24000 + 34000 = = 193.33 100 + 200 300 Combined Standard Deviation Hence X 12 = σ12 = N1σ12 + N2σ 22 + N1d12 + N 2 d 22 N1 + N 2 where N1 = 100, σ1 = 6, σ2 = 8, d1 = ( X 1 − X 12 ) = 240 – 193.3 = 46.7 and d 2 = ( X 2 − X 12 ) = 170 – 193.3 = – 23.3 σ12 = (100)(36) + (200)(64) + (100)(46.7)2 + (200)( −23.3) 2 100 + 200 3600 + 12800 + 218089 + 108578 343, 067 = = 33.81 300 300 Example 26 : From the following frequency distribution of heights of 360 boys in the age-group 10–20 years, calculate the: = (i) arithmetic mean; (ii) coefficient of variation; and (iii) quartile deviation Height (cms) No. of boys Height (cms) No. of boys 126–130 31 146–150 60 131–135 44 151–155 55 136–140 48 156–160 43 141–145 51 161–165 28 Solution : Calculation of X , Q.D., and C.V.. Heights m.p. X f (X – 143)/5 dx fdx fdx2 c.f. 126–130 128 31 –3 –93 279 31 131–135 133 44 –2 –88 176 75 136–140 138 48 –1 –48 48 123 141–145 143 51 0 0 0 174 146–150 148 60 +1 +60 60 234 151–155 153 55 +2 +110 220 289 156–160 158 43 +3 +129 387 332 161–165 163 28 +4 +112 448 360 Σfdx = 182 Σfdx2 = 1618 N = 45 97 Σfdx X = A + N × i where, N = 360, A = 143, i = 5, Σfdx = 182 (i) ∴ X = 143 + (ii) C.V. = 182 × 5 = 143 + 2.53 = 145.53. 360 σ × 100 X 2 σ= = C.V. = (iii) Q.D. = Q1 = Size of 2 1618 182 Σfdx 2 Σfdx − ×i = − ×5 N N 360 360 4.494 − 0.506 × 5 = 2.00 × 5 = 10 10 × 100 = 6.87 per cent 145.53 Q3 − Q1 2 N 360 th observation = = 90th observation 4 4 Q1 lies in the class 136–140. But the real limit of this class is 135.5–140.5. N 4 − cf 0 90 − 75 × i = 135.5 + × 5 = 135.5 + 1.56 = 137.06 48 f Q1 = l1 + 3N 360 th observation = 3 × = 270th observation 4 4 Q3 lies in the class 151–155. But the real limit of this class is 150.5–155.5. Q3 = Size of Q3 = l1 + Q.D. = 3 N 4 − cf 0 270 − 234 × i = 150.5 + × 5 = 150.5 + 3.27 = 153.77. 55 f Q3 − Q 1 153.77 − 137.06 = = 8.355. 2 2 3.7 SUMMARY ● While averages summarize and present data in a single number, variation is studied to get a better idea of the nature of data. ● Variation can be absolute or relative. Absolute variation refers to the amount of variation in a set of data while relative variation serves to compare variability across different sets of data. ● The ‘distance’ measures of variation include range and partial ranges including inter-quartile 98 range and inter-percentile range which are used in addition to or as surrogates for range. Range is commonly used in reporting price movements, quality control, etc. Coefficient of range is a relative measure. ● The measures involving deviations include quartile deviation and its co-efficient: mean deviation and its coefficient; and standard deviation, variance and coefficient of variation. ● Quartile deviation is a quick, inspectional measure of variability and used when there are scattered or extreme values included in the data. ● A measure based on each observation in the data is the mean deviation which is equal to the sum of absolute deviations of the various observations from their mean or median. The relative measure related to this is the coefficient of mean deviation. ● Standard deviation is also based on all observations. It is the best measure of variation as it possesses mathematical properties. ● Coefficient of standard deviation is sometimes used instead of coefficient of variation. ● All coefficients are pure numbers and there are no units associated with them. Hence they are used for making comparisons of variability. ● Graphically, Lorenz curve is used to describe inequalities of income. The extent of departure of the curve of actual distribution of income from the line of equal distribution indicates the degree of inequalities of income. ● Well-defined relationship exists between values of quartile deviation, mean deviation and standard deviation in the case of normal distributions. The relationship works well even for distributions which deviate moderately from normality. 3.8 SELF ASSESSMENT QUESTIONS Exercise 1 : True and False Statements (i) Measures of variation attempt to present in a single number the amount of variation in a set of data. (ii) All measures of relative variation are pure numbers with no units attached to them. (iii) Range cannot be negative. (iv) Range cannot be determined in open-ended frequency distributions. (v) Coefficient of range cannot be greater than 1. (vi) Quartile deviation is same as semi inter-quartile range. (vii) Only absolute values of deviations are considered in the calculation of mean deviation. (viii) Since the sum of absolute deviations measured from median is the minimum, this serves as the most appropriate average for calculating mean deviation. (ix) Since the sum of deviations of a set of values from their mean is equal to zero, it follows that mean deviation from mean would always be equal to zero. (x) Mean deviation can never be negative. (xi) Mean deviation cannot be calculated for distributions with open-ended classes. 99 (xii) The arithmetic mean is used for measuring deviations in calculating standard deviation due to its least squares property. (xiii) Standard deviation cannot be equal to zero. (xiv) Standard deviation can never exceed the arithmetic mean. (xv) Standard deviation is positive or negative depending upon the sign of deviations of various values from their mean. (xvi) Variance is the square root of standard deviation. (xvii) Coefficient of variation is always expressed as a percentage. (xviii) Coefficient of standard deviation is equal to the ratio of standard deviation to arithmetic mean of the data. (xix) Coefficient of variation expresses arithmetic mean as a percentage of standard deviation. (xx) If each of the values of a set of data is increased by 5, the mean and standard deviation would both increase by 5. (xxi) If each of the values of a set of data is multiplied by –5, the standard deviation would also be multiplied by the same number and hence become negative. (xxii) If each of the values of a set of data is increased by K, the coefficient of variation would also increase by K. (xxiii) When each value of a given set of data is multiplied by K, the revised coefficient of variation would be K times the original coefficient value. (xxiv) The combined standard deviation of two sets of data will always lie between the standard deviation values of the two sets. (xxv) The standard deviation of a distribution is approximately equal to 1.25 times the mean deviation and 1.5 times the quartile deviation. (xxvi) Standard deviation is exactly equal to one-sixth of the range. (xxvii) In a normal distribution, the percentage of values included between X − 2σ and X + 2σ is 99.73. (xxviii) Calculating the Z-scores is standardizing the data values. (xxix) The sum total of all Z-scores for a given set of values ranges between ±3. Ans. 1. T, 2. T, 3. T, 4. T, 5. T, 6. T, 7. T, 8. T, 9. F, 10. T, 11. T, 12. T, 13. F, 14. F, 15. F, 16. F, 17. T, 18. T, 19. F, 20. F, 21. F, 22. F, 23. F, 24. F, 25. T, 26. F, 27. F, 28. T, 29. F Exercise - 2 : Questions and Answers (i) What is range? What are its limitations as a measure of variation? Give examples where range can be used satisfactorily for measuring variation. (ii) What are quartiles? How are they used for measuring variation? (iii) Define mean deviation. How does it differ from standard deviation? 100 (iv) Define mean deviation, standard deviation and inter-quartile range of a frequency distribution. Why is standard deviation considered as the most appropriate measure of variation? Give an example in which you would prefer an alternative measure of variation. (v) State and explain the properties of standard deviation and variance. Do you agree that standard deviation is always positive and never negative or zero? (vi) What is coefficient of variation? How is it different from coefficient of standard deviation and variance? (vii) Explain the relationship between quartile deviation. mean deviation and standard deviation in the case of normal distribution. Also, discuss the empirical relationship between mean and standard deviation. (viii) What is Lorenz curve? How is it obtained? Discuss its significance as a tool of studying variation. (ix) Determine (i) weekly and (ii) monthly range of gold prices (per 10 gm) from the following data for a month: Week High Low 1 28,122 27,880 2 29,208 28,890 3 28,890 28,706 4 29,225 28,930 (x) The heights of 11 men are measured as 65, 68, 70, 69, 58, 66, 71, 65, 67, 69 and 73 inches. Calculate the range. If the shortest and the tallest of them are omitted, what is the percentage change in range? (xi) Draw a “less than ogive” from the following data and obtain the lower and upper quartiles there from. Also, calculate the values of quartile deviation and its coefficient. Wages (in Rs.) No. of workers 5,000 or more Nil 4,500 or more 4 4,000 or more 18 3,500 or more 38 3,000 or more 60 2,500 or more 75 2,000 or more 85 1,500 or more 93 1,000 or more 100 101 (xii) The following table shows the percentage of different age groups to the total population of a certain country: Age group (years) Percentage of the total population 0–14 15–19 20–24 25–29 30–39 40–49 50–59 60 and above 42.0 8.7 7.9 7.4 12.6 9.3 6.1 6.0 You are required to find the age limits within which the middle 50 percent of the population lies. Also, calculate (i) inter-quartile range, (ii) quartile deviation, and (iii) co-efficient of quartile deviation. (Note that in census, the age is recorded as on last birthday). (xiii) The distribution of marks of 1200 students appeared in an entrance examination is given below: Marks: 20–30 20–40 20–50 20–60 20–70 20–80 90 210 420 720 945 No. of students: 30 20–90 20–100 1080 1200 (xiv) In the following data, two class frequencies are missing: Class interval : Frequency: Class interval: Frequency: 100–110 110–120 120–130 130–140 140–150 4 7 15 ? 40 150–60 160–170 170–180 180–190 190–200 ? 16 10 6 3 However, it is possible to ascertain that the total frequency is 150 and that the median is equal to 146.25. You are required to find the missing frequencies. Having obtained these, calculate mean, standard deviation and the coefficient of variation. (xv) Find two numbers whose mean is 12 and standard deviation is 4. (xvi) Find the mean and standard deviation of the first 13 natural numbers. (xvii) The mean and variance of the following continuous distribution are 61 and 15.9, respectively. The distribution, after taking step-deviations, is as follows: . d′ –3 –2 –1 0 1 2 3 f 10 15 25 25 10 10 5 You are required to determine actual class intervals. 102 (xviii) The standard deviation of a set of 10 numbers was calculated as equal to 5 while their arithmetic mean was found to be 12. It was discovered later on that an item was recorded as 5 instead of 15. Rectify the error and determine the correct value of standard deviation. (xix) The mean of 5 observations is 4.4 and the variance is 8.24. If three of the observations are 1, 2 and 6, find the other two. (xx) Mean, median and variance of a set of 5 numbers are known to be 12,11 and 9.2, respectively. If two of the numbers are 8 and 16, determine the remaining numbers. (xxi) The following is the record of the number of bricks laid each day for 10 days by two brick layers A and B. Calculate the co-efficient of variation in each case and discuss the relative consistency of the two brick layers. A: 700 675 725 625 650 700 650 700 600 650 B: 550 600 575 550 650 600 550 525 625 600 If each of the values in respect of worker A is decreased by 10 and each of the values in respect of worker B is increased by 50, how will it affect the results obtained earlier? (xxii) A purchasing agent obtained samples of lamps from two suppliers. He had the samples tested in his own laboratory for the length of life, with the following results: Length of life (in hours) Samples from Company A Company B 700 – 900 10 3 900 – 1,100 16 42 1,100 – 1,300 26 12 1,300 – 1,500 8 3 (a) Which company’s lamps have greater average life? (b) Which company’s lamps have more uniform life? (xxiii) A sample of 35 values has mean 80 and standard deviation equal to 4. A second sample of 65 values has mean 70 and standard deviation equal to 3. Find the mean and standard deviation of the combined set of 100 values. (xxiv) Particulars regarding the incomes of employees of two factories are given below: Factory No. of Employees Average Income (Rs.) Variance A 600 475 180 B 500 586 140 (a) In which factory is the variation in income greater? (b) What is the wage bill of each of the-factories? (c) What is the average income of all employees put together? (d) What is the combined standard deviation of incomes of the two sets of employees? 103 (xxv) The number of employees, wages per employee and the variance of wage for two factories are given here : Factory A Factory B No. of employees 50 100 Average wages per day (Rs.) 120 85 9 16 Variance of wages (a) In which factory is there greater variation in the distribution of wages? (b) Suppose in factory B, the wages of an employee were recorded as Rs. 120 instead of Rs.100. What would be the corrected variance for factory B? Will it change the conclusion drawn in (a)? (xxvi) Fill in the blanks: Number Mean Variance Group 1 Group 2 Group 3 Combined 70 30 ? 150 140 ? 146 145 ? 48 56 78 (xxvii) For a set of 100 values, the standard deviation is known to be 14.4 and the co-efficient of variation is 40%. Calculate the arithmetic mean. (xxviii) If a 20 is subtracted from every observation in a data set. then the co-efficient of variation of the resulting data set is 20%. If a 40 is added to every observation of the same data set, then the co-efficient of variation of the resulting set of data is 10%. Find the mean and standard deviation of the original set of data. (xxix) A set of 40 numbers has mean and standard deviation equal to X and σ. respectively. If each of the values of the set is multiplied by 16, the co-efficient of variation works out to be 25% while if each value of the set is increased by 16, the co-efficient of variation becomes 20%. Find the mean and standard deviation of the set of numbers. (xxx) Two groups of workers, consisting of 30 and 50 persons, have the same mean wages but different standard deviations. The respective standard deviations are Rs. 16 and Rs. 12. Obtain the combined standard deviation of their wages. Ans. 10. 15.6, 11. Q1 = 2500, Q3 = 3825, QD = 662.5 CQD = 0.209, 12. 28.214, 14.107, 0.612, 13. 13.575, 0.206, 14. 147.33, 19.198, 13.03%, 15. 8,16, 16. 7, 3.742, 17. 20-40, 40-60 etc. 18. 4.47, 19. 4, 9, 20. 15, 10, 21. 5.568%, 6.38%, 5.562%, 5.876%, 22. 1106.67, 1050, 16.65%, 11.86%, 23. 73.5, 5.85, 24. A, 285000, 293000, 525.45, 56.72, 25. 2.5%, 4.71%, 5.96, 2.88%, 26. 50,155, 38 27. 36, 28. 80, 12, 29. 64, 16, 30. 13.64 104 LESSON-4 SKEWNESS AND KURTOSIS 4. STRUCTURE 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 Objective Tests of Skewness Nature of Skewness Characteristics of Skewness Methods of Skewness 4.4.1 Bowley’s Method 4.4.2 Karl Pearson’s Method 4.4.3 Revisionary Problems Measures of Kurtosis Comparison among Variation, Skewness and Kurtosis Summary Self Assessment Questions 4.0 OBJECTIVE After reading this lesson, you should be able to : (a) Understand the meaning of skewness and kurtosis (b) Distinguish between skewness, variation and kurtosis (c) Compute the skewness and kurtosis (d) Comment upon the nature of distribution with the help of measures of skewness and kurtosis. 4.1 TESTS OF SKEWNESS Measures of Skewness and Kurtosis, like measures of central tendency and dispersion, study the characteristics of a frequency distribution. Averages tell us about the central value of the distribution and measures of dispersion tell us about the concentration of the items around a central value. These measures do not reveal whether the dispersal of value on either side of an average is symmetrical or not. If observations are arranged in a symmetrical manner around a measure of central tendency, we get a symmetrical distribution, otherwise, it may be arranged in an asymmetrical order which gives asymmetrical distribution. Thus, skewness is a measure that studies the degree and direction of departure from symmetry. A symmetrical distribution, when presented on the graph paper, it gives a ‘symmetrical curve’, where the value of mean, median and mode are exactly equal. On the other hand, in an asymmetrical distribution, the values of mean, median and mode are not equal. When two or more symmetrical distributions are compared, the difference in them are studied with ‘Kurtosis’. On the other hand, when two or more symmetrical distributions are compared, it 105 will give different degrees of Skewness. These measures are mutually exclusive i.e. the presence of skewness implies absence of kurtosis and vice-versa. There are certain tests to know whether skewness does or does not exist in a frequency distribution. They are : 1. In a skewed distribution, values of mean, median and mode would not coincide. The values of mean and mode are pulled away and the value of median will be at the centre. In this distribution, mean-Mode = 2/3 (Median – Mode). 2. Quartiles will not be equidistant from median. 3. When the asymmetrical distribution is drawn on the graph paper, it will not give a bell shaped-curve. 4. Sum of the positive deviations from the median is not equal to sum of negative deviations. 5. Frequencies are not equal at points of equal deviations from the mode. 4.2 NATURE OF SKEWNESS Skewness can be positive or negative or zero. 1. When the values of mean, median and mode are equal, there is no skewness. 2. When mean > median > mode, skewness will be postive. 3. When mean < median < mode, skewness will be negative. 4.3 CHARACTERISTICS OF SKEWNESS 1. It should be a pure number in the sense that its value should be independent of the unit of the series and also degree of variation in the series. 2. It should have zero-value, when the distribution is symmetrical. 3. lt should have a meaningful scale of measurement so that we could easily interpret the measured value. 4.4 METHODS OF SKEWNESS Skewness can be studied graphically and mathematically. When we study skewness graphically, we can find out whether skewness is positive or negative or zero. We cannot find out value of coefficient of skewness. This can be shown with the help of a diagram : Mode Median Median Mean POSITIVE SKEWNESS X > MEDIAN > MODE Mode Mean NO SKEWNESS X = MEDIAN = MODE 106 NEGATIVE SKEWNESS X < MEDIAN < MODE Mathematically skewness can be studied as : (a) Absolute Skewness. (b) Relative or coefficient of skewness. When the skewness is presented in absolute item i.e, in units, it is absolute skewness. If the value of skewness is obtained in ratios or percentages, it is called relative or coefficient of skewness. When skewness is measured in absolute terms, we can compare one distribution with the other, if the units of measurement are not different. When it is presented in ratios or percentages, comparison become easy. Relative measures of skewness is also called coefficient of skewness. Mathematical measures of skewness can be calculated on the basis of: (a) Bowley’s Method (b) Karl-Pearson’s Method (c) Kelly’s method 4.4.1 Bowley’s Method Bowley’s method of skewness is based on the values of median, lower and upper quartiles. This method suffer from the same limitations which are in the case of median and quartiles. Wherever positional measures are called for, skewness should be measured by Bowley’s method. This method is, also used in case of ‘open-end series’, where the importance of extreme values is ignored. Absolute skewness = Q3 + Q1 – 2 Median Coefficient of Skewness = Q2 + Q1 − 2 Median Q3 − Q1 Coefficient of skewness lies within the limit ±1. This method is quite convenient for determining skewness where one has already calculated quartiles. For example, if the class intervals and frequencies are given as follows : Class Intervals Frequencies Below 10 5 10 – 20 10 20 – 30 15 30 – 40 10 above 40 5 In this case, if we want to calculate, coefficient of skewness on the basis of this method, then 107 we are required to calculate the values of Median, Q3 and Q1: Calculation of Coefficient of Skewness on the basis of Median and Quartiles Class Intervals Frequency Cumulative frequency Calculations n 2 − c. f . ×i f Below 10 5 5 Median = l1 + 10–20 10 15 n 2= 20–30 15 30 frequency 30, corresponding to class (20–30) 30–40 10 40 ∴ Median = 20 + 40 and above 5 45 = 20 + 45 = 22.5 , It lies in the cumulative 2 22.5 − 15 × 10 15 7.5 × 10 = 25 15 N = 45 45 = 11.25 , lies in the cumulative 4 Q1 = l + 3n 4 − c. f . × i f n 4= Absolute Skewness = Q3 + Q1 – 2 median frequency, corresponding to class interval (10 – 20) where, Q3 = 33.75, Q1= 16.75, Median = 25 Q1 = 10 + ∴ Ab. Skewness = 33.75 + 16.25 – 2(25) Q3 = l + = 50 – 50 = 0 3n/4 = 33.75, that lies in the cumulative Coefficient of Skewness = Q3 + Q1 − 2(Median) Q3 − Q1 11.25 − 5 × 10 = 16.25 10 3n 4 − c. f . ×i f frequency 40, corresponding to group (30–40) Q3 = 30 + Now we have, Q3 = 33.75, Q1 = 16.25, 33.75 − 30 × 10 = 33.75 10 Median 25 ∴ Coefficient of Skewness = 33.75 + 16.25 − 2(25) 0 = =0 33.75 − 16.25 17.5 4.4.2 Karl-Pearson’s Method (Pearsonian Coefficient of Skewness) Karl Pearson has suggested two formulae; 108 (i) where the relationship of mean and mode is established; (ii) where the relationship between mean and median is established. When the values of When the values of Mean and Mode are related Mean and Median are related Absolute skewness = Mean – Mode Absolute skewness = 3(Mean – Median) Coefficient of skewness = Mean – Mode σ Coefficient of skewness = Coefficient of skewness generally lies within ±1 3(Mean – Median) σ Coefficient of skewness generally lies within ±3 Calculation of Coefficient of skewness by using the following formula: Coefficient of skewness = Mean – Mode σ Given X values are = 12, 18, 18, 22, 35, and N = 5 ∴ X = X ΣX 105 = = 21 and Mode = 18 N 5 (X – 21) x x2 12 –9 81 σ= 18 –3 9 ∴ σ= 18 –3 9 22 +1 1 35 + 14 196 Σx 2 where x = ( X − X ) N 296 = 59.2 = 7.7 5 Σx2 = 296 N=5 ∴ Coefficient of skewness = Mean – Mode σ Substitute Mean = 21, Mode =18, Standard deviation = 7.7. 21 − 18 3 = = + 0.4 7.7 7.7 Calculation of Karl-Pearson’s coefficient of skewness by using the following formula: ∴ SK = Coefficient of skewness = 3(Mean – Median) σ 109 For the given data X = 12, 18, 18, 22, 35 Mean = 21, Median = 18, σ = 7.7 ∴ Coefficient of skewness = 3(21 − 18) 3 × 3 9 = = = 1.12 7.7 7.7 7.7 4.4.3 Revisionary Problems Example 1 : Calculate appropriate measure of skewness from the following income distribution: Monthly income (Rs.) Frequency upto–100 9 101–150 51 151–200 120 201–300 240 301–500 136 501–750 33 751–1000 9 above 1000 2 N = 600 Solution : In this problem, the open-ends series is given with inclusive class-intervals. Hence, Bowley’s measure of skewness is better, because it is based on Quartiles and not affected by extreme class intervals. Calculation of coefficient of skewness based on quartiles and median Monthly income (Rs.) Frequency f Cumulative Frequency c.f. Upto 100 09 9 101–150 51 60 151–200 120 180 201–300 240 420 301–500 136 556 501–750 33 589 751–1000 09 598 Above 1000 02 600 N = 600 Coefficient of skewness = Q3 + Q1 − 2 Median Q3 − Q1 110 Median = L1 + n 2 − cf ×i f 600 = 300; It lies in the cumulative frequency 420, which is corresponding to group 2 201 – 300. ∴ N/2 = But the real limits of the class interval are 200.5 – 300.5 ∴ Median = 200.5 + Q1 = L1 + n/4 = 300 − 180 120 × 100 = 200.5 + × 100 = 250.5 240 240 n 4 − cf ×i f 600 = 150. It lies in the cumulative frequency 180,which is corresponding to class interval 4 151–200. But the real limits of this class-interval are 150.5–200.5. ∴ Q1 = 150.5 + Q3 = L1 + 150 − 60 90 × 50 = 150.5 + × 50 = 150.5 + 37.5 = Rs. 188 120 120 3n 4 − cf ×i f where 3n/4 is used to find out upper quartile group. 3 × 600 = 450 . It lies in the cumulative frequency 556, which is corresponding to 4 group 301 – 500. ∴ 3n 4 = The real limits of this class interval are 300.5–500.5 ∴ Q 3 = 300.5 + 450 − 420 30 × 200 = 300.5 + × 200 = 300.5 + 44.12 = Rs. 344.62 136 136 Hence, Coefficient of skewness = 344.62 + 188 − 2(250.5) 532.62 − 501 31.62 = = = + 0.2 344.62 − 188 156.62 156.62 approx. Example 2 : Calculate the appropriate measure of skewness from the following cumulative frequency distribution : Age (under years) : 20 30 40 50 60 70 No. of persons 12 29 48 75 94 106 : 111 Solution: In this problem, we are given the upper limits of classes along with the cumulative frequency. Therefore, we have to find out the lower limits and frequencies for the given data. Age (years) Number of Persons Cumulative Frequency (f) Frequency (c.f.) Below 20 12 12 20–30 17 29 30–40 19 48 40–50 27 75 50–60 19 94 60–70 12 106 N = 106 Because the lower limit of the first group is not given, the appropriate measure of skewness is Bowley’s method. It is based on quartiles and median and is not influenced by extreme classintervals. Q3 + Q1 − 2 Median Bowley’s coefficient of skewness = Q3 − Q1 Thus, we have to calculate the values of Q3, Q1 and median. Median = L1 + Median has n 2 − c. f . ×i f N 106 items or or 53 items below it. 2 2 Therefore, it lies in the cumulative frequency 75, which is corresponding to the class-interval (40–50). Hence, median group is (40–50). where L1 = 40, i = 10, f = 27, ∴ Median = 40 + N = 53, c.f. = 48 2 53 − 48 5 × 10 = 40 + × 10 = 40 + 1.9 = 41.9 27 27 x − c. f Q1 = L1 + 4 ×i f Q1 has N 106 or or 26.5 items below it. 4 4 112 It lies in the cumulative frequency 29, which is corresponding to the-class-interval 20–30. Therefore,Q1 group is 20–30. where L1 = 20, ∴ Q1 = 20 + N = 26.5, i = 10, c.f. = 12, f = 17 4 26.5 − 12 × 10 = 20 + 8.53 = 28.53 17 3N − c. f . Q3 = l1 + 4 ×i f Q3 has 3N 3 × 106 or or 79.5 items below it. 4 4 It lies in the cumulative frequency 94, which is corresponding to the group 50–60. Therefore, Q3 group is 50–60. ∴ Q3 = 50 + 79.5 − 75 4.5 × 10 = 50 + × 10 = 50 + 2.37 = 52.37 19 19 Coefficient of skewness = Q3 + Q1 − 2 Median Q2 − Q1 where Q3 = 52.37, Q1 = 28.53, median = 41.9 ∴ Coefficient of Skewness = 52.37 + 28.53 − 2(41.9) 80.90 − 83.8 −2.90 = = = −0.12 52.37 − 28.53 23.84 23.84 Example 3 : Calculate the Karl-Pearson’s coefficient of skewness from the following data : Marks (above) : 0 10 20 30 40 50 60 70 80 No. of Students: 150 140 100 80 80 70 30 14 0 Solution : Thus formula of Karl-Pearson is applied to find out coefficient of skewness. SK = 3(mean – median) σ n − c. f . Median = L1 + 2 ×i f N 150 or or 75 items below it. It lies in the cumulative frequency 80, which is 2 2 corresponding to the group (40–50). Therefore, median group is 40–50. Median has 113 where, L1 = 40, Median = 40 + N = 76, f = 10, c. f . = 70, i = 10 2 75 − 70 + 10 = 45. 10 Calculation of Mean and Standard Deviation Marks Frequency Mid Deviations from f points Assumed Mean X (X – 45) i = 10 dx2 fdx2 fdx X – 45 10 dx = 0–10 10 5 – 40 –4 16 – 40 160 10–20 40 15 – 30 –3 9 – 120 360 20–30 20 25 – 20 –2 4 – 40 80 30–40 0 35 – 10 –1 1 0 0 40–50 10 45 0 0 0 0 0 50–60 40 55 + 10 +1 1 40 40 60–70 16 65 + 20 +2 4 32 64 70–80 14 75 + 30 +3 9 42 126 Σfdx = –86 Σ fdx2 = 830 N = 150 ∴ Σfd × i , where A = 45, N = 150, i = 10, Σfdx = –86. N X = A+ X = 45 + −86 × 10 = 45 − 5.73 = 39.27 150 2 σ = Σfdx 2 Σfdx − ×i N N where, N = 150, i = 10, Σfdx2 = 830, Σfdx = – 86. 2 ∴ σ = 830 −86 2 − × 10 = 5.5333 − ( −.57) × 10 150 150 or σ = 5.5333 − 0.3249 × 10 = 5.2048 × 10 = 22.8 Coefficient of skewness = 3(Mean – Median) σ where, σ = 22.8, Mean = 39.27, Median = 45. 114 ∴ Coefficient of skewness = 3(39.27 − 45.00) 3( −5.73) −17.19 = = = −0.754 22.8 22.8 22.8 Hence, Coefficient of skewness is – 0.754. Example 4 : (a) In a frequency distribution the coefficient of skewness based on quartiles is 0.6. If the sum of the upper and lower quartile is 100 and median is 38, find the values of lower and upper quartiles. Also find out the value of middle 50% items. (b) In a certain distribution, the following results were obtained: Coefficient of variation = 40% X = 25. Mode = 20. Find out the Coefficient of skewness, by applying Mean – Mode Standard Deviation Solution : (a) Since Bowley’s method is based on quartiles, we shall use the following formula : Coefficient of skewness = Q3 + Q1 − 2(Median) Q3 − Q1 where coeff. of SK. = +0.6, median = 38, (Q3 + Q1) = 100. By substituting the values in the formula, we get + 0.6 = 100 − 2(38) (Q3 − Q1 ) By cross multiplying, we get: 0.6 (Q3 – Q1) = 100 – 76 = 24 or Q3 – Q1 = 24 = 40. 0.6 We can solve the below given simultaneous equations : or Q3 + Q1 = 100 ...(i) Q3 – Q1 = 40 ...(ii) 2Q3 = 140 (adding the equations) Q3 = 70 Since ∴ Q3 + Q1 = 100 Q1 = 100 – 70 = 30. Hence the lower and upper quartiles are 30 and 70. The value of middle 50% items can be obtained with the help of (Q3 – Q1). ∴ The value of middle 50% items is (70 – 30) = 40. (b) In this problem, the value of standard-deviation is missing. We can calculate σ by applying 115 the following formula: C.V. = σ × 100 X We are given, C.V. = 40%, X = 25 ∴ 40 = σ × 100 or σ = 10 25 Coefficient of skewness = Mean – Mode σ or 2Q3 = 140 (adding the equations) Q3 = 70 Since Q3 + Q1 = 100 ∴ Q1 = 100 – 70 = 30 Hence the lower and upper quartiles are 30 and 70. The value of middle 50% items can be obtained with the help of (Q3 – Q1) ∴ The value of middle 50% items is (70 – 30) = 40. (b) In this problem, the value of standard-deviation is missing. We can calculate σ by applying the following formula: C.V. = σ × 100 X We are given, C.V. = 40%, X = 25 ∴ 40 = σ × 100 or σ = 10 25 Coefficient of skewness = Mean – Mode σ We are given, Mean = 25, Mode = 20, σ = 10 Coefficient of skewness = 25 − 20 = 0.5 20 Hence Coefficient of skewness is = + 0.5 Example 5 : What is the relationship between Mean, Median and Mode in: (a) Symmetrical curve. (b) a negatively skewed curve. (c) A positively skewed curve, 116 From the marks obtained by 120 students each in section A and B of a class, the following measures are secured : Section A Section B Mean = 47 Marks Mean = 48 Standard deviation = 15 marks Standard deviation = 15 marks Mode = 52 Mode = 45. Find out the coefficient of skewness and determine the degree of skewness and in which distribution, the marks are more skewed. Solution : The relationship between mean, median and mode, in different cases, can be established as : (a) In a symmetrical curve, there is no skewness. Therefore the value of mean = median = mode. (b) In a negatively skewed curve, the value of mean is less than median is less than mode. In other words, mean < median < mode. (c) In a positively skewed curve, the value of mean is greater than median is greater than mode. In other words, mean > median > mode. In the given problem, for finding out the degree of skewness, we have to compute the coefficient of skewness, where β2 = 3, Mesokurtic Curve β 2 < 3, Platykurtic Curve β 2 > Leptokurtic Curve 4.5 MEASURES OF KURTOSIS Measure of kurtosis is denoted by β2 and in a normal distribution β2 = 3. If β2 is greater than 3, the curve is more peaked and is named as leptokurtic, if β2 is less than 3, the 117 curve is flatter at the top than the normal, and is named as platykurtic. Thus kurtosis is measured by Σfx 4 µ n β2 = 4 = µ 2 Σfx 2 2 n where x = ( X − X ) R.A. Fisher had introduced another notation Greek letter gamma, symbolically, γ2 = β2 – 3 = µ4 = 3. µ 22 In this case of a normal distribution, γ2 is zero. If γ2 is more than zero (positive), then the curve is platykurtic and if γ2 is less than 0 (negative) then the curve is leptokurtic. µ4 Σfx 4 It may be noted that µ 4 = is an absolute measure of kurtosis, but β2 = 2 is a relative n µ2 measure of kurtosis. Larger the value of γ2 in a frequency distribution, the greater is its departure from normality. Skewness and kurtosis. β1 and β2 are measures of symmetry and normality respectively. If β2 = 0, the distribution is symmetrical and if β2 = 3, the distribution curve is mesokurtic. 4.6 COMPARISON AMONG VARIATION, SKEWNESS AND KURTOSIS Dispersion, Skewness and Kurtosis are different characteristics of frequency distribution. Dispersion studies the scatter of the items round a central value or among themselves. It does not show the extent to which deviations cluster below an average or above it. This is studied by skewness. In other words, this tells us about the cluster of the deviations above and below a measure of central tendency. Kurtosis studies the concentration of the items at the central part of a series. If items concentrate too much at the centre, the curve becomes ‘LEPTOKURTIC’ and if the concentration at the centre is comparatively less, the curve becomes ‘PLATYKURTIC’. Example 6 : From the following data given below, calculate the value of kurtosis and find out the nature of distribution: X : 0–10 10–20 20–30 30–40 40–50 f : 5 10 15 10 5 Solution : Calculation of Mean = Calculation of β2 = Σfx 1125 = = 25 N 45 µ4 Σfx 4 µ = = 40000 where, 4 N µ 22 118 ∴ µ2 = Σfx 2 60000 = = 133.33 N 45 β2 = µ4 40000 = =3 2 µ 2 (133.33) 2 Since the value of β2 = 3, the distribution curve is mesokurtic. (CALCULATION OF β 2) X f Classes Frequency Mid Points (X – 25) X fx x x2 x3 x4 fx2 fx3 fx4 0–10 5 5 25 – 20 400 –8000 160000 2000 –40000 800000 10–20 10 15 150 – 10 100 –1000 10000 1000 –10000 100000 20–30 15 25 375 0 0 0 1 0 0 0 30–40 10 35 350 + 10 100 1000 10000 1000 10000 100000 40–50 5 45 225 + 20 400 8000 160000 2000 40000 800000 N = 45 Σ Σfx2 =6000 0 1800000 fx = 1125 4.7 SUMMARY ● Both skewness and kurtosis are related to the shape of the frequency curve. ● Skewness means lack of symmetry, which implies that the mean, median and mode are unequal in such a case. ● Skewness is positive when its longer tail is to the right and negative when it is on the left. ● There are three measures of skewness, given by Karl Pearson which is based on averages and standard deviation; by Bowley which uses median and quartiles; and by Kelly, based on median and the tenth and ninetieth percentiles. ● Kurtosis refers to relative height of the frequency curve. Distributions can be mesokurtic, leptokurtic and platykurtic on this basis. 4.8 SELF ASSESSMENT QUESTIONS Exercise 1 : True or False Statements (i) Since mode is the point corresponding to maximum concentration of frequencies, its value is always higher than the mean and median. (ii) Skewness and kurtosis are both indicative of the nature of dispersion. (iii) All distributions are either positively skewed or negatively skewed. (iv) For a symmetrical distribution, the sum of positive and negative deviations from median is always equal to zero. 119 (v) If the excess of mean over mode is negative, it implies that skewness is negative. (vi) A longer tail to the right indicates positive skewness while a longer tail to the left indicates negative skewness in the data. (vii) For any skewed distribution, Mean – Mode = 3(Mean – Median). (viii) Positive skewness is indicated when X > Me> Mo and negative skewness when X < Me < Mo. (ix) In a highly skewed distribution, the value of second quartile may be different from that of the median (x) In every distribution, the lower and upper quartiles are equidistant from median. (xi) Bowley’s measure of skewness can vary between ±3. (xii) Negatively skewed distributions are usually platykurtic. (xiii) Bowley’s measure of skewness is more appropriate to use in an open-ended distribution. (xiv) A distribution more peaked than normal distribution is called platykurtic distribution. (xv) Kelly’s measure of kurtosis can vary between the limits of – 0.2631 to +0.2369. (xvi) The five-point summary of a distribution includes mean, median, mode, lower quartile and upper quartile. (xvii) A distribution with lower quartile = 127.8, median = 135.2 and upper quartile = 148.8 has negative skewness. Ans. 1. F, 2. T, 3. F, 4. T, 5. T, 6. T, 7. F, 8. T, 9. F, 10. F, 11. F, 12. F, 13. T, 14. F, 15. T, 16. F, 17. F Exercise II. Questions and Answers (i) What is skewness? What are the tests of skewness? Distinguish between positive and negative skewness. Give examples of cases where positively and negatively skewed distributions may be obtained. (ii) Draw rough sketches to show asymmetrical distribution, a negatively skewed distribution and a positively skewed distribution. Also, show the relative location of mean, median and mode in each case. (iii) State the empirical relationship between mean, median and mode for unimodal frequency curves that are moderately skewed. (iv) Explain the measures of skewness given by Karl Pearson and Bowley. (v) What is kurtosis? How is it measured in terms of Kelly’s formula and the beta co-efficient? (vi) “Averages, measures of variation, skewness and kurtosis are complementary in understanding a frequency distribution.” Explain. (vii) For the distribution of daily wages of a factory employing 880 workers, the co-efficient of quartile deviation is 3/5 and the co-efficient of skewness based on quartiles is 1/3. The median wage is known to be Rs. 90. Calculate the lower and upper quartile wages. (viii) In a symmetrical distribution, the mean, standard deviation and range of marks for a group of 20 students are 40, 12 and 60. Find the standard deviation of marks if the students with highest and lowest marks are excluded. 120 (ix) Given that median = 133.5 and mode = 134, obtain the missing frequencies for the following distribution and then calculate Bowley’s co-efficient of skewness: Class Interval Frequency 100–110 110–120 120–130 130–140 140–150 150–160 160–170 8 32 ? ? ? 12 8 Total 460 (x) For a distribution, Bowley’s co-efficient of skewness is 0.6. If the sum of the upper and the lower quartiles is 100 and median is 38, find the values of the upper and lower quartiles. (xi) For a distribution, Bowley’s co-efficient of skewness is – 0.36, lower quartile is 8.6 and median is 12.3. Calculate the co-efficient of quartile deviation for this distribution. (xii) The following table gives the distribution of monthly wages of 500 workers in a factory : Monthly Wages No. of Workers (in Rs.) 1,500 – 2,000 2,000 – 2,500 2,500 – 3,000 3,000 – 3,500 3,500 – 4,000 4,000 – 4,500 10 25 145 220 70 30 Compute average monthly wage, mode, standard deviation and Karl Pearson’s co-efficient of skewness. (xiii) Given that median = 46 and mode = 37, find the missing frequencies of the following distribution and also calculate Karl Pearson’s co-efficient of skewness: Class Interval : 20–30 30–40 Frequency : 12 ? 40–50 50–60 ? ? 60–70 70–80 80–90 Total 12 9 7 100 (xiv) Consider the following data about two distributions: Distribution A Distribution B Mean 120 110 Median 110 120 Standard deviation 10 10 121 Examine the following statements, stating with reasons whether each of them is true or false: (a) Distribution A has the same degree of variation as distribution B has. (b) Distribution A has the same degree of skewness as distribution B has. (xv) Karl Pearson’s co-efficient of skewness of a distribution is 0.40. Its standard deviation is 8 and mean is 30. Find the median and mode of the distribution. (xvi) For a moderately skewed distribution of the retail prices of children’s shoes, it is found that the mean price is Rs. 180 and the median price is Rs. 164. If the co-efficient of variation is 20%, find the Karl Pearson’s co-efficient of skewness. (xvii) For a distribution, mean = 65, median = 70 and co-efficient of skewness = –0.6. Find the mode and co-efficient of variation. (xviii) Using mean and median, calculate Karl Pearson’s co-efficient of skewness for the following distribution: Marks : Frequency : 10-100 10-80 10-60 10-50 10-40 100 88 73 57 35 10-30 10-20 17 5 (xix) Given, mean = 50, co-efficient of variation = 40% and J = –0.4. Find mode, median and standard deviation. (xx) The sum of 20 observations is 300 and the sum of their squares is 5000. Find the coefficient of variation and co-efficient of skewness, given further that median =15. (xxi) Pearson’s co-efficient of skewness for a data distribution is 0.5 and co-efficient of variation is 40%. Its mode is 80. Find the mean and median of the distribution. Ans. 9. –0.115, 10. 70, 30, 11. 0.24, 12. 3.345, 3.167, 503.46, 0.354, 13. 26, 20, 14, 0.7, 14. (a) False CVA < CVB (b) True 15. 38.93, 36.8, 16. 1.33, 17. 80, 38.46%, 18. 0.12, 19. 58, 52.67, 20, 20. 33.33%, 0, 21. 100, 93.33 122 LESSON : 5 MOMENTS 5. STRUCTURE 5.0 Objective 5.1 Calculation of Central Moments 5.1.1 Direct Method 5.1.2 Short Cut Method 5.1.3 Step Deviation Method 5.2 Sheppard’s Correction for Grouping Errors 5.3 Coefficients Based on Moments 5.4 Summary 5.5 Self Assessment Questions 5.0 OBJECTIVE After reading this lesson, you should be able to : (a) Understand the meaning of moments (b) Compute central moments by different methods (c) Comprehend Shepard’s corrections of moments for Grouping errors (d) Comment upon the nature of distribution with the help of α and β. 5.1 CALCULATION OF CENTRAL MOMENTS The concept of moments has crept into the statistical literature from mechanics. In mechanics, this concept refers to the turning or the rotating effect of a force whereas it is used to describe the peculiarities of a frequency distribution in statistics. We can measure the central tendency of a set of observations by using moments. Moments also help in measuring the scatteredness, asymmetry and peakedness of a curve for a particular distribution. Moments refers to the average of the deviations from mean or some other value, raised to a certain power. The arithmetic mean of various powers of these deviations in any distribution is called the moments of the distribution about mean. Moments about mean are generally used in statistics. We use a greek alphabet µ read as mu for these moments. We shall understand the first four moments about mean in the lesson, i.e., µ 1, µ 2, µ 3 and µ 4. We can compute central moments in the following ways : 1. Direct method 2. Short-cut method 3. Step-deviation method. 5.1.1 Direct Method (i) Calculate arithmetic mean (X) 123 (ii) Calculate the sum of deviations (Σx) from arithmetic mean. (iii) Calculate the sum of x2, x3 and x4. In case of frequency distributions multiply the individual value of x2, x3 and x4 with corresponding frequencies and find out the sum of fx2, fx3, and fx4. (iv) Apply the following formulae rule. µ1 = Σx =0 N µ2 = Σx 2 N µ3 = Σx 3 N Σx 4 µ4 = N In case of frequency distribution apply: Σfx µ1 = N µ2 = Σfx 2 N Σfx3 µ3 = N Σfx 4 N Let us take an example to understand the computation of the moments about mean µ4 = Example 1 : Calculate the first four moments about the mean from the following set of numbers 2, 3, 7, 8, 10 Solution : Calculation of Moments X (X – X) x x2 x3 x4 2 –4 16 –64 256 3 –3 9 –27 81 7 1 1 1 1 8 2 4 8 16 10 4 16 64 256 ΣX = 30 0 46 –18 610 124 X = ΣX 30 = = 6, N 5 where N = 5 Moments of the data can be computed by using the values calculated above. µ1 = Σx 0 = =0 N 5 µ2 = Σx 2 46 = = 9.2 N 5 Σx3 −18 = = −3.6 µ3 = N 5 Σx 4 610 = = 122 µ4 = N 5 Therefore, the first four central moments about the mean are : 0, 9.2, –3.6 and 122 respectively. Example 2 : From the marks distribution of 100 candidates, compute the first four moments about mean. Marks Number of Candidates 0–10 10 10–20 15 20–30 25 30–40 25 40–50 10 50–60 10 60–70 5 Solution : Calculation of Moments (Mid Value) fX (X – X) x 10 50 – 26 – 260 6,760 – 1,75,760 15 15 225 – 16 – 240 3,840 – 61,440 9,83,040 20–30 25 25 625 –6 – 150 900 – 5,400 32,400 30–40 35 25 875 4 100 400 1,600 6,400 40–50 45 10 450 14 140 1,960 27,440 3,84,160 50–60 55 10 550 24 240 5,760 1,38,240 33,17,760 60–70 65 5 325 34 170 5,780 1,96,520 66,81,680 N = 100 3100 0 25,400 Marks X 0–10 5 10–20 f 125 fx fx2 fx3 fx4 45,69,760 1,21,200 1,59,75,200 ∴ X = ΣfX 3100 = = 31 marks. N 100 Now, we can calculate the moments about mean as follows : µ1 = Σfx 0 = N 100 µ2 = Σfx 2 25, 400 = = 254 N 100 µ3 = Σfx3 1, 21, 200 = = 1, 212 N 100 µ4 = Σfx 4 1,59,75,200 = = 1,59,752 N 100 Therefore, the Central Moments are : 0, 254, 1212, 159752 respectively. 5.1.2 Short-cut Method If the arithmetic mean is in fractions then, it is difficult to calculate deviations (x) from arithmetic mean. Short-cut method is used in such cases. (i) Take any value as an arbitrary mean (A). (ii) Calculate deviations (d) from A and calculate the first four moments in the similar way as done in direct method. These moments are called moments about an arbitrary origin which are represented by the greek word v read as nu. The formulae for these moments are : v1 = Σ ( X − A) Σd = where d = X – A N Ν Σ ( X − A)2 Σd 2 = v2 = N Ν v3 = Σ ( X − A)3 Σd 3 = N Ν v4 = Σ ( X − A)4 Σd 4 = N Ν v1 = Σf ( X − A) Σfd = N N In case of frequency distribution, 126 v2 = Σf ( X − A) 2 Σfd 2 = N N v3 = Σf ( X − A)3 Σfd 3 = N N v4 = Σf ( X − A) 4 Σfd 4 = N N After calculating moments about an arbitrary origin convert them into Moments about mean by using the following equations: µ 1 = v1– v1 = 0 µ 2 = v2 – v12 = σ2 µ 3 = v3 – 3v2v1 + 2v13 µ 4 = v4 – 4v3.v1 + 6v2.v12 – 3v14 We can calculate the Moments about an arbitrary origin from Moments about the mean by this relationship: v1 = µ 1+d where d is the difference between the v2 = µ 2 + d2 mean and origin about which the Moments v3 = µ 3 + 3µ 2d + d3 are to be calculated. v4 = µ 4 + 4µ 3d + 6µ 2d2 + d4 ∴d=X–A Example 3 : We are given the following set of numbers 1, 3, 7, 9, 10. Calculate the first four moments about the origin 4. Solution : Calculation of First Four Moments about A = 4 X d = (X – A) d2 d3 d4 1 –3 9 – 27 81 3 –1 1 –1 1 7 3 9 27 81 9 5 25 125 625 10 6 36 216 1296 N=5 10 80 340 2084 v1 = Σd 10 = =2 N 5 Σd 2 80 = = 16 v2 = N 5 127 v3 = Σd 3 340 = = 68 N 5 Σd 4 2084 = = 416.8 v4 = N 5 Therefore the Moments about an arbitrary origin are 2, 16, 68 and 416.8 respectively. Example 4 : Calculate first four moments about mean for the distribution of heights of the following 100 students. Heights (Inches) 61 64 67 70 73 Number of Students 5 18 42 27 8 Solution : Calculation of Central Moments (short-cut method) Heights No. of students A = 67 f×d fd × d fd2 × d fd3 × d X f d = (X – 67) fd fd2 fd3 fd4 61 5 –6 –30 180 –1,080 6,480 64 18 –3 –54 162 – 486 1,458 67 42 0 0 0 0 0 70 27 +3 81 243 729 2,187 73 8 +6 48 288 1,728 10,368 45 873 891 20,493 N = 100 Now we can substitute the calculated values in the formulae v1 = Σfd 45 = = 0.45 N 100 v2 = Σfd 2 873 = = 8.73 N 100 Σfd 3 891 = = 8.91 v3 = N 100 v4 = Σfd 4 20493 = = 204.93 N 100 Moments about mean can be calculated as follows : µ 1 = v1 –v1 = 0 = 0.45 – 0.45 = 0 µ 2 = v2 – v12 = σ2 = 8.73 – (0.45)2 = 8.73 – 0.2025 = 8.5275 128 µ 3 = v3 – 3v2v1 + 2v13 = 8.91 – 3(8.73 × 0.45) + 2(0.45)3 = 8.91 – 11.7855 + 0.18225 = – 2.6932 µ 4 = v4 – 4v3. v1 + 6v2.vl2 – 3v14 = 204.93 – 4 × 8.91 × 0.45 + 6 × 8.73 × (0.45)2 – 3 × (0.45)4 = 204.93 – 16.038 + 10.60695 – 0.1230 = 199.3759 Hence the Central Moments are : 0,8.5275, –2.6932 and 199.3759. If we are given the values of central moments and were interested in finding the Moments about an arbitary origin (A = 67). Then we can calculate as follows : X = A+ Σfd 45 = 67 + = 67.45 N 100 d = ( X − A) = 67.45 − 67 = 0.45 v1 = µ1 + d = 0 + 0.45 = 0.45 v2 = µ 2 + d 2 = 8.5275 + (0.45) 2 = 8.5275 + .2025 = 8.73 v3 = µ3 + 3µ 2 d + d 3 = −2.6932 + 3 × 8.5275 × (0.45) + (0.45)3 = −2.6932 + 11.512125 + 0.091125 = 8.91. v4 = µ 4 4µ3d + 6µ 2 d 3 + d 4 = 199.3759 – (4 × –2.69325 + 0.45) + 6 × 8.5275 × (0.45)2 + (0.45)4 = 199.3759 – 4.84785 + 10.36091 + 0.041006 = 204.93 ∴ Moments about an arbitrary origin (67) are : 0.45, 8.73, 8.91 and 204.93. 5.1.3 Step-Deviation Method It is the most appropriate method to calculate central moments in problems of continuous frequency distributions with equal class-intervals. Step-deviation method is similar to short cut method. The only difference is that in case of step-deviation method, we take a common factor from among the deviations (d) which are taken from assumed mean (A). (i) Calculate deviations (d) from arbitrary origin (A ). (ii) Take a common factor from all the values of d and find out the sum of d′. (iii) Find out the values of d′2, d′3 and d′4 and their aggregates. (iv) Calculate the value of v1, v2, v3 and v4 by using the formulae. (v) Convert the calculated Moments about an arbitrary origin into Moments about the mean with the help of these relationship: µ 1 = v1 − v1 = 0 µ 2 = v2 − v12 = σ 2 129 µ 3 = v3 − 3v2v1 + 2v13 µ 4 = v4 − 4v3 .v1 + 6v2 .v12 + 3v14 We shall understand the computation of the first four moments about an arbitrary origin by step deviation method. Example 5 : Calculate first four moments about mean with the help of moments about an assumed mean 35 from the following data : Class Frequency 0–10 4 10–20 10 20–30 21 30–40 32 40–50 21 50–60 7 60–70 5 Solution : Calculation of Moments about arbitrary Mean A = 35, C = 10 (M id Points) Class X f X – A d′ = C 0–10 5 4 –3 – 12 36 – 108 324 10–20 15 10 –2 – 20 40 – 80 160 20–30 25 21 –1 – 21 21 – 21 21 30–40 35 32 0 0 0 0 0 40–50 45 21 1 21 21 21 21 50–60 55 7 2 14 28 56 112 60–70 65 5 3 15 45 135 405 –3 191 3 1043 N= 100 fd′ 1 fd′ 2 fd′ 3 fd′ 4 First of all, we shall calculate the first four moments about an arbitrary mean by substituting the values. v1 = Σfd ′ −3 ×C = × 10 = −0.3 N 100 130 v2 = 191 Σfd ′ 2 × 100 = 191 × C2 = 100 N 3 Σfd ′ 3 × 103 = 30 × C3 = v3 = 100 N v4 = 1043 Σfd ′ 4 × 104 = 1, 04,300 × C4 = 100 N From these we get the central moments as below : µ1 = v1 − v1 = −0.3 − ( −0.3) = 0 µ 2 = v2 − v12 = σ2 = 191 − ( −0.3) 2 = 191 − 0.09 = 190.91 µ3 = v3 − 3v2v1 + 2v13 = 30 – 3 × 191 (–0.3) + 2 (–0.3)3 = 30 + 171.9 – 0.054 = 201.846 µ 4 = v4 − 4v3 .v1 + 6v2 .v12 − 3v14 = 1, 04,300 − 4 × 30( −0.3) × 6 × 191( −0.3) 2 − 3( −0.3) 4 = 1, 04,300 + 36 + 103.14 − 0.0243 = 1, 04, 439.12 5.2 SHEPPARD’S CORRECTION FOR GROUPING ERRORS When data are grouped into frequency distributions, the individual values lose their identity. While calculating moments, it is assumed that the frequencies are concentrated at the mid-points of the classes for a continuous frequency distribution. Let us understand by assuming a class of 10–20 whose relative frequency is 20. To compute Moments or arithmetic mean, we always take the mid-point of the class 10–20 10 + 20 = 15. But in reality it may be just possible that more than half the values for the 2 class 10–20 are more than 15. Due to this assumption the grouping errors enter into the calculation of Moments. To remove these errors W.F. Sheppard introduced some corrections which are known as Sheppard’s corrections. These are : which is The first and third Moment needs no corrections. µ 2 (corrected) = µ 2 (uncorrected) – 1 2 i 12 µ 4 (corrected) = µ 4 (uncorrected) – 7 4 1 2 i µ 2 (uncorrected) i + 240 2 where i is the class-interval. 131 Sheppard’s corrections are applied only under these conditions : (i) When the frequency distribution is continuous. (ii) When frequency tapers off to zero in both directions. (iii) When the frequencies are not less than 1000. (iv) When the frequency distribution is not J-shaped or U-shaped or skewed, (v) When the class-interval is uniform. Let us understand with the help of an example. Example 6 : Applying Sheppard’s corrections, find out the corrected values of the moments where the class interval is 10 and µ1 = 0, µ 2 = 254, µ 3 = 1212 and µ 4 = 1,59,752. Solution : We are given all the values of four moments and class-interval. µ1 and µ3 needs no correction. µ2 (corrected) = µ 2 − 1 2 1 100 i = 254 − × 102 = 254 − = 245.667 12 12 12 1 7 4 1 7 2 i = 1,59752 − × 254 × 10 2 + × 104 µ4 (corrected) = µ 4 − µ 2i + 2 240 2 240 = 1,59752 – 12,700 + 291.667 = 147,343.667. Therefore, the corrected values of four moments are 0,245.667, 1,212 and 147,343.667 respectively. 5.3 COEFFICIENTS BASED ON MOMENTS There are three coefficients which are used in practice. They are α (Alpha), β (Beta), γ (Gamma) coefficients. These coefficients are calculated on individual relationships of various Moments. The formulae are : Alpha-Coefficients α1 = α3 = µ1 = 0, σ µ3 = µ3 α2 = , and α 4 = σ Beta-Coefficients 3 µ 32 2 µ2 =1 σ2 µ4 µ4 = σ 4 µ 22 µ4 µ32 = α 22 and β 2 = 2 = α 4 3 µ2 µ2 Gamma-Coefficients β1 = γ 1 = β 1 = α 3 and γ 2 = β 2 − 3 = µ 4 − 3µ 22 µ 22 132 Beta-Coefficients (β1 and β2) are used to find the skewness and kurtosis of a distribution. Let us take an illustration to understand coefficients. Example 7 : The values of µ1 , µ2 , µ3 , and µ 4 , are 0, 9.2, 3.6 and 122 respectively. Find out the skewness and kurtosis of the distribution. Solution : To comment upon skewness and kurtosis of the distribution, we shall calculate the values of β1 and β2. µ32 (3.6)2 12.96 = = 0.0166 β1 = 3 = 2 778.688 µ 2 (9.2) µ4 122 122 β2 = µ 2 = (9.2)2 = 84.64 = 1.4 2 Hence the distribution is positively skewed and the curve is platykurtic or flat at the top. Example 8 : Calculate the first four Moments about an arbitrary origin. Convert them into Moments about mean. Applying Sheppard’s corrections, calculate corrected Moments and beta coefficients from the following data: Experience No. of Employees (years) 0–1 15 1–2 22 2–3 45 3–4 35 4–5 30 5–6 20 6–7 16 7–8 10 8–9 5 9–10 2 Solution : Calculation of Moments Experience Mid Points No. of Employees (Years) X f Let A = 4.5 d = (X – A) fd fd2 fd3 fd4 0–1 0.5 15 –4 – 60 240 – 960 3,840 1–2 1.5 22 –3 – 66 198 – 594 1,782 2–3 2.5 45 –2 – 90 180 – 360 720 3–4 3.5 35 –1 – 35 35 – 35 35 133 4–5 4.5 30 0 0 0 0 0 5–6 5.5 20 1 20 20 20 20 6–7 6.5 16 2 32 64 128 256 7–8 7.5 10 3 30 90 270 810 8–9 8.5 5 4 20 80 320 1,280 9–10 9.5 2 5 10 50 250 1,250 – 139 957 – 961 9,993 200 We can find out v1 = Σfd −139 = = −0.695 N 200 Σfd 2 957 = = 4.785 v2 = N 200 v3 = Σfd 3 −961 = = −4.805 N 200 Σfd 4 9993 = = 49.965 v4 = N 200 Computed moments are moments about an arbitrary point, 4.5. The central moments are calculated below: µ1 = v1 – v1 = – 0.695 – (–0.695) = 0 µ2 = v2 – v12 = 5.985 – (–0.695)2 = 5.985 – 0.483 = 5.502 µ3 = v3 – 3v2v1 + 2v13 = – 4.805 – 3 (5.985) × (– 0.695) + 2(–0.695)3 = – 4.805 + 12.479 – 0.671 = 7.003 µ4 = v4 – 4v3.v1 + 6v2.v12 – 3v14 = 49.965 – 4 (– 4.805) × (– 0.695) + 6 × 5.985 (–0.695)2 – 3(– 0.695)4 = 49.965 – 13.358 + 17.345 – 0.700 = 53.252. Applying Sheppard’s corrections, we have 1 2 i 12 1 2 = 5.502 − × (1) = 5.502 − 0.083 = 5.419 12 µ 2 (corrected) = µ 2 − µ 4 (corrected) = µ 4 – 1 7 µ 2i 2 + × i4 12 240 = 53.252 − 1 7 × 5.502 × (1) 2 × × (1) 4 2 240 134 = 53.252 − 2.751 + 7 = 53.252 − 2.751 + 0.029 = 50.53 240 Beta Constants µ 32 (7.003) 2 49.042 = = 0.31 β1 = 3 = µ 2 (5.419)2 159.132 µ4 50.33 50.33 β2 = µ 2 = (5.419)2 = 29.336 = 1.714 2 Therefore the central moments after correction are 0, 5.419, 7.003 and 50.53. β1 = 0.31 and β2 = 1.714. 5.4 SUMMARY ● Moments provide a useful method to study various characteristics of a set of data. Moments calculated about mean are called central moments. There can also be moments about any given value A. When A = 0, moments calculated are called moments about origin. ● It is possible to convert moments about A into central moments and vice-versa. ● The first moment about zero is equal to mean and the first moment about mean is equal to zero. ● The second central moment is the variance of the distribution. ● The beta and g-statistics are calculated with central moments and are used to learn about skewness and kurtosis. β1 and g1are measures of skewness while β2 and g2 measure kurtosis. ● If the third central moment is positive, the skewness is positive while if it is negative, the skewness is negative. For a symmetrical distribution, it is equal to zero. β1 is always positive but the greater its value, the more the skewness. ● If β2 = 3, the distribution is mesokurtic; if β2 is less than 3, the distribution is platykurtic while if it is greater than 3, the distribution in question is leptokurtic. 5.5 SELF ASSESSMENT QUESTIONS Exercise 1 : True and False Statements (i) The first moment about mean is always equal to zero. (ii) The moments of even order can never assume negative values. (iii) The fourth moment about mean always exceeds the third moment which, in turn, always exceeds the second moment. (iv) The second moment about origin is equal to the variance of the distribution. (v) A negative value of the third moment about an arbitrary point does not necessarily mean that the distribution has negative skewness. 135 (vi) Sheppard’s corrections aim at removing the effect of grouping error. (vii) The corrected values of moments, obtained after applying Sheppard’s corrections, are always higher than the corresponding uncorrected values of the moments. (viii) All symmetrical distributions are not mesokurtic, but all mesokurtic distributions are symmetrical. (ix) A look at the value of the fourth central moment reveals whether the distribution is mesokurtic or not. (x) When β1 > 0, the skewness is positive and when β1 < 0, the skewness is negative. Ans. 1. T, 2. T, 3. F, 4. F, 5. T, 6. F, 7. F, 8. F, 9. F, 10. F Exercise II : Questions and Answers (i) Define moments. Establish the relationship between moments about arbitrary origin and central moments and vice versa. (ii) Explain how moments help in determining the shape of a frequency distribution. In this context, explain the calculation and interpretation of beta and gamma statistics. (iii) Explain clearly the difference between skewness and kurtosis. Is it correct to say that a distribution which has β2 equal to 3 must be symmetrical? (iv) “A distribution, for which β1 equal to 0, is necessarily a normal distribution.” Do you agree? Explain. (v) What are Sheppard’s corrections for moments? State the conditions under which they are applied. (vi) The following is the distribution of heights of the students of a class: Height (cms) No. of Students 120–125 3 125–130 7 130–135 25 135–140 30 140–145 25 145–150 7 150–155 3 (a) Calculate Karl Pearson’s co-efficient of skewness using (i) mean and mode, and (ii) mean and median. (b) Calculate Bowley’s co-efficient of skewness. (vii) Calculate (a) central moments, (b) moments about A = 6, and (c) moments about origin for the following data: 9, 4, 6, 9, 11, 13, 8, 4 136 (viii) Calculate four central moments for the following distribution: Class Interval Frequency 10–20 20–30 30–40 40–50 50–60 60–70 8 12 15 8 5 2 Total 50 (ix) Given, n = 10, Σ(X – 10) = – 40, Σ(X – 10)2 = 2,870 and Σ(X – 10)3 = – 480. Calculate the following: (a) Arithmetic mean (b) Variance (c) Third moment about 10 (d) A measure of relative skewness based on moments (x) For the following distribution, calculate first four (i) moments about 45 and (ii) central moments. Also, comment on the shape of the distribution using gamma statistics. Class Interval Frequency 10–20 2 20–30 5 30–40 12 40–50 20 50–60 8 60–0 2 70–80 1 (xi) Given the following distribution: Class Interval Frequency 10–20 2 20–30 6 30–40 12 40–50 20 50–60 12 60–70 6 70––80 2 137 (a) Calculate mean, median and mode and also find the value of Karl Pearson’s coefficient of skewness. (b) Calculate the first four central moments and the beta co-efficient of skewness. (c) Comment on the results. (xii) You are given the following frequency distribution: Class Interval Frequency 80–100 12 100–120 15 120–140 20 140–160 38 160–180 60 180–200 33 200–220 14 220–240 8 (a) Calculate moments about 170. (b) Convert these to central moments and obtain the values of beta and gamma statistics. Also, comment on the shape of the distribution. (xiii) For the following distribution, calculate first four moments about A = 150 and obtain central moments from these. Also, calculate beta co-efficients and comment on the skewness and kurtosis. Class Interval Frequency 80–100 3 100–120 5 120–140 20 140–160 16 160–180 10 180–200 4 200–220 2 (xiv) The first four moments of a distribution are 2, 20. 40, and 500, respectively. Comment on the shape of the distribution. (xv) The first four moments of a distribution are calculated as 6; 235; 1,248; and 24,680. You are required to examine the skewness and kurtosis of the distribution. 138 (xvi) From the following information about two distributions, find which is more skewed? Distribution Second Moment Third Moment A 16 –15.7 B 40 25.8 (xvii) For a mesokurtic distribution, the fourth central moment is 768. Obtain its standard deviation. (xviii) The first four moments of a distribution about the value 4 of the variable are –1.5, 17, –30 and 108. Its mean is given to be 2.5. You are required to (a) Calculate the central moments and moments about origin. (b) Determine the co-efficient of variation and variance of the distribution. (c) Examine the shape of the distribution. (xix) For a mesokurtic distribution, β1 = 0.004 and µ3 = 16. Calculate the value of its fourth central moment. (xx) For a mesokurtic distribution, co-efficient of variation = 40% and arithmetic mean = 40. Find the value of its fourth central moment. (xxi) If variance = 42, then what values of µ4 would make a distribution (i) mesokurtic, (ii) platykurtic, and (iii) leptokurtic? (xxii) The first two moments about 40 for a set of 25 values were calculated as equal to 65 and 2,985, respectively. Test if the calculations are consistent. (xxiii) You are given here the results of calculations in respect of a negatively skewed distribution: N = 100, mean = 14, variance = 230, β1 = 0.8 and β2 = 2.4. It was discovered later on that an item 12 was wrongly recorded as 2. Find the corrected values of mean, variance and the two beta constants. (xxiv) For a mesokurtic distribution, it is known that the first moment about 32 is 28 while the fourth moment about 60 is 62,208. Determine the values of mean and co-efficient of variation. (xxv) Apply Sheppard’s corrections to the following moments : First moment = – 7 Third moment = 873 Second moment = 193 Fourth moment = 87,504 Width of class interval = 10 Ans. 6. 0, 0, 7. 0, 9, 2.25, 154.5, 2, 13, 64.25, 404.5, 8, 73, 730.25, 7778.5 8. 0, 179.4, 951, 80, 354, 9. 6, 271, –48, 0.733 10. –2.6, 150, –1, 100, 75,000, 0, 143.24, 11. 45, 45, 45, 0, 0, 180, 0, 90000, 0; 12. –8.6, 1212, –40, 400, 44, 40000, 0, 1, 138.04, –10, 402.5, 35, 71,667, 0.0734, –0.271, 2.758, –0.242; 13. –5, 740, – 6000, 1544000, 0, 715, 4850, 15.33, 125, 0.0644, 3; 14. 1, 2.39; 15. Low positive skewness, platykurtic; 16. A, 17. 4; 18. 0, 14.75, 39.75, 142.3125, 2.5, 21, 166, 1132, 183.3%, 21, 0.702, –2.35; 19. 4800; 20. 1,96,608; 21. 5.292, < 5.292, > 5.292; 22. No, 0; 23. 14.1, 228.59, 0.771, 2.042; 24. 60, 20%; 25. 1.357, 154.578 ❑ 139 UNIT-2 PROBABILITY, PROBABILITY DISTRIBUTIONS AND DECISION THEORY LESSON-1 : THEORY OF PROBABILITY 1. STRUCTURE 1.0 Objective 1.1 Probability Foundations 1.2 Events 1.2.1 Mutually Exclusive and Overlapping Events 1.2.2 Complementary events 1.2.3 Independent and Dependent Events 1.3 Methods of Assigning Probability 1.3.1 Classical Approach 1.3.2 Relative Frequency Approach 1.3.3 Personalistic Approach 1.4 Computation of Probability 1.5 Laws of Probability 1.5.1 Addition Rule 1.5.2 Conditional Probability 1.5.3 Multiplication Rule 1.6 Bayes’ Theorem 1.7 Expected Value 1.8 Summary 1.9 Self Assessment Questions 1.0 OBJECTIVE After reading this chapter, you should be able to : (a) Define probability (b) Understand the concept of experiment, sample space, events and their relationships (c) Describe the classical, relative frequency and personalistic approaches of probability (d) Compute probabilities with the help of addition, multiplication, conditional and Bayes’ theorem 1.1 PROBABILITY FOUNDATIONS Most of the decision-making situations in business management involve uncertainty. Since uncertainty is present and is an important aspect in determining the consequences of various alternative courses of action, it is imperative to get proper appreciation of it, draw a mathematical picture of it and attempt to measure it in numerical terms. There are many advantages in having a numerical measure for uncertainty. Besides facilitating understanding and allowing analysis, it helps in communication between executives. Verbally, a manager at a meeting might indicate that he is “fairly sure” about the success of a particular project. This phrase might mean something quite 140 The theory of probabitity takes on practical value when it is defined in relation to an experiment. Such an experiment might be tossing a coin, taking out a card out of a standard deck of playing cards, tossing a six-faced die, observing the number of defectives in a lot of electric bulbs, tossing a pair of dice, drawing a ball from an urn containing balls,.. and so on. Once the experiment has been defined, all possible outcomes from the experiment are identified. This exhaustive set of outcomes constitutes the sample space, S. The sample space is a key concept and an important base of probability theory. One of the simplest sample spaces can be the set of outcomes when a pair of coins is tossed. It consists of four outcomes which can be conveniently represented as : Second Die different to the other executives at the meeting. ‘Fairly sure’ might mean that success will occur 9 out of 10 times to one decision maker (implying that he is 90 percent sure) while the same phrase might indicate 7 times out of 10 to another. Numbers remove such confusion. Besides this, an important advantage of a numerical measure is the ability to use mathematics for analysis. Uncertainty is expressed in numerical terms by the theory of probability as probability is at once the language and measure of uncertainty. In this lesson, we are going to study as to how the probability also provides a foundation for the whole of the analytical statistics that we are going to learn in the course of these lessons. 6 5 4 3 2 1 1 2 3 4 5 6 First Die Sample space for pair of dice experiment Figure 1 S = {HH, HT, TH, TT} where H denotes a head and T denotes a tail. We can consider the case of a manufacturer who produces electric bulbs in large batches. From each batch, a sample of 80 items is selected at random, and the number of defective items are recorded. Although the number of defectives in any sample cannot be predicted with certainty, all of the possible outcomes may be known. The number of defective items in a sample can be any integer from 9 to 80. Here the sample space is : S = {0, 1, 2, 3, .......80} In the same manner, when a pair of dice is tossed, a total of 36 outcomes are possible. This can be represented as shown in figure. 1 In all the three examples, the number of outcomes from the experiment are known to be finite. While in most cases it is so but it is not a rule. The number of outcomes can be infinite as well. For example, it we consider the experiment of observing the life- time of an electric bulb in hours, the outcome can be any real, non-negative number. Thus, this sample space contains an infinite number of sample points. 141 1.2 EVENTS An event refers to any set of possible outcomes in a sample space. If the sample space for an event has the elements S1, S2, S3,...Sn, an event in the sample space S would be any one, or collection of S1, S2, S3;...Sn. In a sample space, every combination of sample points may be defined as an event. In an experiment of counting defective items in a sample, the set of all possible outcomes having less than 10 defective items can be represented by the event A. Each sample point does not have to identify a separate event. The faces of a die provide a sample space of six outcomes. If the occurrence of each face identifies a different event, there are six possible events. On the other hand, suppose that an even number represents a gain of Rs. 100 to a person X and an odd number represents a loss of Rs. 100 to the person Y. In this case, there are six outcomes but only two events – gain of Rs. 100 to X, and loss of Rs. 100 to Y. Every event A is a subset of the sample space and every event is a collection of the elements of a sample space. Events can be classified as being elementary or compound. Elementary events are said to be those which have a single sample point whereas compound events are those which contain more than one sample point. Thus, whereas the compound events can be decomposed, the elementary events cannot be. The appearance of 3 on a die is an elementary event while the appearance of an even number on the die is a compound event (as it contains three sample points). Now, keeping in mind the definitions of experiment, sample space and events, we introduce some more concepts. 1.2.1 Mutually Exclusive and Overlapping Events Two events are mutually exclusive if the occurrence of one event precludes the occurrence of the other. For example, the events that (i) an employee would be late, and (ii) the employee would be absent, on a particular day, are mutually exclusive since both cannot occur simultaneously. An employee cannot be both late and absent on a particular day. Similarly, suppose we consider a box in which 20 cards, marked 1 to 20 are placed and a card is drawn at random. If A be the event that the number on the card is divisible by 3 and B be the event that the number would be divisible by 7, then the events A and B are mutually exclusive. This is because for A to occur, the number would be one of 3, 6, 9, 12, 15 or 18, and B to occur, it should be one of 7 or 14. Since no number is common to them, they are mutually exclusive. On the other hand, two or more events which are not mutually exclusive are called overlapping events. In the above example of cards, suppose A represents the event that the number on the card chosen is divisible by 3 and B represents the event that the number is divisible by 5, then for A to occur the number must be either 3,6,9,12,15 or 18, and for B to occur, it must be one of 5,10,15 and 20. Note that if the number 15 is obtained, it implies that both A and B have taken place. Thus A and B are not mutually exclusive. We can use Venn diagram to depict mutually exclusive and overlapping events. This is shown in figure 2. Part (a) of the figure shows the mutually exclusive events A and B, each of them defined over the sample space S. Note that A and B have no sample points in common. On the other hand part (b) of the figure shows overlapping events A and B, as they have some common sample points. 142 S S A∩B A A B B (a) Mutually Exclusive Events (b) Overlapping Events Figure 2 1.2.2 Complementary Events Events are said to be complementary when the sample space is partitioned into the segment that represents the occurrence of an event A, and the segment that is not a part of A. Thus, the complement of an event A is the collection of all possible outcomes that are not contained in event A. For example, in the toss of a coin, appearance of head and tail are complementary to each other. Complementary events are shown in figure 3. Here A and A are complementary to each other. The events of a person being able to hit a target, and not being able to hit the target are complementary, and so are the events of the appearance of a head and a tail on tossing a coin. Complementary Events Figure 3 1.2.3 Independent and Dependent Events Two events are said to be independent if the occurrence of one event in no way influences the occurrence of the other event. For example, we toss a six-faced die and call the event of appearance of an even number as the event A and the appearance of an odd number as the event B. Now, suppose that in the first toss we get an even number. If we toss the die the second time, we can still get an even or an odd number and their chances are not influenced by the result of the first trial. Thus, the appearance of an even number in the first trial and the appearance of an even number in the second trial is an example of independent events. Similarly, if we pick a card at random from a deck of playing cards, note its suit and put it back and then draw one card from the deck, then the chances of a king card, for example, in the second trial is not at all affected by the card we had drawn at the first trial. But if we take out a card and do not replace it back, then the chances of drawing a king card in the second trial are certainly affected by the card we had drawn in the first trial. If it were a king card the first time, then only 3 king cards remain in the 51 cards while if a nonking card was drawn then we would have 4 king cards in the lot of 51 cards. So the chances of a king card in the second trial are dependent upon the results of the first trial. This time, the events of a king card in the first trial and king card in the second trial are not independent because the outcome in one trial is in some way influenced by the outcome of the previous trial. 143 1.3 METHODS OF ASSIGNING PROBABILITY There are three methods of assigning probability to an event. They are : (i) Classical approach, (ii) Relative frequency approach, and (iii) Personalistic approach. We now disucss these methods : 1.3.1 Classical Approach The classical approach to determine probability is the oldest one. It originated with the games of chance. According to this theory, if there are n outcomes of an experiment which are mutually exclusive and equally likely to occur, then the probability of each sample point is 1/n. Thus, if a fair die is tossed, each of six numbers 1, 2,... 6, is equally likely to occur and the probability that a given number, say 5, would occur is 1/6. From this, the classical interpretation of probability is: if the sample space of an experiment has n(S) equally likely outcomes and if an event A, defined on this sample space has n(A) sample points, then the probability that event A would occur is the ratio of n(A) to n(S). Thus, P(A) = n(A) n(S) To illustrate, we consider the following examples. Example 1 : A six-faced die is tossed once. Find the probability that the number obtained on tossing is (i) and odd number, (ii) a number greater than 2. Solution : Let A : the event that the number is an ood number, and B : the event that the number is greater than 2. From the given information, n(S) = 6 (as there are six possible outcomes) n(A) = 3 (being numbers 1, 3 and 5), and n(B) = 4 (being numbers 3, 4, 5 and 6) ∴ P(A) = P(B) = n(A) 3 1 = = ; and n(S) 6 2 n(A) 4 2 = = n(S) 6 3 Example 2 : A card is drawn from a deck of playing cards at random. Find the chance that (i) it is a face card, (ii) it is a black ace card. Solution : Let A : the event that the card is a picture card B : the event that the card is black ace card we have, n(S) = 52 (there being 52 cards) n(A) = 12 (there being 4J, 4Q and 4K cards with faces) 144 n(B) = 2 (there being 2 black aces) ∴ P(A) = P(B) = 12 3 = , and 52 13 2 1 = 52 26 Example 3 : Find the probability that a leap year selected at random shall contain 53 Sundays. Solution : Like every year, a leap year would have 52 full weeks. The remaining two days of the years could be: Sunday and Monday, Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, or Saturday and Sunday. We observe here that n(S) = 7. Since two of the obove combinations have a Sunday included, we have n(A) = 2. Therefore, P(A) = n(A) 2 = . n(S) 7 The classical theory, under the assumption of equally likely outcomes, depends on logical reasoning. It does very well when we are concerned with balanced coins, perfect dice, wellshuffled pack of cards and all those situations where all outcomes are equally likely. However, problems are immediately encountered when we have to deal with the unbalanced coins, loaded dice and so on. In such situations, we have to depend on the relative frequency approach. 1.3.2 Relative Frequency Approach It is based on the actual observation. For example, if we were interested in the probability of 50 or fewer customers arriving at a super market before 10 a.m. we would pick a trial number of days (n) and count how often 50 or fewer actually did arrive before 10 a.m. Our probability assessment would be the ratio of days when 50 or fewer customers arrived to the total number of days observed. Similarly, to have an idea of the probability that a head would appear on tossing a coin, we may actually toss the coin a number of times, say 1000, and find the number of times a head appears. If 536 times a head has appeared, then the probability of head to occur will be taken to be the ratio of the two : 536/1000. Naturally, if the coin is a fair one then the ratio will approach to 0.5 as we continuously increase the number of trials. And if the coin is not a fair one then the chances of a head would tend to approach the true probability of the head occurring on this coin depending upon how biased is the coin. Formally, the probability assessment for event A using the relative frequency approach is given by : P(A) = Number of times A occurs Number of trials, n It can be easily visualised that when the number of trials increases, we get better and better 145 estimate of the true probability of the event in question. Both the classical and the relative frequency approaches to probability are objective in nature. The classical definition is objective in the sense that it is based on deduction from a set of assumptions while the relative frequency approach is objective because the probability is derived from repeated empirical observations. However, both the theories fail when we are dealing with unique events. For example to determine the probability that a certain student will succeed in a particular examination, we can apply none of the two. This is because it cannot be ruled that for every student the events of succeeding and not succeeding are equally likely. Similarly, we cannot subject the candidate to appear in the examination several times to estimate his probability of success. In such cases, we have the personalistic approach to probability. 1.3.3 Personalistic Approach The approach views that the probability of an event is a measure of the degree of belief that an investigator has in the happening of it. It grants that the probability of the same event may be assigned differently by different investigators according to the confidence each one has in its happening. Thus, whereas the chances of a candidate succeeding in an examination may be placed at 80 percent by one person, another might estimate the chances to be 95 percent. Accordingly, the two would assign a probability 0.80 and 0.95 respectively for the event to happen. In may be mentioned that the three approaches to probability definitions are not competitive rather they are complementary in nature. 1.4 COMPUTATION OF PROBABILITY As we have already seen, the probability of an event is defined as the ratio of the number of favorable outcomes (for the event) to the total number of possible outcomes. Little difficulty is experienced when the total and favorable outcomes are small in number but when they are large, we may require the use of counting techniques to identify their number. Therefore, we first state the method of obtaining permutations and combinations. (1) If a job can be done in m ways and another job can be done in n ways, then the total number of ways in which both of them can be done is m × n. This is the fundamental multiplication rule. Example 4 : A man can go from city A to cirty B by three routes and come back by any of four routes, in how many ways can he perform his to and fro journey. Solution : He can perform the journey in a total of 3 × 4 = 12 different ways. Example 5 : Three balanced dice are tossed. Find the chance that the sum of digits on the two would be equal to 10. Solution : Total number of ways in which three dice can fall = 6 × 6 × 6 = 216. Total number of ways in which a total of 10 can appear = 27 (as shown below) (1, 3, 6), (1, 4, 5), (1, 5, 4), (1, 6, 3), (2, 2, 6), (2, 3, 5), (2, 4, 4), (2, 5, 3), (2, 6, 2), (3, 1, 6), (3, 2, 5), (3, 3, 4), (3, 4, 3), (3, 5, 2), (3, 6, 1), (4, 1, 5), (4, 2, 4), (4, 3, 3), (4, 4, 2), (4, 5, 1), (5, 1, 4), (5, 2, 3), (5, 3, 2), (5, 4, 1), (6, 1, 3), (6, 2, 2), (6, 3, 1) 146 Accordingly, P(total of 10) = 27 1 = 216 8 (2) The total number of arrangements of n distinct objects considered all at a time is equal to n! Thus, nPn = n ! Example 6 : In how many ways can the letters in the word DELHI be arranged ? Solution : Since all the 5 letters are different, they can be arranged in 5! = 120 ways. (3) The total number of arrangements of n distinct objects taken r at a time equals. n Pr = n! (n − r ) Example 7 : A car dealer has 4 places in his showroom. It has just received a consignment of 10 cars of different shades. In how many ways can he arrange cars in the showroom ? Solution : nP r = 10 P4 = 10! 10 × 9 × 8 ×7 ×6! = =5040 (10 − 40)! 6! (4) If out of n objects, k1 are alike, k2 are alike, k3 are alike....and so on such that k1 + k2 + k3 + ......... = n, the number of arrangements of the n objects would be equal to n Pk1, k 2, k3 .... = n! k1 ! k 2 ! k3 !,.... Example 8 : In how many ways can the “letters in the word STATISTICS be arranged ? Solution : Here n = 10, k1 (S) = 3, k2(T) = 3, k3(I) = 2, k4(A) = 1 and k5(C) = 1 Accordingly 10P 3, 3, 2, 1, 1 = 10! = 50400. 3!3!2!1!1! (5) Out of a total of n distinct objects, the number of combinations of r objects can be obtained as follows : nC r= n! or n !/ (n − r )!r ! (n − n)!r ! Example 9 : In how many ways can a committee of 3 persons be chosen out of a total of 10 persons ? Solution : Here n = 10 and r = 3. The total number of committees would be : nC r = 10 C3 = 10! 10 × 9 ×8 ×7! = =120 (10 − 3)!3! 7! × 3 × 2 147 Example 10 : A committee of four is to be selected randomly out of a total of 10 executives, 3 of which are chartered accountants. Find the probability that the committee would include exactly 2 CAs. Solution : The committee of 4 executives can be selected out of a total of 10 executives in 10C4 ways. The number of ways in which 2C As can be selected out of 3 is equal to 3C2 while the number of ways in which 2 executives out of a total of 7 executives is equal to 7C2. C 2 × 7 C2 63 = ∴ P(committee includes exactly 2 CAs) = 10 210 C4 3 Example 11 : Two cards are drawn at random from a well-shuffled deck of cards. Find the probability that both are ace cards. Solution : No. of ways in which 2 cards can be selected out of 52 cards = 52 C2 = 52 × 51 ×50! =1326 50!2 4 No. of ways in which 2 aces can be selected out of 4 ace cards = C 2 = ∴ P(2 ace cards) = 4! =6 2!2! 6 1 = 1326 221 1.5 LAWS OF PROBABILITY The probability associated with any event represents the likelihood of that event occurring on a particular trial of an experiment. This probability also measures the perceived uncertainty about whether the event will occur. If we are not uncertain at all, we assign the event a probability of zero or one. If the event be A, then P(A) = 0 means that event A would not occur, while P(A) = 1 indicates that event A would definitely occur. Thus, for any event, the probability would range between zero and one. Probability is non-negative concept. Symbolically, Rule 1 : 0 ≤ P (A) ≤ 1 Example 12 : (i) Determine the probability that a 7 would appear on a six-faced die tossed once, (ii) Determine the probability that an even or an odd number would appear on tossing a die. Solution : (i) P(7) = 0 = 0 (since a 7 does not exist, therefore there is no question of its occurrence) 6 (ii) P(even or odd number) = 6 = 1 (since the number appearing must be an even or an odd 6 one) Rule 2 : The probability of the complement of event A is one minus the probability of event A. 148 Symbolically, P(A) = 1–P(A) To be able to hit and not be able to hit a target for example, are complementary events. If the probability of a person to hit a target is given to be 3/5, then the probability that he would not be able to hit the target would be : P(not hitting the target) = 1 − 3 2 = 5 5 1.5.1 Addition Rule When making a decision involving probabilities, we often need to combine event probabilities with some event of interest. Here we first consider the calculation of probability that event A or B, each of them being defined on the sample space would occur. We use the addition rules of probability for this purpose. Rule 3 : When the events are mutually exclusive, the probability of occurrence of either of them is given by the sum of their individual probabilities. For two events A and B which are mutually exclusive, P(A or B) = P(A) + P (B) P(A ∪ B) = P(A) + P(B) Alternatively, where (A ∪ B) reads A union B and means A or B. Thus, for the mutually exclusive events, the probability that either one of them would occur is given by the sum of their individual probabilities. This rule is known as the special rule of addition. In general terms, the rule is P(A ∪ B ∪ C ∪...K) = P(A) + P(B) + P(C) + ...+ P(K) Example 13 : A box contains 20 discs numbered 1 to 20. A disc is selected at random. Find the probability that the number on the disc is divisible by 5 or 7. Solution : Let A be the event that the number is divisible by 5, and B be the event that the number is divisible by 7. Since there is no number which is common in these, the events A and B are mutually exclusive. Accordingly. P(A) = 4 2 , P(B) = 20 20 and P(A ∪ B) = 4 2 6 3 + = = 20 20 20 10 Rule 4 : When the events are overlapping : when two events A and B are overlapping, then the probability that either A or B or both of them would occur is given by the sum of individual probabilities of events A and B to occur minus the probability of their joint occurrence. Symbolically, P(A or B or Both) = P(A) + P (B) – P (A and B) Alternatively, P (A ∪ B) = P(A) + P (B) – P (A ∩ B) When they are three overlapping events A, B, and C, we have , P(A ∪ B ∪ C) = P(A) + P (B) + P (C) – P (A ∩ B) – P (A ∩ C) – P (B ∩ C) + P (A ∩ B ∩ C) Example 14 : A box contains 20 discs numbered 1 through 20. A disc is selected at random. Find 149 the chance that its number is divisible by 3 or 5. Solution : Let A be the event that the number is divisible by 3, and B the event that the number is divisible by 5. Here six numbers 3, 6, 9, 12, 15 and 18 are divisible by 3 and four numbers 5, 10, 15 and 20 are divisible by 5. We notice that the number 15 is included in both the lists. Thus, we have, P(A) = 6 4 1 , P(B) = , and P(A ∩ B) = 20 20 20 Accordingly, P(A ∪ B) = 6 4 1 9 + − = 20 20 20 20 Example 15 : A survey conducted to know the smoking habits of 500 persons yielded the following results : Cigarette Brand No. of smokers A 140 B 175 C 100 A and B 45 A and C 38 B and C 44 A and B and C 18 Find the probability that a person selected at random from the above group would be (i) a smoker of brand A or B, (ii) a smoker of A or B or C. (iii) a non-smoker. Solution: From the given information, P(A) = P(A ∩ C) = 140 175 100 45 , , P(B) = , P(C) = , P(A ∩ B) = 500 500 500 500 38 44 18 , P(B ∩ C) = , and P(A ∩ B ∩ C) = 500 500 500 Accordingly, (i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 140 175 45 270 + − = =0.54 500 500 500 500 (ii) P (A ∪ B ∪ C) = P (A) + P(B) + P(C) – P(A ∩ B) – P (A ∩ C) – P (B ∩ C) + P (A ∩ B ∩ C) = 140 175 100 45 38 44 18 306 + + − − − + = 500 500 500 500 500 500 500 500 150 0.612 = (iii) P (non-smoker) = 1 – P (smoker) = 1 − 306 194 = =0.388 500 500 1.5.2 Conditional Probability In dealing with probability we often need to determine the chances of two or more events occurring either at the same time or in succession. For example, a quality control manager for a manufacturing company may be interested in the probability of selecting two successive defectives from an assembly line. In other instances, the decision maker may know that an event has occurred and may want to know the chances of a second event occurring. For example, the market research organization engaged by a company may give a favorable report for a high sales figure for a new product to be introduced by the company. The company managing director might well be interested to know the probability of making high sales given a favourable report. These situations require tools different from those presented above in context of addition rules. Specifically we need to understand the rules for conditional probability and multiplication of probabilities. To understand, suppose that the employees of an organization are cross-classified according to sex and rank as follows : Officer Clerk Total Males 400 300 700 Females 200 1000 300 Total 600 400 1000 If an employee is selected at random then the probability that the employee would be a male = 700/1000. since out of the total of 1000 employees, a total of 700 are males. This probability is unconditional in the sense that we are not given any information about the type of employee selected. Now, if it is given that an employee is selected at random and he is an officer, then the probability that he would be a male shall be equal to 400/600, because the focus would be only on the officer employees which are 600 in all and of which there are 400 who are males. This probability is conditional. If we let the event A to represent the event that the employee would be a male, event A to represent that an employee would be an officer, we can write the conditional probability as : P(A/B) = 400 2 = 600 3 where (A/B) reads as event A given that event B has occurred. Upon a closer look we can represent the probability as: P(A/B) = P(A ∩ B) P(B) In this, P(A ∩ B) represents the probability that both events A and B would occur and P(B) is the probability for the event B to occur. Thus, P(A/B) is the conditional probability of A given B and is defined provided P(B) > 0. For the above example, P(A ∩ B) = 400/1000 and P(B) = 600/1000. As such, 151 P(A/B) = 400 /1000 400 2 = = 600 /1000 600 3 Example 16 : Consider an experiment in which two successive draws are to be made from an urn containing three white balls and five black balls. Assume that the balls are drawn at random and that the ball chosen on the first draw is not replaced. Find the probability that (i) the first ball drawn is white, and (ii) the second one is black. Solution : (i) Let A be the event that the first ball drawn is white and B be the event that the second 3 ball drawn is black. From the given information, P(A) = , since there are three white balls in a 8 total of eight balls. (ii) To determine the conditional probability of B given A, P (B/A), which is the probability of drawing a black ball on the second draw after drawing a white ball for the first draw, it should be noted that if A has already occurred, then there is a total of seven balls remaining and 5 five of them are black. Thus, P(B/A) = . 7 1.5.3 Multiplication Rule From the conditional probability defined in the preceeding paragraphs, since P(A/B) = P(A ∩ B) P(B) Rule 5 : P(A ∩ B) = P (B) × P (A/B) Also P(A ∩ B) = P (A) × P (B/A) This is called the multiplication rule for the non-independent events A and B, and states that the joint probability of the events A and B is given by the probability of the event A multiplied by the probability for event B given that event A has occurred (or the probability of event B, multiplied by the probability for event A given that event B has occurred). Similarly, for events A, B, and C which are not independent, we have P(A ∩ B ∩ C) = P (A) × P (B/A) × P (C/A ∩ B) Example 17 : Two balls are selected one after the other from an urn containing 7 black and 8 green balls. The first ball is not replaced before the second one is drawn. Find the probability that both would be green. Solution : Let A be the event that the first ball drawn is green and B be the event that the second ball drawn is green. From the given information. P(A) = 8 7 , P(B/A) = (if a green ball is taken out there would be 7 green balls in a total of 15 14 14 balls) 8 7 4 × = 15 14 15 Rule 6 : If the events A and B are independent, the probability that both events occur can be ∴ P(A ∩ B) = P (A) × P(B/A) = 152 determined by using P(A) and P(B). As mentioned earlier, two events are independent if the occurrence of one has no effect upon the occurrence of the other. More formally, if A and B are independent, P(A/B) = P (A), and P(B/A) = P (B). If A and B are independent, the conditional probability of A, given B, is the same as P(A), since the occurrence of the event B does not affect the occurrence of the event B; P(A/B) = P(A). The joint probability of independent events may be seen as the product of the probabilities of the events A and B, since P(A/B) = P(A ∩ B) = P(A) P(B) and P(A ∩ B) = P(A) × P(B) To generalize, for independent events A, B, C ... we have P(A ∩ B ∩ C) = P(A) × P(B) × P (C) × .... Example 18 : Two balls are selected one after the other from an urn containing 7 black and 8 green balls. The first ball is replaced before the second one is drawn. Find the probability that both would be green. Solution : Let A and B be the events that the first and the second ball, respectively, would be green. From the given information, P(A) = 8 8 and P(B) = 15 15 Accordingly, P(A ∩ B) = P(A) × P(B) = 8 8 64 × = 15 15 225 If it is significant to note that the condition P(A ∩ B) = P(A) × P(B) is satisfied then the events A and B are said to be independent, just as when they are independent then this relation is satisfied. This condition can be employed to determine whether the given events A and B are independent. Example 19 : For the data given in example, test whether the events of an employee selected being male and an employee selected being clerk are independent. Solution : Let A be the event that an employee selected is male and B be the event that an employee selected would be a clerk. From the given information, the number of employees who are males = 700 the number of employees who are clerks = 400 the number of employees who are males and clerks = 300 Accordingly P(A) = 700 , 1000 P(B) = 400 , 1000 and P (A ∩ B) = 153 300 1000 Here since 700 400 300 × ≠ , therefore the events A and B are not independent. 1000 1000 1000 Now we shall discuss the theorem of total probability, also called as the theorem of elimination. Rule 7 : If H1, H2, ....Hn be n mutually exclusive events, each with a non-zero probability, and E be an event defined on the same sample space and can be associated with either of them, the total probability of event E to occur is given by : P(E) = P(H1) × P(E/H1) + P(H2) × P(E/H2) +....+ P(Hn) × P(E/Hn). Alternatively, n P(E) = ∑ [P(Hi ) × P(E/Hi )] i=1 In this formulation, of course, P(Hi) and P(E/Hi) must all be given. To illustrate the application of this theorem, consider the following example. Example 20 : Two sets of candidates are competing for the positions of board of directors of a company. The chances for the first set to win are 60% while the chances for the second set are 40%. If the first set wins, the probability that a product will be introduced is 0.80 while if the second set wins, the probability for the product to be introduced is 0.30. Determine the probability that the product will be introduced. Solution : If H1 is the event that the first set wins, H2 is the event that the second set wins, and E is the event that the product is introduced, then P(H1) = 0.60, P(E/H1) = 0.80, P(H2) = 0.40, P(E/H2) = 0.30 Accordingly, P(E) = P(H1) × P(E/H1) + P(H2) × P(E/H2) = 0.6 × 0.8 + 0.4 × 0.3 = 0.48 + 0.12 = 0.60 Thus, there is a sixty per cent chance that the product shall be introduced. 1.6 BAYES’ THEOREM Conditional probabilities provide a lot of good information for decision makers. For instance, say the medical researchers are interested in determining the probability of getting cancer by a person supposing he was exposed to hazardous chemicals. That is, P(cancer/hazardous chemicals). In such cases we use the conditional probability rule P(A/B) = P(A ∩ B) P(B) However, in many practical applications, decision makers may know that an event has occurred but do not know what the chances were of that event before the fact. This cannot be known by the use of conditional probability rule directly. In such cases we employ an extension of conditional 154 probability called Bayes’ Theorem. This theorem deals with the conditional probability of an event Hi, given the probability of E, where E may have elements in each of the events H1, H2, H3, ....Hn with no element of E in more than one Hi. The Venn diagram of the figure displays such a condition. H2 H1 H3 H4 Bayes’ Theorem Figure 4 We shall illustrate the concept with an example and then make a generalization. Example 21 : Box 1 contains 5 white balls and 3 red balls. Box 2 contains 4 white balls and 4 red balls. A box is selected at random and one ball is randomly taken from that box. If the ball is white, what is the probability that it came from box 1 ? box 2 ? Solution : Let H1 : the box 1 is selected, H2 : the box 2 is selected, and E : the ball is white. From the given information, P(H1) = 1 1 5 4 , P(H2) = , P(E/H1) = , and P(E/H2) = . 2 2 8 8 Here we wish to calculate P(H1/E) P(H2/E) From the theorem of conditional probability, P (H E ) = P(HP(E)∩ E) and P (H E ) = P(HP(E)∩ E) 1 1 2 2 P(H1 ∩ E) is the probability of first selecting box 1 and then selecting one white ball from it. P(H1 ∩ E) = P(H1) × P(E/H1) = 1 5 5 × = 2 8 16 P(H2 ∩ E) is the probability of first selecting box 2 and then selecting one white ball from it. P(H2 ∩ E) = P(H2) × P(E/H2) = 1 4 4 × = 2 8 16 Since the ball selected can be from box 1 or box 2, we have, P(E) = P(H1 ∩E) + P(H2 ∩ E) = P(H1) × P(E/H1) + P(H2) × P(E/H2) 1 5 1 4 5 4 = × + × = + 2 8 2 8 16 16 155 9 = . 16 Accordingly, P ( ) H1 5 P(H1 ∩ E) P(H1 ∩ E) = = = 16 E P(E) P(H1 ∩ E) + P(H 2 ∩ E) 9 16 =5 9 ( ) 4 4 P(H 2 ∩ E) Also, P H 2 = = 16 = E 9 P(E) 9 16 Notice here that naturally either box 1 or box 2 would have been selected. When no information about the colour of the ball is known, the probability that box 1 is selected is 1/2 and so is the probability that box 2 is selected. Thus, P(H1) = 1/2 and P(H2) = 1/2 are the prior probabilities. Having known later on that the ball selected is of the white colour, we have revised these probabilities of P(H1/E) = 5/9 and P(H2/E) = 4/9. These probabilities are known as posterior probabilities. Thus the prior probabilities are transformed into posterior probabilities by incorporating the additional information, with the help of conditional and joint probabilities. The information in the above stated example can be restated as follows : Event Prior Prob. Conditional Prob. Joint Prob. Posterior Prob. (Hi) P(Hi) P(E/Hi) P(Hi ∩ E) P(Hi/E) H1 1/2 5/8 5/16 5/9 H2 1/2 4/8 4/16 4/9 Total, P(E) = 9/16 We can formally state the Bayes’ theorem now as follows : If H1, H2, ...Hn be mutually exclusive and collectively exhaustive events and E be an event which is arbitrarily defined on this sample space such that P(E) > 0, then the Bayes’ Therom states that P ( ) Hi P(H i ∩ E) = E P(E) n where in P (E) ∑ P(Hi ∩ E) i =1 Example 22 : A company has two suppliers of raw materials used in making cement. Vendor A supplies 30 per cent of raw materials while vendor B supplies 70 per cent. Tests have shown the 40 per cent of vendor A’s materials are poor quality whereas 5 per cent of vendor B’s materials are poor quality. The cement company’s manager has just found that there is a poor quality material in inventory. Which company most probably supplied the material ? Solution : Let H1 be the event that the material is supplied by vendor A, H2 be the event that the material is supplied by vendor B, E be the event that the material is of the poor quality. Given: Prior probabilities: P(H1) = 0.30, 156 P(H2) = 0.70 Conditional probabilities: P(E/H1) = 0.40, OP(E/H2) = 0.05. P(H1 ∩ E) = P(H1) × P(E/H1) = 0.30 × 0.40 = 0.120 Joint probabilities: P(H2 ∩ E) = P(H2) × P(E/H2) = 0.70 × 0.05 = 0.035 Total probability, P(E) = P(H1 ∩ E) + P (H2 ∩ E) = 0.120 + 0.035 = 0.155. Posterior probabilities: P (H E ) = P(HP(E)∩ E) =0.120 0.155 1 =0.77 P 0.035 (H E ) = P(HP(E)∩ E) =0.155 =0.23 1 2 2 Thus, vendor A most likely supplied the poor quality material. 1.7 EXPECTED VALUE An important concept, which has its origin in gambling and to which the probability is applied is the expected value. According to this, if an experiment has n outcomes that are assigned the payoffs x1, x2,............ xn occurring with probabilities p1, p2,....pn respectively, then the expected value is given by E(x) = x1 × p1 + x2 × p2 + ............. + xn × pn Example 23 : A player is engaged in the experiment of rolling a fair die. The player recovers an amount of rupees equal to the number of dots on the face that turns up, except when face 5 or 6 turns up in which case the player will lose Rs. 5 or Rs. 6 respectively. What is the expected value of the game to the player ? Solution : From the given information we have : Outcome: 1 2 3 4 5 6 Probability: 1/6 1/6 1/6 1/6 1/6 1/6 Payoff: 2 3 4 –5 –6 1 1 1 × 5− × 6 − × 6 6 6 1 = − 6 1 The expected value of the games is : E(x) = 1 × 1 1 1 + 2 × +3 × 6 6 6 4+ Thus, the player would expect to lose on an average Re 1/6 or 17 p. on each throw. Example 24 : An oil company may bid for only one of two contracts for oil drilling in two different areas A and B. It is estimated that a net profit of Rs. 4,00,000 would be realized from the first field and Rs. 5,00,000 from the second field. Legal and other costs of bidding for the first oil field are Rs. 1,02,500 and for the second one are Rs. 1,05,000. The probability of discovering oil in the first field is 0.60 and in the second is 0.70. The manager of the company wants to know as to for which oil 157 field should the manager bid ? Solution : The expected values for the two contracts are calculated below : Calculation of Expected Value Investment A Outcome Amount Probability Expected Value Success 4,00,000 0.6 2,40,000 (1,02,500) 0.4 (41,000) Failure Total 1,99,000 B Success Failure 5,00,000 0.7 3,50,000 (1,05,000) 0.3 (31,500) Total 3,18,500 Example 25 : A box contains 4 white, 8 green and 8 red marbles. A player selects one marble at random. The player wins Rs. 6 if the marble he selects is white, Rs. 2 if it is green, but must pay if it is red. How much should he pay for a red marble if this is to be a fair game ? Solution: Probability of selecting a white ball = 4 , 20 Probability of selecting a green ball = 8 , 20 Probability of selecting a red ball = 8 . 20 For a game to be fair, its net pay off must be equal to zero. We have, Colour of Ball Payoff Probability Expected Value White 6 4/20 24/20 Green 2 8/20 16/20 Red –x (suppose) 8/20 –8x/20 0 ∴ 24 16 8 x 40 8 x + − = or ⇒ 20 20 20 20 20 or x = 40/8 = Rs. 5. 1.8 SUMMARY ● Probability is the likelihood that something will happen. When we calculate the probability of an event, we assign it a number between zero and one, depicting how likely it is to happen. ● There are three approaches to calculate, probability of an event. These are: (i) classical approach, 158 where the probability of an event is the ratio of number of favourable outcomes to the number of total possible outcomes; (ii) relative frequency approach, where an estimate of probability is given by the ratio of the number of favourable outcomes to the number of trials made; and (iii) personalistic approach, where the probability to an event is assigned by an individual depending on his degree of belief in the occurrence of the event. ● There are several theorems of probability, which are used to calculate probabilities in different situations. ● Theorem of complementary events : This is used to determine the probability of an event happening by subtracting the probability of the event not happening from 1. ● Theorem of addition: It deals with the probability of occurrence of either of the events when they are mutually exclusive or when they are overlapping. According to this, the probability that either of the events will happen is equal to the sum of their individual probabilities less the probability of their joint occurrence. ● Theorem of multiplication: This theorem deals with the calculation of the probability when our interest is in the occurrence of the events jointly. For independent events, it uses multiplication of individual probabilities while for events which are not independent, it uses conditional probability. A conditional probability, is the likelihood that an event will happen, given that another event has already happened. ● A probability tree provides a useful way of the handling and analysing conditional probabilities occurring at multiple levels. It represents the given information through various branches on a set of chance nodes. ● Bayes’ theorem: This theorem provides a method of revising given probabilities on the basis of additional information. This involves transforming prior probabilities into posterior probabilities with the help of conditional and joint probabilities. 1.9 SELF ASSESSMENT QUESTIONS Exercise 1. True or False Statements (i) The outcomes of an experiment are known as sample space. (ii) Two events are said to be mutually exclusive if the happening of one does not affect the probability of . happening of the other. (iii) Mutually exclusive events may or may not be collectively exhaustive. (iv) Overlapping events are same as non-independent events. (v) Appearance of a heads and appearance of a tails in single trial of a coin represent independent events. (vi) Two mutually exclusive events are not necessarily complementary events but two complementary events are mutually exclusive. (vii) Overlapping events are those which can occur severally and jointly. (viii) The classical approach defines probability of an event as the ratio of number of favourable outcomes to the total number of trials. 159 (ix) Personalistic approach can be used to obtain probabilities of unique events only. (x) All three approaches to the definition of probability have one thing in common, the probability is expressed as a ratio not exceeding 1. (xi) For an experiment involving a toss of three coins, n(S) is equal to 6. (xii) If P(A/B) = P(B). then A and B are said to be independent. (xiii) In a toss of a die, the probability of getting a 5 shall be same as the probability of getting 5 given that the number is odd. (xiv) In the statistical sense, E1, and E2 are independent only when P(E1 ∩ E2) = P(E1) × P(E2). (xv) Bayes’ theorem is used to calculate revised probabilities called posterior probabilities from prior probabilities, using conditional and joint probabilities. (xvi) The sum of posterior probabilities is equal to 1 as is the sum of prior probabilities, in a given problem. Ans. 1. T, 2. F, 3. T, 4. F, 5. F, 6. T, 7. T, 8. F, 9. F, 10. T, 11. F, 12. F, 13. F, 14. T, 15. T, 16. T Exercise 2 : Questions and Answers (i) What is probability? Explain the calculation of probability under the classical approach. (ii) Which probability approach would you use to calculate the following probabilities? Give reasons also. (a) The next toss of a fair coin will land on heads. (b) India will win the next match with England. (c) The sum of the faces of two dice will be eight. (d) The success of a new product launched in the market. (iii) “Complementary events are mutually exclusive but mutually exclusive events may not be complementary.” Discuss with examples. (iv) Distinguish between mutually exclusive and overlapping events. How is the theorem of addition applied in both these cases? (v) Distinguish clearly between mutually exclusive and independent events. Can two events be mutually exclusive and independent simultaneously? Do you agree that on tossing a coin once, the appearance of heads and appearance of tails represent independent as well as mutually exclusive events? (vi) In each of the following cases, examine whether events are mutually exclusive, overlapping, complementary, independent or not-independent: (a) On a single toss of a die. appearance of 5 or 6 or appearance of a number smaller than 4. (b) A bank employee being an assistant manager or being a female. (c) A claim adjuster in an insurance company being a male or above 50 years of age. (d) An employee being a clerk or a sportsman. 160 (e) A person in a hospital being a heart specialist or over 45 years of age or a lab technician. (f) A two-shift factory employee working in morning shift or evening shift. (g) A teacher in a college working in the commerce department or the chemistry department. (h) A college employee being a teaching faculty or a member of non-teaching staff. (i) A factory employee being male or being a trade union member. (j) Appearance of an odd number or appearance of a number greater than 4 on a single toss of a die. (k) Getting two defectives one after another from a lot of 50 units of an item. (vii) Explain the meaning of marginal, joint and conditional probability. How can we obtain marginal probability of event E from the given joint probabilities of events A and E, B and E, C and E, and D and E, where A, B, C and D are the events to which E is related? (viii) What is statistical independence? How can we ascertain whether events A and B are statistically independent? (ix) State and explain Bayes’ Theorem. (x) A, B and C are competing for the award of a building contract. It is believed that the chances of A’s getting the contract are one-half of the combined chances of B and C’s getting it. Further B’s chances are believed to be one-half of C’s chances. What is the probability of each one getting the contract? (xi) A survey on MBA students provided the following data for 2,018 students: Age group Whether applied to more than one school Yes No 23 or under 297 201 24–26 209 379 27–30 185 268 31–35 66 193 36 and over 51 169 (a) What is the probability that a randomly selected applicant is 23 or under? (b) What is the probability that a randomly selected applicant is older than 26? (c) What is the probability that a randomly selected applicant applied to more than one school? (d) What is the probability that a randomly selected applicant is above 36 and has not applied to more than one school? (e) What is the probability that a randomly selected applicant is under 27 and has applied to more than one school? 161 (xii) There are twenty tickets in a bag, numbered consecutively from 1 to 20. A ticket is selected at random. Find the chance that the number on the ticket is: (a) Greater than 14 (b) Divisible by 4 (c) Between 8 and 15, both inclusive. (xiii) There are four shops and four applicants each of whom applies for one shop at random. Find the probability that (a) Each of them applies for a different shop. (b) Each of them applies for the same shop. (xiv) A committee of 6 is to be chosen randomly from a group of 8 men and 4 women. Determine the probability that it shall be composed of (a) 4 men and 2 women (b) 6 women (c) 2 men and 4 women (d) 6 men (xv) Among the 90 pieces of mail delivered to an office, 50 are addressed to the accounting department and 40 are addressed to the marketing department. If two of these pieces of mail are delivered to the managers’ office by mistake, and the selection is random, what are the probabilities that (a) Both of them should have been delivered to the accounting department; (b) Both of them should have been delivered to the marketing department; (c) One should have been delivered to the accounting department and the other to the marketing department? (xvi) A bag contains 6 green, 7 blue and 2 red balls. Three balls are chosen at random. Find the probability that (i) all of them are green, (ii) both the red balls are included and (iii) the balls are all of different colours. (xvii) Two unbiased dice are tossed. What is the probability that the total of numbers on them would be a multiple of 3? (xviii) A pack contains 30 tickets numbered consecutively from 1 to 30. A ticket is chosen at random from this. Find the chance that the number on this would be (i) a multiple 6 or 7 and (ii) a multiple of 3 or 5. (xix) Five candidates A, B, C, D and E appear for an interview. Two candidates D and E are eliminated in the first round of the interview. A has twice the chance of being selected than B, and B has twice the chance as C, in the final interview. D bets that either A or B will be selected and E bets that either B or C will be selected. Who is likely to win the bet? (xx) Given the following probability table of television viewing frequencies (X) and the income levels (Y): Viewing Income levels (Y) frequency (X) High Middle Low Regular Occasional Rarely Total 0.10 0.10 0.05 0.25 0.15 0.20 0.05 0.40 0.05 0.10 0.20 0.35 162 Total 0.30 0.40 0.30 1.00 (a) What is the probability that a person is a low income individual and views TV regularly? (b) If an individual is at low income level, what is the probability that he/she views TV regularly? (c) What is the probability that given an individual does not have high income, he/she rarely watches TV? (d) If an individual occasionally watches TV, what is the probability that he/she is a high income earner or a middle income earner? (e) Is viewing TV regularly independent of earning high income? Explain. (xxi) The probability that a contractor will not get a plumbing contract is 2/3 and the probability that he will get an electric contract is 5/9. If the probability of getting at least one contract is 4/5. what is the probability that he will get both the contracts? (xxii) An unbiased die and a biased die are tossed together. Find the probability that the sum of digits obtained on them is even, given that on the biased die, it is thrice as likely to show an even number as an odd one when tossed once. (xxiii) A six-faced die is so biased that the digits 1, 3 or 5 on it are thrice as likely as the digits 2,4 or 6, when tossed once. Find the probability that in two tosses of this die, the sum of digits would be odd. (xxiv) A husband and wife appear in an interview for two vacancies for the same post. The probability of the husband’s selection is 1/7 and that of wife’s selection is 1/5. What is the probability that (a) Both of them will be selected? (b) Only one of them will be selected? (c) None of them will be selected? (xxv) (a) In rolling a pair of dice, what is the probability of rolling a total of 21 on the first two rolls? (b) Given that P(A) = 0.65. P(B) = 0.80, P(A/B) = P(A) and P(B/A) = 0.85. Is this a consistent assignment of probabilities? (xxvi) An MBA applies for one job in two firms X and Y. The probability of his being selected in firm X is 0.7 and his being rejected in firm Y is 0.5. The probability of at least one of his applications being rejected is 0.6. What is the probability that he will be selected in one or both of the firms? (xxvii) During a survey of road safety, it was found that 60 percent of accidents occur at night, 52 percent are alcohol related, and 37 percent are alcohol related and occur at night. (a) What is the probability that an accident was alcohol related given that it occurred at night? (b) What is the probability that an accident occurred at night given that it was alcohol related? 163 (xxviii) An advertising executive is studying television-viewing habits of married men and women during prime time hours. On the basis of past viewing records, the executive has determined that during prime time, husbands are watching television 60% of the time. It has also been determined that when the husband is watching television, 40% of the time the wife is also watching. When the husband is not watching television, 30% of the time the wife is watching television. Find the probability that, (a) If the wife is watching television, the husband is also watching television. (b) The wife is watching television during prime time. (xxix) In a small factory, machines A, B and C manufacture 35%, 25% and 40% respectively of the total output. Of their output, respectively, 0.5, 4 and 2 percent are defective. One item is drawn and found to be defective. What are the respective probabilities that it was produced by machines A, B and C? (xxx) Reliance Industries Limited is determining whether it should submit a bid for oil exploration contract. In the past, main competitor of RIL, ONGC has submitted bids 66 percent of the time. If ONGC does not bid for oil exploration contract, the probability that RIL will get the contract is 0.45. If ONGC does bid for oil exploration contract, the probability that RIL will get the contract is 0.25. (a) If Reliance Industries gets the contract, what is the probability that ONGC did not bid? (b) What is the probability that Reliance Industries will get the contract? Ans. 10. 0.33, 0.22, 0.44, 11. 498/2018, 932/2018, 808/2018, 169/2018, 507/2018 12. 0.3, 0.25, 0.4 13. 3/32, 1/64 14. 0.4545, 0, 0.0303, 0.0303 15. 0.306, 0.195, 0.4993 16. 20/455, 13/455, 84/455 17. 0.33 18. 9/30, 14/30 19. D is likely to win 20. 0.0575, 0.01429, 0.333, 0.75, No 21. 14/45 22. 0.5 23. 0.375 24. 1/35, 10/35, 24/35 25. 20/1296, Not consistent 26. 0.8 27. 0.617, 0.712 28. 0.67, 0.36 29. 0.362, 0.406, 0.232 30. 0.519, 0.318 164 LESSON-2 PROBABILITY DISTRIBUTIONS 2. STRUCTURE 2.0 Objective 2.1 Binomial Distribution 2.1.1 Properties of Binomial Distribution 2.2 Poisson Distribution 2.3 Normal Distribution 2.3.1 Properties of Normal Distribution 2.3.2 Calculation of Probabilities 2.4 Summary 2.5 Self Assessment Questions 2.0 OBJECTIVE After reading this lesson, you should be able to : (a) Understand the concept of probability distribution and random variables (b) Understand the characteristics and procedure of computing probabilities using binomial and Poisson distribution (c) Comprehend the difference between discrete and continuous probability distribution. (d) Understand normal distribution, properties of a normal curve and computation of probabilities using z-values. (e) Analyse the situations under which a Poisson distribution is treated a binomial or normal distribution. 2.1 BINOMIAL DISTRIBUTION Refer to a set of mathematical models of the relative frequencies of a finite number of observations of a variable. It is systematic arrangement of probabilities of mutually exclusive and collectively exhaustive elementary events of an experiment. Observed frequency distributions are based upon actual observation and experimentation. We can deduce mathematically a frequency distribution of certain population based on the trend of the known values. This kind of distribution on experience or theoretical considerations is known as theoretical distribution or probability distributions. These distributions may not fully agree with actual observations or the empirical distributions based on sample observations. If the number of experiments is increased sufficiently the observed distributions may come closer to theoretical or probability distributions. Theoretical distributions are useful for situations where actual observations or experiments are not possible. Moreover, it can be used to test the goodness of fit. They provide decision makers with a logical basis for making decisions and are useful in making predictions on the basis of limited information or theoretical considerations. 165 There are broadly three theroetical distributions which are generally applied in practice. They are : 1. Binomial distribution 2. Poisson distribution 3. Normal distribution We shall discuss them in detail one by one. It is a discrete distribution. The binomial distribution was discovered by James Bernoulli in 1700 to deal with dichotomous classification of events. It is a probability distribution expressing the probability of one set of dichotomous alternatives, i.e., success or failure. The binomial probability distribution is developed under some assumptions which are: (i) An experiment is performed under similar conditions for a number of times. (ii) Each trial shall give two possible outcomes of the experiment success or failure. S = {failure, success] (iii) The probability of a success denoted by p remains constant for all trials. The probability of a failure denoted by q is equal to (1 – p). (iv) All trials for an experiment are independent. If a trial of an experiment can result in success with probability p and failure with probability q = (1 – p), the probability of exactly successes in n trials is given by P(x) = nCx pxqn-x where x = 0, 1, 2...n where P(x) = Probability of x successes n Cx = n! (! is termed factorial) x !(n − x)! The entire probability distribution of x = 0, 1, 2,....n can be written as follows : Binomial Probability Distribution Number of success Probability x P(x) n 0 1 n 2 n : C0 p 0q n C1 p1q n −1 C2 p 2q n− 2 : n x : Cx p xq n − x : n n 166 Cn p nq n − n We should note that the variable x (number of successes) is discrete. It can take integer values 0, 1, 2, ..., n. The probabilities specified in the above table are in fact successes terms of the Binomial Expansion of (p + q)n, which is (q + p) n = nC0 q n p 0 +nC1q n −1 p1 +nC2 q n − 2 p 2 +C3q n −3 p3 ...+nCn q n − n p n n 2.1.1 Properties of Binomial Distribution (i) The shape and location of binomial distribution changes as p changes for a given n or n changes for given p. If n increases for a fixed p, the binomial distribution moves to the right, flattens and spreads out. The mean of the distribution (np) increases as n increases for constant value of p. (ii) The mode of the binomial distribution is equal to the value of x which has the largest probability. (iii) If n is large and p and q are not close to zero, the binomial distribution can be approximated by a normal distribution with standardised variable. z = X − np npq (iv) The mean and the standard deviation of the Binomial distribution is np and npq respectively. (v) The other constants of the distribution can be calculated. µ 2 = npq µ 3 = npq (q – p) µ 4 = 3n2p2q2 + npq (1 – 6pq) We can calculate the value of β1 and β2 to measure nature of the distribution. µ32 n 2 p 2 q 2 (q − p) 2 (q − p 2 ) = β1 = 3 = npq µ2 n3 p 3q 3 β2 = µ 4 3n2 p 2 q 2 + npq(1 −6 pq) 1 −6 pq = =3 + 2 2 2 2 npq n p q µ2 The binomial distribution is useful in describing variety of real life events. Binomial distribution is useful to answer questions such as : If we conduct an experiment n times under the stated conditions, what is the probability of obtaining exactly x successes? For example, if 10 coins are tossed simultaneously what is the probability of getting 4 heads ? We shall explain the usefulness of binomial distribution by certain examples. Example : A coin is tossed eight times. What is probability of obtaining 0, 1, 2, 3, 4, 5, 6, 7 and all heads ? 167 Solution : Let us denote the occurrence of head as success by p. So that p = 1 2 ∴ q=1–p= 1 2 and n = 8 (given) We can calculate various probabilities by expanding the binomial theorem. ( q + p )8 = 8C0 q8 p 0 +8C1q 7 p1 +8C2 q6 p 2 +C3q5 p 3 +C4 q 4 p 4 8 8 C +5q3 p 5 8 + 8C7 q1 p 7 + 8Cn q 0 p8 Therefore the probability of obtaining 0 heads = 8 C0 q8 p 0 = 8! 8!× 0! 8 1 1 × 2 2 0 8 8! 1 1 = × = 2 8! 2 8 1 = Ans. 256 The probability of obtaining 1 head = 8C1q7p1 7 8! 1 1 1 × × =8 × = 2 7!× 1! 2 2 7 1 × 2 8 = Ans. 256 The probability of getting 2 heads = 8C2q6p2 6 2 8! 1 1 8 × 7 × 6! 1 = × = × 2!× 6! 2 2 2! × 6! 2 8 8 ×7 1 = × 2 2 8 28 = Ans. 256 The probability of getting 3 heads = 8C3q5p3 5 8! 1 1 × × = 3!× 5! 2 2 3 = 8 × 7 ×6 ×5! 1 × 3 ×2 ×1 ×5! 2 8 56 = Ans. 256 The probability of getting 4 heads = 8C4q4p4 4 8! 1 1 × × = 4!× 4! 2 2 4 = 8 × 7 × 6 ×5 ×4! 1 × 4 ×3 ×2 ×1 ×4! 2 8 70 = Ans. 256 The probability of getting 5 heads = 8C5q3p5 3 8! 1 1 × × = 3!× 5! 2 2 5 = 8 × 7 ×6 ×5! 1 × 3 ×2 ×1 ×5! 2 8 56 = Ans. 256 The probability of getting 6 heads = 8C6q2p6 2 8! 1 1 × × = 2!× 6! 2 2 6 8 × 7 × 6! 1 = × 2 ×1 ×6! 2 8 168 28 = Ans. 256 C+6 q 2 p 6 8 The probability of getting 7 heads = 8C7q1p7 1 8! 1 1 × × = 7!× 1! 2 2 7 8 × 7! 1 = × 7! 2 8 8 = Ans. 256 The probability of getting 8 heads = 8C8q0p8 8! 1 1 × × = 8!× 0! 2 2 8 8! 1 = × 8! 2 8 1 = Ans. 256 We can also calculate the probability of 4 or more heads or maximum 6 heads. Probability of 4 or more heads = 8 C4 q 4 p 4 + 8C5 q 3 p 5 + 8C6 q 2 p 6 +8C7 q1 p 7 +8Cn q 0 p8 70 = 256 56 + 256 28 + 256 8 1 7 8 0 8 Probability of getting more than 6 heads = C7 q p + Cn q p = 8 + 256 1 + 256 163 = Ans. 256 8 1 9 + = Ans. 256 256 256 Example 1 : A box contains 100 transistors, 20 of which are defective, 10 are selected at dom. Calculate the probability that (i) all 10 are defective, (ii) all 10 are good. Solution : Let x represent the number of defective transistors selected. Then the value of x would be, x = 0, 1, 2,.... 10. Let us put p as the probability of a defective transistor. ∴ p= 20 1 4 = and q = 1 – p = , n = 10 100 5 5 Using the formula for binomial expansion, the probability of x defective transistors is P(x) = nCxpxqn–x (i) Probability that all 10 are defective = 10C10 p10q10 – 10 = 10! 10!× 10! 1 × 5 10 0 4 1 × = 5 5 10 1 =10 Ans. 5 (ii) Probability that all 10 are good = 10C0 p0q10 0 10 10 10! 1 4 4 × × = = Ans. 10! 5 5 5 2.2 POISSON DISTRIBUTION This is also a discrete distribution. It was originated by a French mathematician Simeon Denis Poisson in 1837. The Poisson distribution is the limiting form of binomial distribution as n becomes infinitely large (n > 20) and p approaches zero (p < 0.05) such that np = m remains fixed. The 169 possion distribution is useful for rare events. Suppose in the binomial distribution. (a) p is very small (b) n is so large that np = m is constant Then, we would get the following distribution x 0 Probability e–m 1 e− m 2 m 1! e−m m2 2! ............ x 3 e− m m3 3! .............. Total e−m mx x! It is a Poisson distribution. Under these conditions the probability of getting x successes is P(x) = e −m mx x! Sum of the probabilities of 0, 1, 2, 3 successes is m m 2 m3 m x + + .... + e − m 1 + + 1! 2! 3! x! e=m .em 1= where e is a constant whose value is 2.7183 and m is the parameter of the distribution i.e. the average number of occurrences of an event. A classical example of the Poisson distribution is given by road accidents. As we know the number of people travelling on the road is very large i.e. n is large. Probability that any specific individual runs into an accident is very small. However. np = average number of road accidents is a finite constant on any particular day. Therefore, x (number of road accidents on a particular day) follows Poisson distribution. The various parameters of Poisson Distribution are : Mean (m) = np (variance) = np = m σ = np µ 2 = np = m µ3 = m µ 4 = m + 3m2 ∴ β1 = µ 32 µ 32 = m2 1 = m3 m µ 4 m + 3m2 1 =3 + β2 = 2 = 2 m µ2 m 170 Example 2 : If one house in 1000 has a fire in a district per year. What is the probability that exactly 5 houses will have fire during the year if there are 2000 houses ? Solution : We shall apply Poisson distribution m = np where n = 2000, p = ∴ m = np = 2000 × P(x) = e–m 1 = 2. 1000 mx × when x = 5 and e = 2.7183 x! P(5) = 2.7183–2 × = Reciprocal (AL (2 log 2.7183)) 1 1000 2 × 2 × 2 ×2 ×2 4 25 = (2.7183)–2 × = 2.7183–2 × 5 × 4 × 3 ×2 ×1 15 5! 4 4 = Reciprocal (7.389) 4 = 0.1352 × = 0.036 Ans. 15 15 15 Example 3 : If 3% of the bulbs manufactured are defective, calculate the probability that a sample of 100 bulbs-will contain no defective and one defective bulb using Poisson distribution. Solution : Given number of defective bulbs are 3% (3/100). ∴ m = np = 100 × 3 = 3. 100 Probability of no defective bulb in a sample of 100 is P(x) = e–m × mx where m = 3, and e = 2.7183 x! P(o) = 2.7183–3 = 0.05 Ans. Probability of one defective bulb in a sample of 100 is P(1) = e–m × m1 = 2.7183–3 × 3 = 0.15 Ans. 1! 2.3 NORMAL DISTRIBUTION The most important continuous probability distribution used in the entire field of statistics is normal distribution. The normal curve is bell-shaped that extends infinitely in both directions coming closer and closer to the horizontal axis without touching it. The mathematical equation of normal curve was developed by De Moivre in 1733. A continuous random variable x is said to be normally distributed if it has the probability density function represented by the equation of normal curve. 1 e y= σ 2π − ( x −µ )2 2 σ2 171 , – ∞ ≤x ≤ + ∞ Where µ and σ are mean and standard deviation which are two parameters and e = 2.7183, p = 3.1416 are constants. It may be understood that the normal distributions can have different shapes depending upon values of µ and σ but there is one and only one normal distribution for any given pair values of µ and σ. 2.3.1 Properties of Normal Distribution 1. If the parameters µ and σ of the normal curve are specified, the normal curve is fully determined and we can draw it by obtaining the value of y corresponding to different values of x (the abscissa). 2. The normal curve tends to touch the x-axis only at infinity i.e. the x-axis is an asymptotic to the normal curve. It is a continuous curve stretching from – ∞ to + ∞. 3. The mean, median and mode of the normal distribution are equal. 4. The height of the normal curve is maximum at x = µ. Hence the mode of the normal curve is x = µ. 5. The two quartiles Q1 and Q3 are equidistant from the median. Q1 = µ – 0.6745 σ Q3 = µ + 0.6745 σ Hence Quartile Deviation = Q3 − Q1 = 0.6745 σ 2 4 σ or MD = 0.7979 σ 5 6. Mean deviation about mean is 7. The points of inflexion of the normal curve occur at x = µ + σ and x = µ – σ 8. The tails of curve extend to infinity on both sides of the mean. The maximum ordinate at X = µ is given by y = 9. 1 σ 2π Approximately 100% of the area under the curve is covered by µ + 3σ. Distance from the mean % of total area under the ordinate in terms of ± σ normal curve Mean ± 3σ 68.27 Mean ± 2σ 95.45 Mean ± 3σ 99.73 10. All odd moments are equal to zero. µ1 = µ3 = 0 β1 = 0 and β2 = 3. Thus the curve is mesokurtic. 11. The normal distribution is formed with a continuous variable. 172 12. The fourth moment is equal to 3σ4 for a normal distribution. The equation of the normal curves gives the ordinate of the curve corresponding to any given value of x. But we are interested in finding out the area under the normal curve rather than its ordinate (y). A normal curve with 0 mean and unit standard deviation is known as the standard normal curve. With the help of a statistical table which gives the area and ordinates of the normal curve are given corresponding to standard normal variate. x−µ and not corresponding to x. σ Let us see the normal curve area under x-scale and z-scale. z = Fig. 1 2.3.2 Calculation of Probabilities Now we discuss the method of calculating probabilities where the distribution follows the normal pattern.In fact,the probability forthe variable to assum e a value w ithin a given range,say X1 and X2, is equal to the ratio of the area under the curve in that range to the total area under the curve. To obtain the relevant areas, we first transform a given value of the variable X into standardized variate Z as follows : X −µ Z= σ Then we consult the normal area table. This table is constructed in a manner such that the areas between mean (µ) and the particular values of Z are given. The first column of this table contains values of Z from 0.0 to 3.0, while the top row of the table gives values 0.00; 0.01; 0.02; ....0.09. To find the area (from mean) to a specific value of Z, we look up in the first column for Z-value upto its first decimal place while its second decimal place is read from the top row. To illustrate, if we want to find the area between mean and Z = 1.42, then we look for 1.4 in the first column and 0.02 in the top row. Corresponding to these, the value in the table reads 0.4222. Similarly, it can be verified that area upto Z = 0.10 is equal to 0.0398 while for Z = 2.59, it is 0.4952. Let us understand few more things : (i) The area under the curve from Z = 0 (when X = µ) to a particular value of Z gives the proportion of the area under this part of the curve to the total area under the curve. Thus, Z = 0 to Z = 1.42 the value 0.4222. Naturally, this is taken as the probability that the variable in question will assume a value within these limits. 173 (ii) Since the normal curve is symmetrical with respect to mean, the area between µ(Z = 0) and particular value of Z to its right will be same as the value of Z to its left. Thus, area between Z = 0 and Z = 1.5 is equal to area between Z = 0 and Z = –1.5. Remember that for values of X greater than µ, the Z value will be positive while for X < µ, the value of Z would be negative. (iii) The general procedure for calculating probabilities is like this : (a) specify clearly the relevant area under the curve which is of interest. (b) determine the Z value (s). (c) obtain the required area (s) with reference to the normal area table. Example 4 : Find the area under the normal curve : (i) between Z = 0 and Z = 1.20 (ii) between Z = 1.0 and Z = 2.43 (iii) to the right of Z = 1.37 (iv) between Z = –1.3 and Z = 1.49 (v) to the right of Z = –1.78 Solution : For each of these, the relevant portions under the normal curve are shown shaded and the areas determined with reference to the normal area table. Fig. 2 (i) Area between Z = 0 and Z = 1.20 is 0.3849. Fig. 3 (ii) Area between Z = 0 and Z = 1.0 is 0.3413 Area between Z = 0 and Z = 2.43 is 0.4925 ∴ Area between Z = 1.0 and Z = 2.43 is 0.4925 – 0.3413 = 0.1512. Fig. 4 174 (iii) Area between z = 0 and Z = 1.37 is 0.4147. Total area under the curve being equal to 1, the area to the right of Z = 0 is 0.5, as is the area to the left of it ∴ Area beyond Z = 1.37 is 0.5000 – 0.4147 = 0.0853. Fig. 5 (iv) Area between Z = 0 and Z = 1.3 is 0.4032. Area between Z = 0 and Z = 1.49 is 0.4319. ∴ Area between Z = 1.3 and Z = 1.49 is 0.4032 + 0.4319 = 0.8351. Fig. 6 (v) Area between Z = 0 and Z = – 1.78 is 0.4625. Area between Z = 0 is 0.5. ∴ Area to the right of Z = – 1.78 is 0.4625 + 0.5 = 0.9625. Example 5 : Balls are tested by dropping from a certain height of bounce. A ball is said to be fast if it rises above 36 inches. The height of the bounce may be taken to be normally distributed with mean 33 inches and standard deviation of 1.2 inches. If a ball is drawn at random, what is the chance that it would be fast ? Solution : The given information is depicted in figure 7. Here we have to calculate the probability that the height of the bounce, X, would be greater that 36. This is shown shaded in figure 7. Fig. 7 We have, X = 36, µ = 33 and σ = 1.2 Z= X − µ 36 − 33 = =2.5 σ 1.2 175 From the normal area table, area between Z = 0 and Z = 2.5 is equal to 0.4938. So area beyond Z = 2.5 is 0.5 – 0.4938 = 0.0062. Therefore, P(X > 36) = 0.0062, the chance of getting a fast ball. Example 6 : The life (x) of electric bulbs in hours is supposed to be normally distributed as 1 ( x −155)2 e 722 19. 2π What is the probability that the life of a bulb will be : (i) Less than 117 hours (ii) more than 193 hours (iii) between 117 and 193 hours. Solution : Given µ = 155 and σ = 19 117 − 155 = −2 Therefore, corresponding to x = 117 the standard normal variate is z = 19 Fig. 8 We have to obtain the area to the left of Z = –2 [Pr(Z< – 2)]. From the table we see the area z = 0 and z = –2 and subtract it from 0.5. ∴ 0.5 – .4772 = 0.0228 Hence the probability of life of bulbs more than 193 hours is 0.0228. To obtain the probability that the life of the bulb is more than 193 hours, we obtain the corresponding standard normal variate 193 − 155 = +2 z= 19 Fig. 9 And the area between 117 hours and 193 hours shall be 117 Fig. 10 176 where Z = + 2. Hence Pr(–2 < Z < +2) = Pr (117 < x < 193) = .4772 + .4772 = .9544 Ans. Example 7 : The results of a particular examination are given below in a summary form : Result Percentage of Candidates Total Passed 80 Passed with distinction 10 Failed 20 It is known that a candidate fails if he obtains less than 40 marks (out of 100), while he must obtain at least 75 marks in order to pass with distinction. Determine the mean and the standard deviation of marks assuming distribution of marks to be normal. Solution : According to the given information, Percentage of students getting marks less than 40 = 20, Percentage of students getting marks between 40 and 75 = 70, and Percentage of students getting marks above 75 = 10. The relevant area is shown in figure 15. Fig. 11 Here P(X < 40) = 0.20, P (40 < X < 75) = 0.70 and P(X > 75) = 0.10 Let µ and σ represent the mean and standard deviation of the distribution. We have, area between µ and X = 40 equal to 0.30, and area between µ and X = 75 as equal to 0.40. Now we have, 40 − µ , and σ 75 − µ , For X = 75, Z = σ Corresponding to the area 0.30 in the normal area table, Z = 0.84. Thus, for X = 40, we have Z = – 0.84 (Since the value of 40 lies to the left of µ). Similarly, for the area equal to 0.40, we have Z = 1.28. For X = 40, Z = We have, then 40 − µ σ = – 0.84 75 − µ σ = – 1.28 177 and ....(i) ....(ii) Rearranging the above equations, we get µ – 0.84 σ = 40 and µ + 1.28 σ = 75 ....(iii) ....(iv) Subtracting equation (iii) from equation (iv), we get 2.12 σ = 35 σ = 35/2.12 = 16.51 or Substituting the value of σ in equation (iii) and solving for µ, we get µ – (0.84) (16.51) = 40 µ = 40 + 13.87 = 53.87 or Thus, Mean = 53.87 marks and standard deviation = 16.51 marks. Example 8 : There are 900 students in B.Com (Hons.) course of a college and the probability of a student needing a particular book on a day is 0.10. How many copies of the book should be kept in the library that there should be at least 0.90 chance that a student needing that book will not go disappointed ? Assume normal approximation to the binomial distribution. Solution : According to the given information, n = 900, p = 0.10, and q = 1 – p = 1 – 0.10 = 0.90. Therefore, mean = np = 900 × 0.10 = 90, and σ = npq = 900 ×0.10 ×0.90 =9. Here we are required to determine X, to the right of which 10 per cent of the area under the curve lies. Area between µ = 90 and X would be equal to 0.50 – 0.10 = 0.40. Now, Z value corresponding to the area 0.40 equals 1.28. Thus, X − 90 = 1.28 9 Solving for X, we get X = 1.28 × 9 + 90 = 11.52 + 90 = 101.52. Z= Therefore 102 books should be kept in the library to meet the demand of students. Table-1 Table of values of e–µ (0 ≤ µ ≤ 1) µ 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0 1.0000 0.9048 0.8187 0.7408 0.6703 0.6065 0.5488 0.4966 0.4493 0.4066 1 0.9900 0.8958 0.8106 0.7334 0.6636 0.6005 0.5434 0.4916 0.4449 0.4025 Note: e–0.4 = 0.6703, 2 0.9802 0.8869 0.8025 0.7261 0.6570 0.5945 0.5379 0.4868 0.4404 0.3985 3 0.9704 0.8781 0.7945 0.7189 0.6505 0.5886 0.5326 0.4819 0.4360 0.3946 4 0.9608 0.8694 0.7866 0.7118 0.6440 0.5827 0.5273 0.4771 0.4317 0.3906 e–0.37 = 0.6907, 178 5 0.9512 0.8607 0.7788 0.7047 0.6376 0.5770 0.5220 0.4724 0.4274 0.3867 6 0.9418 0.8521 0.7711 0.6977 0.6313 0.5712 0.5169 0.4677 0.4332 0.3829 e–0.99 = 0.3716 7 0.9324 0.8437 0.7634 0.6907 0.6250 0.5655 0.5117 0.4630 0.4190 0.3791 8 0.9231 0.8353 0.7558 0.6839 0.6188 0.5599 0.5066, 0.4584 0.4148 0.3703 9 0.9139 0.8207 0.7443 0.6771 0.6126 0.5543 0.5016 0.4538 0.4107 0.3716 Table of values of e–µ (0 ≤ µ ≤ 10) µ 1 2 3 4 5 6 0.36788 0.13534 0.04979 0.01832 0.006738 7 0.00091 0.00248 Note: 8 9 10 0.00012 0.00033 0.00040 e–1.5 = e–1 × e–0.5 = 0.36788 × 0.6065 = 0.2231 e–4.34 = e–4 × e–0.34 = 0.01832 × 0.7118 = 0.01304 Table-2 Normal Curve Z-score An entry in the table is the area under the curve between Z = 0 and a positive value of Z. Areas for negative values of Z arc obtained by symmetry. Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.1 0.2 0.3 0.4 0.0000 0.0398 0.0793 0.1179 0.1554 0.0040 0.0438 0.0832 0.1217 0.1591 0.0080 0.0478 0.0871 0.1255 0.1628 0.0120 0.0517 0.0910 0.1293 0.1664 0.0160 0.0557 0.0948 0.1331 0.1700 0.0199 0.0596 0.0987 0.1368 0.1736 0.0239 0.0636 0.1026 0.1406 0.1772 0.0279 0.0675 0.1064 0.1443 0.1808 0.0319 0.0714 0.1103 0.1480 0.1844 0.0359 0.0753 0.1141 0.1517 0.1879 0.5 0.6 0.7 0.8 0.9 0.1915 0.2257 0.2580 0.2881 0.3159 0.1950 0.2291 0.2611 0.2910 0.3186 0.1985 0.2324 0.2642 0.2939 0.3212 0.2019 0.2357 0.2673 0.2967 0.3238 0.2054 0.2389 0.2703 0.2995 0.3264 0.2088 0.2422 0.2734 0.3023 0.3289 0.2123 0.2454 0.2764 0.3051 0.3315 0.2157 0.2486 0.2794 0.3078 0.3340 0.2190 0.2517 0.2823 0.3106 0.3365 0.2224 0.2549 0.2852 0.3133 0.3389 1.0 1.1 1.2 1.3 1.4 0.3413 0.3642 0.3849 0.4032 0.4192 0.3438 0.3665 0.3869 0.4049 0.4207 0.3461 0.3686 0.3888 0.4066 0.4222 0.3485 0.3708 0.3907 0.4082 0.4236 0.3508 0.3729 0.3925 0.4099 0.4251 0.3531 0.3749 0.3944 0.4115 0.4265 0.3554 0.3770 0.3962 0.4131 0.4279 0.3577 0.3790 0.3980 0.4147 0.4292 0.3599 0.3810 0.3997 0.4162 0.4306 0.3621 0.3830 0.4015 0.4177 0.4319 1.5 1.6 1.7 1.8 1.9 0.4332 0.4452 0.4554 0.4641 0.4713 0.4345 0.4463 0.4564 0.4649 0.4719 0.4357 0.4474 0.4573 0.4656 0.4726 0.4370 0.4484 0.4582 0.4664 0.4732 0.4382 0.4495 0.4591 0.4671 0.4738 0.4394 0.4505 0.4599 0.4678 0.4744 0.4406 0.4515 0.4608 0.4686 0.4750 0.4418 0.4525 0.4616 0.4693 0.4756 0.4429 0.4535 0.4625 0.4699 0.4761 0.4441 0.4545 0.4633 0.4706 0.4767 2.0 2.1 2.2 2.3 2.4 0.4772 0.4821 0.4861 0.4893 0.4918 0.4778 0.4826 0.4864 0.4896 0.4920 0.4783 0.4830 0.4868 0.4898 0.4922 0.4788 0.4834 0.4871 0.4901 0.4925 0.4793 0.4838 0.4875 0.4904 0.4927 0.4798 0.4842 0.4878 0.4906 0.4929 0.4803 0.4846 0.4881 0.4909 0.4931 0.4808 0.4850 0.4884 0.4911 0.4932 0.4812 0.4854 0.4887 0.4913 0.4934 0.4817 0.4857 0.4890 0.4916 0.4936 0.4945 0.4946 0.4959 0.4960 0.4669 0.4970 0.4948 0.4961 0.4971 0.4949 0.4962 0.4972 0.4951 0.4963 0.4973 0.4952 0.4964 0.4974 2.5 0.4938 0.4940 2.6 0.4953 0.4955 2.7 0.4965 0.4966 0.4941 0.4943 0.4956 0.4957 0.4967 0.4968 179 2.8 0.4974 0.4975 2.9 0.4981 0.4982 3.0 0.4987 0.4987 0.4976 0.4977 0.4982 0.4983 0.4987 0.4988 0.4977 0.4978 0.4984 0.4984 0.4988 0.4989 0.4779 0.4985 0.4989 0.4979 0.4985 0.4989 0.4980 0.4986 0.4990 0.4981 0.4986 0.4990 2.4 SUMMARY ● Probability distributions are obtained primarily on theoretical considerations and describe how the outcomes of an experiment are expected to vary. ● A probability distribution may involve a discrete or a continuous random variable. ● The two basic measures to describe a probability distribution are expected value and standard deviation. For a discrete probability distribution, expected value, µ = Σpx and standard deviation, σ= Σp( x −x ) 2 . ● The binomial, and Poisson distributions involve discrete random variables while normal distribution involve continuous random variables. ● A binomial distribution involves n independent trials, each of which can result in only two possible outcomes, called success and failure. The probabilities of various number of successes are given by the binomial expansion (q + p)n. The binomial formula is used to calculate the probability of x successes in n trials. ● The mean and standard deviation of a binomial distribution are given by np and npq , respectively. The distribution is symmetrical if p = q and skewed if p ≠ q. For a given number of trials, the greater the difference between p and q, more the skewness. ● The Poisson distribution is normally used to analyse phenomena that produce rare occurrences. So, it is called the distribution of rare events. It is defined by a single parameter m, which is its mean value. Its mean and variance are equal. ● A Poisson distribution is positively skewed and skewness decreases as m increases. ● The Poisson distribution can also be used as approximation to binomial distribution when the number of trials is large and the probability of success in a trial is very small. ● The normal distribution is an all important probability distribution. It involves a continuous variable, has a curve that is unimodal. symmetrical and asymptotic to the x-axis. ● The normal distribution has two parameters – mean and standard deviation. For every pair of values, there is a distinct normal distribution. The distribution with mean µ = 0 and standard deviation σ = 1 is called standardised normal distribution. ● The proportion of area lying in a given interval to the total area under the normal curve gives the probability that the variable in question will take a value within that interval. ● To obtain probabilities, the given values are transformed into standard normal distribution. This is called z-transformation. It is defined as z = (x – µ)/σ. The normal area table is consulted to determine the relevant area. ● Normal distribution can be used as an approximation to different discrete probability distributions like binomial, and Poisson distributions under appropriate conditions. 180 2.5 SELF ASSESSMENT QUESTIONS Exercise 1 : True and False Statements (i) A variable is said to be a random variable if it takes different values as a result of the outcomes of a random experiment. (ii) A probability distribution can also be obtained using historical data. (iii) A discrete random variable can assume any value in a given range whereas a continuous random variable can assume only isolated values. (iv) The expected value, E(x), of a probability distribution is obtained by summation of the products of the values of x and their corresponding probabilities. (v) The two parameters of a binomial distribution are n and p. (vi) A binomial distribution involves infinite number of trials. (vii) A binomial distribution with n trials involves a total of n number of successes. (viii) A binomial distribution can be completely identified by n and p. (ix) If mean and standard deviation values are given, we can fit a binomial distribution. (x) The variance of a binomial distribution can be smaller than, equal to or greater than its mean value. (xi) Regardless of the value of n, a binomial distribution is symmetrical if p = q. (xii) A binomial distribution is positively skewed when p > q and negatively skewed when p < q. (xiii) A Poisson distribution is positively skewed and the skewness decreases when m increases. (xiv) The mean and standard deviation of a Poisson distribution are always equal. (xv) A binomial distribution with p = 0.5 results in a uniform distribution involving discrete variable. (xvi) Binomial approximation to Poisson distribution is used when n is very large and p is very small. (xvii) A normal distribution is defined by mean and standard deviation. (xviii) The standard normal distribution has µ = 1 and σ = 0. (xix) The smaller the standard deviation, the greater the height of the normal curve. (xx) For a normal distribution with µ = 110 and σ = 10, the area included between x = 120 and x = 130 is equal to the area between x = 90 and x = 100. (xxi) The curve of normal distribution is symmetrical and mesokurtic. (xxii) µ ± 3σ covers 99.27 percent area under the normal curve. (xxiii) In using normal approximation to the binomial distribution, the mean and standard deviation are taken to be equal to np and npq, respectively. Ans. 1. T, 2. T, 3. F, 4. T, 5. T, 6. F, 7. F, 8. T, 9. T, 10. F, 11. T, 12. F, 13. T, 14. F, 15. F, 16. F, 17. T, 18. F, 19. F, 20. T, 21. T, 22. F, 23. F 181 Exercise 2 : Questions and Answers (i) Explain the concept of probability distributions. Give two examples of how a probability binomial distribution can be obtained. (ii) Distinguish between discrete and continuous probability distributions. (iii) What are the conditions under which a binomial distribution is used? How are the probabilities calculated in case~of a binomial distribution? Discuss the conditions under which a binomial distribution can be approximated as (i) a normal distribution and (ii) a Poisson distribution. (iv) Write a note on the binomial distribution. In particular, mention its assumptions, its expected value and variance, and the shape. (v) Write a note on Poisson distribution. Under what conditions is it used as an approximation to the binomial distribution? (vi) State and explain the properties of a normal curve. Show that the height of a normal curve at mean is the highest. (vii) What are the parameters of (i) binomial distribution, (ii) Poisson distribution, (iii) hypergeometric distribution, and (iv) uniform distribution? (viii) Determine the probability of getting three heads in 6 tosses of a fair coin. (ix) The probability that a student will graduate next year is 0.4. Determine the probability that out of five students, each of which has the same chance of graduating, (i) none, (ii) one, (iii) at least one, (iv) no more than one, and (v) all will graduate. (x) The incidence of occupational disease in an industry is such that the workmen have a 30 percent chance of suffering from it. What is the probability that out of 8 workmen, 6 or more will contact the disease? (xi) A factory finds that, on an average, 20 percent bolts are defective from one machine. If 10 bolts are selected at random, find the probability that (a) Exactly two bolts are defective, (b) Less than two bolts are defective. (c) More than two bolts are defective. (d) More than eight bolts are defective. (xii) A firm produces a product and finds that 10 percent of its output is defective. A small sample of 5 items is taken from the production line. Find the probability of getting each of the following number of defective items in the sample: 0, 1, 2, 3, 4, and 5. (xiii) Find the probability of getting a 5 or 6 thrice in five tosses of an unbiased die. (xiv) A sign on the gas pumps of a chain of gasoline stations encourages customers to have their oil checked, claiming that one out of five cars needs to have oil added. If this is true, what is the probability of the following events ? (a) One of the next four cars needs oil. (b) Two out of the next eight cars need oil. (xv) The mean of a binomial distribution is 20 and its standard deviation is 4. Calculate the values of n. p and q. (xvi) Assuming the binomial distribution applies, find the expected value, variance, and standard deviation of the distribution determined by n = 80 and p = 0.6. 182 (xvii) Calculate the probabilities of 0, 1. 2, 3, 4 and 5 heads on the toss of a set of five balanced coins. Also, obtain the mean and standard deviation of the distribution. (xviii) Bring out the fallacy, if any, in the following statements: (a) The mean of a binomial distribution is 6 and its standard deviation is 3. (b) The mean of a binomial distribution is 3 and its variance is 4. (xix) For a binomial variable x, it is given that n = 8, and P(x – 2) = 16 P(x = 6). Determine the values of p and q. (xx) If the probability of a defective bolt is 0.2, find the mean and standard deviation of the number of defective bolts in a total of 900 such bolts. (xxi) (a) In a binomial distribution involving 5 independent trials, probabilities of 1 and 2 successes are 0.4096 and 0.2048, respectively. Find the parameter p of the distribution. (b) In a binomial distribution w|th 6 independent trials, the probabilities of 3 and 4 successes are found to be 0.2457 and 0.0819, respectively. Find mean and variance of this distribution. (xxii) The administrative officer of a nursing home reports that the number of patients admitted to the ICU on any day follows a Poisson probability law, with a mean of 7. What is the probability that on a given day (a) No patient will be admitted? (b) Exactly seven patients will be admitted? (c) No more than three patients will be admitted? (xxiii) Assume that the number of network errors experienced in a day on a local area network (LAN) is distributed as Poisson random variable. The average number of network errors experienced in a day is 2.4. What is the probability that in any given day (a) Exactly one network error will result? (b) Two or more network errors will result? (xxiv) Assume the mean height of soldiers to be 68.22 inches with variance of 10.8 inches, how many soldiers in a regiment of 1400 would you expect to be taller than 72 inches? (xxv) In an examination 10% of students passed with distinction, 60% passed and 30% failed. If it is given that a candidate needs 40% marks to pass and 75% marks to pass with distinction, determine mean and standard deviation of the distribution of marks assuming the marks are distributed normally. (xxvi) A project yields an average case flow of Rs. 550 lakh and standard deviation of Rs. 110 lakh. Compute the following probabilities : (a) Cash flow will be more than Rs. 675 lakh, (b) Cash flow will be less than Rs. 450 lakh and (c) Cash flow will be between Rs. 425 and Rs. 750 lakh. Ans. 8. 0.3125, 9. 0.0778, 0.2592, 0.92224, 0.3370, 0.01024, 10. 0.1129, 11. 0.302, 0.3758, 0.3222, 0.0000042, 12. 0.5905, 0.3281, 0.0729, 0.0881, 0.00045, 0.00001, 13. 0.165, 14. 0.4096, 0.1536, 15. 100, 0.2, 0.8, 16. 48, 19.2, 4.38 17. 1/32, 5/32, 10/32, 10/32, 5/32, 1/32, 2.5, 1.12, 18. q > 1 in each case 19. 1/3 and 2/3 20. 180, 12 21. 0.25, 24/13, 216/169, 22. 0.0009, 0.149, 0.0818, 23. 0.2177, 0.6916, 24. 175, 25. 50.17, 19.4, 26. 0.1271, 0.1814, 0.8385 183 LESSON-3 STATISTICAL DECISION THEORY 3. STRUCTURE 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 Objective Probability in Decision Making Decision Making Process Decision Under Uncertainty 3.3.1 Maximax Criterion 3.3.2 Maximin Criterion 3.3.3 Laplace Criterion 3.3.4 Minimax Regret Criterion Decision Under Risk 3.4.1 Expectation Criterion 3.4.2 Expected Opportunity Loss Expected Value of Perfect Information (EVPI) Decision Tree Summary Self Assessment Questions 3.0 OBJECTIVE After reading this lesson, you would be able to : (a) Understand the steps of decision-making process (b) Comprehnd the concepts of states of nature and courses of action and compute payoff and regret tables (c) Learn about various decision-making rules (d) Compute EVPI and take decisions with the help of Decision tree analysis 3.1 PROBABILITY IN DECISION MAKING Statistical techniques are being widely used to solve business problems. These techniques are being used to solve problems for which information is incomplete, uncertain or in some cases almost completely lacking. This new area of statistics is known as statistical decision theory. In decision theory we must decide among alternatives by taking into account the monetary considerations of our actions. A manager who wants to select from among a number of available investment alternatives should consider the profit or loss that might result from each alternative. Decision theory involves selecting an alternative and having a reasonable idea of the economic consequences of choosing that action. Decision theory may be applied to problems whether the time span is one day or five years, whether it involves financial management or a plant assembly line. Most of these problems have 184 common characteristics. The elements common to most decision theory problems are : (i) An objective, (ii) Several courses of actions, (iii) A calculable measure of the benefit or worth of various alternatives, (iv) Events beyond the control of the decision maker, and (v) Uncertainty about which outcome or state of nature will actually happen. Most complex managerial decisions are made with some uncertainty. Managers authorize substantial capital investments with incomplete knowledge about product demand. When decisions are made under uncertain future conditions, use of probabilities provides us with a rational technique for making choices. Example 1 : A bakery provides cakes at a cost of Rs. 6 and sells them for Rs. 10 each. A cake not sold on a particular day is worthless. The baker’s problem is to determine the optimum number of cakes to be made each day. On days when his stock is more than his sales his profits are reduced by the cost of unsold cakes. On days of demand exceeding his stock he loses sales and makes smaller profits than he could have. He has kept a record of his sales for past 100 days to tell him about the historical pattern of sales. Daily sales of cakes No of days sold 300 15 400 20 500 45 600 15 700 5 100 Solution : On the basis of above information we can assign probabilities of sale of cakes in different quantities. For example a probabiltiy of 0.45 is assigned to the sale figure of 500 cakes. The table assigning probabilities to various quantities can be prepared. Daily sales (events) No. of days sold Probabilities of each number being sold E1300 15 0.15 E2400 20 0.20 E3500 45 0.45 E4600 15 0.15 E5 700 5 0.05 100 1.00 We can make a table of net benefits that accrue to the decision maker from a given combination of Act and Event. This is known as pay off matrix. For the given example we can make the pay off table or matrix. When the baker sells one cake he get a profit of Rs. (10 – 6) = Rs. 4 and when a cake remains unsold he loses Rs. 6. The pay off matrix is given below : 185 Production (Decision maker’s alternative) A1 A2 A3 A4 A5 Units 300 400 500 600 700 E1 300 Rs. 1200 Rs. 600 E2 400 1200 E3 500 0 (600) (1200) £ 1600 * 1000 400 (200) £ 1200 1600 2000 * 1400 800 £ E4 600 1200 £ 1600 2000 2400 * 1800 E5 700 1200 £ 1600 2000 2400 2800 * We have calculated the profits or losses for various situations, when demand and production both match, there is maximum profit but when production exceeds demand profit reduces because the unsold quantity is thrown and even the cost not recovered e.g., when both production and demand is 300 units, there is a payoff (profit) of 300 × 4 = Rs. 1200. When demand is 400 but production is 300 units then also there is profit of 1200 because he can sell only 300 units and earn (300 × 4 = Rs. 1200). But when production is 400 units and demand 300 units, he earns 300 × 4 = Rs 1200 but looses 100 × 6 = Rs 600 on account of unsold quantity of 100 cakes and net pay off is just Rs. 600 i.e. (1200 – 600). When production is 500 units and sales 300 units he neither earns nor loses because by selling 300 units he makes a profit of 300 × 4 = Rs 1200. He loses Rs. 1200 cost of 200 units that remained unsold and the net result is zero profit. But when he produces 600 cakes and sells only 300 he incuss a loss of Rs. 600 [(300 × 4) – (300 × 6) = 1200 – 1800 = – 600]. 3.2 DECISION-MAKING PROCESS As indicated, the decision theory is used to determine optimal strategy where a decision-maker is faced with several decision alternatives and an uncertain pattern of future events. All decisionmaking situations are characterised by the fact that two or more alternative courses of action are available to the decision-maker to choose from. Further, a decision may be defined as the selection by the decision-maker of an act, considered to be best according to some pre-designated standard, from among the available options. The decision-making process uses the following steps: (a) Identification of the various possible outcomes, called states of nature or events. The events are beyond the control of the decision-maker. (b) Identification of all the courses of action, Aj, or the strategies which are available to the decision-maker. The decision-maker has control over choice of these. (c) Determination of the pay-off function which describes the consequences resulting from different combinations of the acts and events. (d) Choosing from among the various alternatives on the basis of some criterion, which may involve the information given in step (c) only or some additional information. Example 2 : A toy making company is bringing out a new type of toy. It is considering whether to 186 bring out a full, partial or minimal product line. The company has three levels of product acceptance. The management will make its decision on the basis of anticipated profit from the first year of production. The relevant data are given in the table below : Anticipated Profit Product Product Line Acceptance Full Partial Minimal Good Rs. 80,000 Rs. 70,000 Rs. 50,000 Fair Rs. 50,000 Rs. 45,000 Rs. 40,000 Poor Rs. –25,000 Rs. –10,000 Rs. 0 Solution : Take optimal decision using various decision criteria. Let us first analyse the given information. States of nature : The states of nature or events here are in terms of the product acceptance which may be good, fair or poor. The management has no control over this aspect. Courses of actions : The management has to accept one of the product lines: full, partial or minimal. These are the choices available out of which one has to be adopted. The courses of action are also called simply acts or strategies. The decision-maker has a control over these. Pay-off table: A pay-off table explains the economics of the given problem. A pay-off is a conditional value or profit/loss, or a conditional cost. It is conditional in the sense that a certain profit/loss is associated with each course of action. Thus, the profit or loss resulting from the adoption of a certain strategy is dependent upon particular event that may occur. To illustrate, if the management goes in for full product line and the product acceptance is good, the company would make a profit of Rs. 80,000. The given pay-off values are shown in Table 1 with additional information. Regret or Opportunity Loss Table: The outcomes of various combinations of acts and events can also be expressed in terms of regret or opportunity loss values. The regret values are also conditional as each one of them results from a certain combination of act and event. Regret is defined as the amount of pay-off foregone by not adopting the optimal course of action – that which would yield the highest pay-off, for each given event. To illustrate, in the context of our example, in the event of good product acceptance, the full product line would yield a profit of Rs. 80,000; a partial product line would result in a profit of Rs. 70,000; while a minimal product line would give a profit of Rs. 50,000. The optimal course of action in this case would be full product line. Thus, at the end of the year, if the product acceptance was good, the management would not regret if it had gone for full product line. In case it had decided for partial product line, it would regret somewhat while a decision of minimal product line would cause a greater regret. The regret would be Rs. 10,000 (= 80,000 – 70,000) and Rs. 30,000 (= 80,000 – 50,000), respectively, with these two policies. 187 Table 1 : Decision-making Using Different Criteria Anticipated Profit (in Rs.) Act Event Full Product Line Partial Product Line Minimal Product Line Good Product Acceptance 80,000 70,000 50,000 Fair Product Acceptance 50,000 45,000 40,000 Poor Product Acceptance –25,000 –10,000 0 Maximum 80,000 70,000 50,000 Minimum –25,000 –10,000 0 35,000 35,000 30,000 Average Table 2 : Conditional Regret Table Act Event Full Product Line Partial Product Line Minimal Product Line Good Product Acceptance 0 10,000 30,000 Fair Product Acceptance 0 5,000 10,000 Poor Product Acceptance 25,000 10,000 0 The elements of a decision process are: A decision-maker. A set of possible outcomes, or events, in the decision situation. A set of courses of action available to the decision-maker. A set of conditional pay-offs corresponding to various possible combinations of events and actions. Selection of a particular course of action based on some criterion. After setting up the pay-off table and the regret table we proceed to take a decision. There are several rules, or criteria, on the basis of which decision may be taken.The selection of an appropriate criterion depends on factors such as the nature of decision situation, attitude of the decision-maker etc. We shall first discuss the decision rules for taking decisions in conditions of uncertainty and then for conditions of risk. 3.3 DECISIONS UNDER UNCERTAINTY The decision situations where there is no way in which the decision-maker can assess the probabilities of the various states of nature are called decisions under uncertainty. In such situations, the decisionmaker has no idea as to which of the possible states of nature would occur nor has he a reason to believe why a given state is more, or less, likely to occur as another. With probabilities of the various outcomes not known, the actual decisions are based on specific criteria. The several principles which may be employed for taking decisions in such conditions are discussed below. 188 3.3.1 Maximax Criterion An optimist believes that whatever course of action he chooses from all possible outcomes would be the best. This rule suggests, that for each strategy, the maximum pay-off should be considered. Further, the maximum of these pay-offs should be chosen for decision. For our example, the maximum profit associated with different courses of action is as follows: Full product line: Rs. 80,000; Partial product line: Rs. 70,000; and Minimal product line: Rs. 50,000. The highest of these corresponds to the full product line. Therefore, optimal decision is to go for full product line. 3.3.2 Maximin Criterion This principle is adopted by pessimistic decision-makers who are conservative in their approach. A pessimist is one who believes that whatever course of action he chooses would be the worst out of all the possible outcomes. Using this approach, the minimum pay-offs from various strategies are considered and the maximum one is selected. It therefore chooses the best (the maximum) profit from the set of worst (the minimum) profits. The minimum pay-off associated with the full, partial and minimal product line is Rs. (–) 25,000, Rs. (–) 10,000 and Rs. 0, respectively. According to this criterion the management would go for minimal product line. 3.3.3 Laplace Criterion The Laplace principle is based on the simple rule that if we are uncertain about various events, then we may treat them as equally probable. Therefore the expected value of pay-off for each strategy is calculated and the strategy with the highest mean value is adopted. The expected pay-offs for various courses of action are calculated as : Full product line : (80,000 + 50,000 – 25,000)/3 = Rs. 35,000 Partial product line : (70,000 + 45,000 – 10,000)/3 = Rs. 35,000 Minimal product line : (50,000 + 40,000 + 0)/3 = Rs. 30,000 Since the highest expected pay-off is shared by the strategies of full product line and partial product line, both could be adopted by the management. 3.3.4 Minimax Regret Criterion The Minimax Regret principle is based on the concept of regret. According to the principle the course of action that minimises the maximum regret will be selected. It is known as savage principle. First the regret matrix is derived from the pay-off matrix. Then the maximum regret value corresponding to each of the strategies is determined and the strategy which minimises the maximum regret is chosen. The principle of choice is also conservative in approach and is very close to the minimax principle applied to the original matrix containing pay-off values. From the regret matrix given in Table 2 we get the following maximum regret values associated with the various couses of action : 189 Full product line : Rs. 25,000 Partial product line : Rs. 10,000 Minimal product line : Rs. 30,000 The maximum regret value for the strategy of partial product line is minimum and will be the optimal choice. When probabilities of various events are not given, the decision criteria could be: 1. Considering best of the best pay-off for every act. 2. Considering best of the worst pay-off for every act. 3. Considering best of the weighted average of the best and worst pay-offs under every act. 4. Considering best of the simple average of all pay-offs under every act. 5. Considering act with least of maximum regret values associated with all acts. 3.4 DECISION UNDER RISK The decision situations wherein the decision-maker chooses to consider several possible outcomes and the probabilities of their occurrence can be stated are called decisions under risk. The probabilities of various outcomes may be given or they may be determined from the past records. Under conditions of risk, there are generally two criteria to choose from. (a) Expectation Criterion (b) Expected opportunity loss or expected regret 3.4.1 Expectation Criterion Decision making in situations of risk is on the basis of the expectation principle with the event probabilities assigned. The expected pay off for each strategy is calculated by multiplying the pay off values with their respective probabilities and then adding up these products. The strategy with n the highest expected pay off represents the optimal choice. Symbolically, Σ Pi = pi aij where aij ⇒ i =1 the pay off resulting from the combinations of ith event and jth act. pi represents the probability of ith event. In the example, we are given probabilities of various events and we can determine the expected pay offs. Events Probabilities A1 A2 A3 A4 A5 (Pi) 300 400 500 600 700 E1 300 0.15 1200 600 0 –600 –1200 E2 400 0.20 1200 1600 1000 400 200 E3 500 0.45 1200 1600 2000 1400 800 E4 600 0.15 1200 1600 2000 2400 1800 E5 700 0.05 1200 1600 2000 2400 2800 190 The calculation of expected pay off values EP for various acts is shown below :for A1, EP1 = 0.15 × 1200 + 0.20 × 1200 + 0.45 × 1200 + 0.15 × 1200 + 0.05 × 1200 = Rs. 1200 for A2, EP2 = 0.15 × 600 + 0.20 × 1600 + 0.45 × 1600 + 0.15 × 1600 + 0.05 × 1600 = Rs. 1450 for A3, EP3 = 0.15 × 0 + 0.20 × 1000 + 0.45 × 2000 + 0.15 × 2000 + 0.05 × 2000 = Rs. 1500 for A4, EP4 = 0.15 × –600 + 0.20 × 400 + 0.45 × 1400 + 0.15 × 2400 + 0.05 × 2400 = Rs. 1100 for A5, EP5 = 0.15 × –1200 + 0.20 × –200 + 0.45 × 800 + 0.15 × 1800 + 0.05 × 2800 = Rs. 550 Since maximum expected pay off is associated with strategy A3, the best course of action is to produce 500 cakes. 3.4.2 Expected Opportunity Loss The expected opportunity loss or expected regret criterion is another basis on which a decision may be taken. For this purpose opportunity loss matrix alongwith probability distribution can be reproduced as given below: Calculation of expected Regret: Acts Event Probability A1 A2 A3 A4 A5 (Pi) 300 400 500 600 700 E1 300 0.15 0 600 1200 1800 2400 E2 400 0.15 400 0 600 1200 1800 E3 500 0.15 800 400 0 600 1200 E4 600 0.15 1200 800 400 0 600 E5 700 0.15 1600 1200 800 400 0 Now we can determine expected regret (ER) for various strategies for A1, ER1 = 0.15 × 0 + 0.20 × 400 + 0.45 × 800 + 0.15 × 1200 + 0.05 × 1600 = Rs. 4700 for A2, ER2 = 0.15 × 600 + 0.20 × 0 + 0.45 × 400 + 0.15 × 800 + 0.05 × 1200 = Rs. 450 for A3, ER3 = 0.15 × 1200 + 0.20 × 600 + 0.45 × 0 + 0.15 × 400 + 0.05 × 800 = Rs. 400 for A4, ER4 = 0.15 × 1800 + 0.20 × 1200 + 0.45 × 600 + 0.15 × 0 + 0.05 × 400 = Rs. 800 for A5, ER5 = 0.15 × 2400 + 0.20 × 1800 + 0.45 × 1200 + 0.15 × 600 + 0.05 × 0 = Rs. 1350 Under this criterion, the optimal strategy is the one which minimizes the expected regret. Since the minimum value occurs at A3 it represents the optimal decision. This is same as under expected pay off criterion. 3.5 EXPECTED VALUE OF PERFECT INFORMATION (EVPI) Assuming that we can obtain a perfect prediction about the future and also the cost of this information, we can compare the cost of that information with the additional profit we would realize as a result of having the information. The difference between expected cost with optimal policy and expected cost with perfect information is known as EPVI. 191 In our example if the bakery shop knows that next year’s demand is going to be 300 units with probability 0.15 and a profit of Rs. 1200. We calculate expected profit 1200 × 0.15 = Rs. 180. When the shop manager knows that demand will be 400 units he will earn Rs. 1600 and since the probability of this event is 0.20 his expected profit will be 0.20 × 1600 = Rs. 320. Similarly the expected pay off for each level of demand can be obtained and aggregated. These expected profits we get are EVPI. EVPI = 0.15 × 1200 + 0.20 × 1600 + 0.45 × 2000 + 0.15 × 2400 + 0.05 × 2800 = Rs. 1900 Alternatively, EVPI = EPj + ERj For A1, EP1 = 1200 and ER1 = 700 and EVPI = 1200 + 700 = Rs. 1900 For A2, EP2 = 1450 and ER2 = 450 and EVPI = 1450 + 450 = Rs. 1900 For A3, EP3 = 1500 and ER3 = 400 and EVPI = 1500 + 400 = 1900 and similarly for A4 and A5. Example 3 : A grocery shop is faced with the problem of how many cakes to buy in order to meet the day’s demand. The left over cakes are a total loss. If the customer’s demand is not satisfied, the sales will be lost. The shopkeeper has got the information regarding past sales for past 200 days : Sales per day No. of days Probability 25 20 0.10 26 60 0.30 27 100 0.50 28 20 0.10 (i) Prepare the payoff matrix and opportunity loss (regret) matrix. (ii) Find the optimal number of cakes that should be bought each day. (iii) Find EVPI. The cost of a cake is Rs. 8 and it is sold for Rs. 10 each. Solution : (i) Profit = (Cakes sold × selling price) – (Cakes unsold × cost price) Opportunity loss (regret) = Maximum profit in a row – Profit under each column in that row. Pay off Table (Rs.) 25 26 27 28 Probability 25 50 42 34 26 0.10 26 50 52 44 36 0.30 27 50 52 54 46 0.50 28 50 52 54 56 0.10 EMV 50 51 49 42 192 Regret Table (Rs.) 25 26 27 28 Probability 25 0 8 16 24 0.10 26 2 0 8 16 0.30 27 4 2 0 8 0.50 28 6 4 2 0 0.10 EOL 3.20 2.20 4.20 11.20 (ii) Now we can calculate expected monetary value and expected opportunity loss EMV = Σ (Profit Column × Probability column) EOL = Σ (Regret Column × Probability column) Max EMV is 52 corresponding to 26 cakes and min. EOL is 2.20 corresponding to 26 cakes ∴ optimum number of cakes to be purchased is 26 (iii) EVPI= (Max. Pay off in each row × Corresponding Probability) – Max. EMV = (50 × .10 + 52 ×.30 + 54 × .50 + 56 × .10) – 51 = (5 + 15.60 + 27 + 5.60) – 51 = 53.20 – 51 = Rs. 2.20 Example 4 : The payoff (in Rs.) of three acts A1, A 2 and A3 and the possible states of nature S1, S2 and S3 are given below: Acts States of Nature A1 A2 A3 S1 –20 –50 200 S2 200 –100 –50 S3 400 600 300 The probabilities of the states of nature are 0.3, 0.4 and 0.3 respectively. Determine the optimal act using the expectation principle. Solution : Acts States of Nature A1 A2 A3 Probabilities S1 –20 –50 200 0.3 S2 200 –100 –50 0.4 S3 400 600 300 0.3 Expected profit for Act A1 = – 20 × 0.3 + 200 × 0.4 + 400 × 0.3 = – 6 + 80 + 120 = Rs. 194 Expected profit for Act A2 = – 50 × 0.3 – 100 × 0.4 + 600 × 0.3 = – 15 – 40 + 180 = Rs. 125 193 Expected profit for Act A3 = 200 × 0.3 – 50 × 0.4 + 300 × 0.3 = 60 – 20 + 90 = Rs. 130 ∴ Act A1 is the Optimal Act. Example 5 : Each unit of a product produced and sold yields a profit of Rs. 50 but a unit produced but not sold results in a loss of Rs. 30. The probability distribution of the number of units demanded is as follows : No. of Units Demanded Probability 0 0.20 1 0.20 2 0.25 3 0.30 4 0.05 How many units be produced to maximise the expected profits? Also calculate EVPI. Solution : Given : Profit for units produced and sold = Rs. 50 Loss for units produced and not sold = Rs. 30 Pay Off Table (Production) Demand Probability 0 EMV 1 0 1 2 3 4 0.2 0.2 0.25 0.2 0.05 0 0 0 0 0 0 0 0 0 0 Total (30) 50 50 50 50 0 EMV 2 EMV 3 (6) (60) (12) (90) 10 20 4 (10) 12.50 100 17.50 70 15 100 45 150 2.50 100 7.50 150 34 EMV 4 EMV (18) (2) 17.50 45 7.50 (120) (40) 40 120 120 (24) (8) 10 36 10 52 50 42 Note : Values in brackets are negative. We should produce 2 units because EMV = Rs. 52 (maximum). Further, EVPI = EPPI – EMV EPPI Table Demand Probability Max. pay off 0 0.2 0 0 × 0.2 = 0 1 0.2 50 50 × 0.2 = 10 2 0.25 100 100 × 0.25 = 25 3 0.30 150 150 × 0.30 = 45 4 0.05 200 200 × 0.05 = 10 Total EPPI 90 ∴ EVPI = 90 – 52 = Rs. 38 194 Example 6 : A physician purchases a particular vaccine on Monday of each week. The vaccine must be used within the week following, otherwise it becomes worthless. The vaccine costs Rs. 2 per dose and the physician charges Rs. 4 per dose. In the post 50 weeks, the physician has administered the vaccine in the following quantities: Doses per Week Number of Weeks 20 5 25 15 50 25 60 5 On the basis of EMV, find how many doses the physician must purchase each week to maximise his profit ? Solution : Given : Cost = Rs. 2.00, Price = Rs. 4.00 Profit = Rs. 4 – Rs. 2 = Rs. 2.00 per dose We shall calculate pay off for different actions : P11 (Demand/Dose = 20) P12 P13 P14 P21 P22 P23 P24 P31 P32 P33 P34 P41 P42 P43 P44 = = = = = = = = = = = = = = = = 20 × 2 = Rs. 40 40 – 10 = 30 40 – 60 = –20 40 – 80 = – 40 40 – 0 = 40 50 – 0 = 50 50 – 50 = 0 50 – 70 = –20 40 – 0 = 40 50 = 50 50 × 2 = 100 100 – 20 = 80 20 × 2 = 40 25 × 2 = 50 50 × 2 = 100 60 × 2 = 120 Pay Off Table Probability Doses per week A1 A2 A3 A4 Demand 20 25 50 60 (5/50) = 0.1 20 40 30 –20 –40 (15/50) = 0.3 25 40 50 0 –20 (25/50) = 0.5 50 40 50 100 80 (5/50) = 0.1 60 40 50 100 120 195 n ∴ EMV for A1 = ∑ pi xi x =1 = (0.1 × 40) + (0.3 × 40) + (0.5 × 40) + (0.1 × 40) = 4 + 12 + 20 + 4 = Rs. 40 EMV for A2 = (0.1 × 30) + (0.3 × 50) + (0.5 × 50) + (0.1 × 50) = 3 + 15 + 25 + 5 = Rs. 48 EMV for A3 = (0.1 × –20) + (0.3 × 0) + (0.5 × 100) + (0.1 × 100) = – 2 + 0 + 50 + 10 = Rs. 58 EMV for A4 = (0.1 × –40) + (0.3 × –20) + (0.5 × 80) + (0.1 × 120) = – 4 – 6 + 40 + 12 = Rs. 42 The physician should purchase 50 doses each week because EMV for A3 is Rs. 58 (Maximum). 3.6 DECISION TREE It is quite useful to represent the structure of a decision problem under uncertainty by a ‘decision tree diagram’ or by “decision tree.” A decision tree is a graphic representation of decision process. We can introduce probabilities into the analysis of complex decisions involving (i) many alternatives, and (ii) future conditions that are not known but can be specified in terms of a set of probabilities. The decision tree analysis helps in making decision concerning a wide variety of problems such as project management, personnel, new product strategies, acquisition or disposal of physical properties, investments, etc. Decision trees have standard symbols, squares symbolize decision points and circles represent chance events. From each square and circle branches are drawn. These represent each possible outcome or state of nature. Steps in Decision Tree Analysis In a decision tree analysis, the decision-maker follows the following six steps* : 1. Define the Problem in Structured Terms. First of all, the factors relevant to the solution should be determined. Then probability distributions that are appropriate to describe future behaviour of those factors are estimated. 2. Model the Decision Process. A decision tree that illustrates all the alternatives in a problem is constructed. The entire decision process is presented in an organised step-by-step procedure. 3. Apply the Appropriate Probability Values and Financial Data. To each of the branches and sub-branches of the decision tree the appropriate probability values and financial data are applied. This will help us to distinguish between the probability value and conditional monetary value associated with each outcome. 4. “Solve” the Decision Tree. Using the method explained above locate that particular branch of the tree that has the largest expected value or that maximises the decision criteria. 5. Perform Sensitivity Analysis. Determine how the solution reacts to changes in inputs. Changing probability value and conditional financial values, enables the decision maker to test the magnitude and the direction of the reaction. 196 6. List the Underlying Assumptions. The accounting, cost finding and other assumptions used to arrive at a function should be explained. This will help others to know what risks they are taking when they use the results of decision tree analysis. Advantage of the Decision Tree Approach The decision tree analysis is important because of the following: 1. It structures the decision process enabling decisions to be made in an orderly manner. 2. It requires the decision-maker to examine all possible outcomes – desirable and undesirable. 3. It communicates the decision-making process to others. 4. It allows a group to discuss alternatives by focusing on each financial figure, probability value and underlying assumptions – one at a time to arrive at a consensus decision instead of debating that decision entirely. 5. It can be used with a computer so that many different sets of assumptions can be simulated and their effects on the final outcomes can be analysed. Example 7 : There is 40% chance that a patient admitted to hospital is suffering from cancer. A doctor has to decide whether an operation should be performed or not. If the patient is suffering from cancer and the serious operation is performed, the chance that he will recover is 70%, otherwise it is 35%. On the other hand, if the patient is not suffering from cancer and the operation is performed, the chance that he will recover is 20% otherwise it is 100%. Assume that recovery and death are the only two possible outcomes. What decision should the doctor take? Solution : The chance of recovery after operation = 0.28 + 0.12 = 0.40 The chance of recovery without operation = 0.14 + 0.60 = 0.74 Since, the chances of recovery without operation > chances of recovery with operation Therefore, the doctors should not undertake operation. Example 8 : A businessman has two independent investments A and B available to him but he lacks the capital to undertake both of them simultaneously. Investment A requires capital of Rs. 30,000 and investment B Rs. 50,000. Market survey shows: high, medium and low demands with corresponding probabilities of 0.4, 0.4 and 0.2 respectively in case of investment A and 0.3, 0.4 and 0.3 for investment B. Returns from investment A are Rs. 75,000, Rs. 55,000 and Rs. 35,000 and corresponding figures for investment B are likely to be Rs. 100,000, Rs. 80,000 and Rs. 70,000 for 197 high, medium and low demand respectively. What decision should the company take? Decide by constructing an appropriate decision-tree. Solution : Expected net profit for investment A = 0.4 × 75,000 + 0.4 × 55,000 + 0.2 × 35,000 – 30,000 = Rs. 29,000 Expected net profit for investment B = 0.3 × 100,000 + 0.4 × 80,000 + 0.3 × 70,000 – 50,000 = Rs. 33,000 Since the expected net profit for investment B is more than A, the businessman should invest in B. Example 9 : A person has two independent investments A and B available to him, but he can undertake only one at a time due to certain constraints. He can choose A first and then stop, or if A is successful, then take B or vice-versa. The probability of success of A is 0.6 while for B it is 0.4. Both investments require an initial capital outlay of Rs. 10,000 and both return nothing if the venture is unsuccessful. Successful completion of A will return Rs. 20,000 (over cost) and successful completion of B will return Rs. 24,000 (over cost). Draw a decision-tree and determine the best strategy. 198 EVALUATION OF DECISION POINTS Decision point D3 (i) Accept A Outcome Probability Conditional values Expected values Success 0.6 20,000 12,000 Failure 0.4 –10,000 –4,000 8,000 (ii) Stop D2 (i) Accept B — — — — Success 0.4 24,000 9,600 Failure 0.6 –10,000 –6,000 3,600 (ii) Stop D1 (i) Accept A, then B — — — — Success 0.6 20,000 + 3,600 14,160 Failure 0.4 –10,000 –4,000 10,160 (ii) Accept B, Success 0.4 24,000 + 8,000 12,800 then A Failure 0.6 –10,000 –6,000 6,800 EMV is highest when he accepts investment A and on success invests money in B. 3.7 SUMMARY ● Decision theory is concerned with decision-making under conditions of uncertainty. The decision process involves the steps of identification of states of nature, courses of action and pay-off table depicting consequences resulting from their interaction, and then choosing the appropriate action on the basis of given criterion, pay-offs, the outcome of event-action combinations can also be expressed in terms of regret. ● Among the rules for taking decisions are (a) maximax/minimin: where we select best of the best; (b) maximin/minimax: selecting best of the worst; (c) Hurwicz criterion: choosing the best using weighted average of the best and the worst; (d) Laplace principle: selecting the act with best average; and (e) Savage principle: involving selection of act with least of the maximum regret values. ● Where probabilities are used, the decision rules include (a) maximum likelihood principle where the best act of the most probable event is selected; (b) expectation principle: in which the act with the best expected value is chosen; and (c) expected regret criterion, where the act with least expected value is selected, leading to identical decision as under expectation principle. ● In addition to the optimal course of action, expected value of perfect information, EVPI, is also calculated. It equals the expected regret value of the optimal act and represents the maximum amount that a decision-maker would be willing to pay in case he is provided with perfect information. 199 ● The decision analysis using expectation principle can be extended to cover situations where given probabilities of various states of nature are revised on the basis of some additional information about them (the states). The prior probabilities are converted into posterior probabilities using the conditional, joint and total probabilities. Such an analysis is called posterior analysis. ● The decision-tree approach to decision-making is used in situations where multi-stage decisions are needed. The sequences of action-event combinations available to the decision-maker are presented graphically in the form of a decision tree. In analysing such situations, alternatives are evaluated by proceeding in a backward manner – by evaluating the best course at later stages to decide the best action at the earlier stages. The decision criterion is expected monetary value, EMV. 3.8 SELF ASSESSMENT QUESTIONS Exercise 1 : True and False Statements (i) Decision theory is concerned with determining optimal strategies where a decision-maker is faced with a number of alternatives and a risky pattern of future events. (ii) Regret is the amount of money paid for not adopting the optimal course of action. (iii) Laplace principle is based on the premise of equal chances of occurrence of possible events. (iv) Minimax is an optimist’s choice while Minimjn is a pessimist’s criterion. (v) In Hurwicz criterion, the maximum pay-off is multiplied by α, and the minimum by 1– α, where α may be any number between zero and 100. (vi) In situations of decisions under uncertainty, the Laplace criterion is the least conservative while the minimax criterion is the most conservative. (vii) In case of pay-offs represented as profits, the Savage criterion for selecting optimal course of action will be based on the maximin principle. (viii) Expected pay-off and expected regret criteria would both lead to indentical decisions in a given case. (ix) All criteria for decisions under risk inherently assume that the particular (optimal) decision reached will be repeated a large number of times. (x) EPPI is the expected pay-off under certainty. (xi) EVPI is the expected regret value of any strategy. (xii) In Bayesian approach to decision making, the optimal strategy is determined using posterior probabilities. (xiii) Posterior probabilities are obtained by modifying prior probabilities taking into consideration the information from a sample. (xiv) The decision tree approach to decision-making is appropriate in those situations where a sequence of decisions is involved. (xv) In the decision trees, no more than two alternative courses of action can emanate from a decision node. 200 (xvi) In decision trees, the probabilities of all events at chance nodes and the monetary evaluations of different alternatives must all be known in advance. (xvii) The probabilities of various outcomes at each chance node should always add up to one. (xviii) A decision taken on the basis of expected monetary value would always prove to be the right decision. Ans. 1. T, 2. F, 3. T, 4. F, 5. F, 6. T, 7. F, 8. T, 9. T, 10. T, 11. F, 12. T, 13. T, 14. T, 15. F, 16. T, 17. F Exercise 2 : Questions and Answers (i) Describe the steps involved in the process of decision-making. (ii) What are pay-off and regret functions? How can entries in a regret table be derived from a pay-off table? (iii) Explain and illustrate the following principles of decision-making: (a) Laplace, (b) Maximax, (c) Maximin, (d) Hurwicz, and (e) Savage. (iv) How are maximum likelihood and expectation principles of choice differentiated? Do they always lead to same decisions? (v) Define the term EPPI. How is it calculated? What does it signify? (vi) What do you understand by EVPI? How is it calculated? (vii) Explain the procedure of analysing a decision tree. (viii) The research department of Hindustan Lever has recommended the marketing department to launch a shampoo of three different types. The marketing manager has to decide one of the types of shampoo to be launched under the following estimated pay-offs (in millions of Rs.) for various levels of sales: Type of Shampoo Estimated level of sale (Units) 15,000 10,000 5,000 Egg shampoo 30 10 10 Clinic shampoo 40 15 5 Deluxe shampoo 55 20 3 What will be the marketing manager’s decision if (a) Maximin, (b) Maximax, (c) Laplace, and (d) Minimax regret criterion is applied? (ix) A food products company is contemplating the introduction of a revolutionary new product with new packaging to replace the existing product at much higher price (S1) or a moderate change in the composition of the existing product with a new packaging at a small increase in price (S2) or a small change in the composition of the existing product except the word ‘new’ with negligible increase in price (S3). The three possible states of nature are: (i) high increase in sales (N1), (ii) No change in sales (N2) and (iii) decrease in sales (N3). The marketing department of the company worked out pay-offs in terms of yearly net profits for each of the three strategies of these events. 201 Pay-offs in Rs. Strategies States of Nature N1 N2 N3 S1 7,00,000 3,00,000 1,50,000 S2 5,00,000 4,50,000 0 S3 3,00,000 3,00,000 3,00,000 Which strategy should the executive concerned choose on the basis of: (a) Maximin criterion; (b) Maximax criterion; (c) Minimax criterion; and (d) Laplace criterion? (x) The oil company of India is interested in acquiring a piece of land which is considered likely to contain oil deposits. The company has the option of (a) buying the land outright, (b) obtaining an option to buy, drill for oil and if found exercise the option, and (c) not buying or obtaining option. There are three possibilities on such land: large oil reserves may be found; minor reserves may be found, or there may be no oil. The pay-offs (in lacs of Rs. resulting from various combinations of acts and events are tabulated below: Acts Buy land Obtain option No action Large Reserves 40 28 0 Minor Reserves 10 1 0 No Oil –25 –2 0 What action should be taken by the company when the decision criterion is: (a) Laplace, (b) Maximin, (c) Maximax. (d) Minimax Regret, and (e) Expected pay-off (when the probabilities of obtaining large, minor, and no reserves are estimated to be 0.2, 0.5 and 0.3, respectively)? (xi) A firm is considering the purchase of some complex equipment from either of the two suppliers S1 and S2. Supplier S1 is capable of supplying the equipment on time to meet a certain desired deadline. The price chargeable by S1 is, however, considerably higher than that of S2. It is felt by the management of the firm that S 2 may deliver the equipment or may not be able to deliver on time. It is even suspected that supplier S 2 may never be able to deliver the equipment to the specifications. However, the management believes that if it waits for some months, it may get better information on S2’s capabilities of supplying the equipment. The management is considering three alternative courses of action. A1 : Order from S1. If later on it is clear that S2 can supply, order from S1 can be cancelled. Of course, delay would be caused when the order is given to S2. A2: Order from supplier S2. If it is known later on that S2 cannot supply the equipment, the order may be switched to S1. 202 A3: Wait till the time information on S′2s capabilities is known. This would obviously cause delay. The outcomes (profits) in the various possible situations are: Event Course of action A1 A2 A3 E1 250 100 200 E2 250 125 300 E3 250 625 450 E1 : S2 fails to deliver E2 : S2 delivers late E3 : S2 delivers on time What would be the management’s decision according to each of the following criteria: (a) Laplace, (b) Maximin, (c) Hurwicz (with α = 0.5), and (d) Minimax Regret. (xii) An investor is given the following investment alternatives and~percentage rates of return: States of Nature (Market Conditions) Low Medium High Regular shares 2% 5% 8% Risky shares –5% 7% 15% Property –10% 10% 20% Over the past 300 days, 150 days have been medium market conditions and 60 days have had high market increases. On the basis of these data, state the optimal investment strategy for the investor. (xiii) The manager of a small departmental store must place order every week for an item which costs Rs. 15 and sells for Rs. 25. Units not sold during the week are disposed of for Rs. 10 each. The demand for this item is estimated to be as follows: Demand (Units): 25 Probability: 0.10 26 27 28 29 30 0.15 0.30 0.20 0.15 0.10 (a) Determine the optimal number of units to be produced, using the expected monetary value criterion. (b) Determine the expected value of perfect information. (xiv) Three types of souvenirs can be sold outside a stadium. From the following conditional pay-off table, construct the opportunity loss table. (Sales are dependent on the winning team.) Types of Souvenir I (Rs.) II (Rs.) III (Rs.) Team A wins 1,200 800 300 Team B wins 250 700 1,100 203 Point out which type of souvenir should be bought if probability of Team A’s winning is 0.6. (xv) Chemical Products Ltd. produces a compound which must be sold within the month it is produced, if the normal price of Rs. 100 per drum is to be obtained. Anything unsold in that month is sold in a different market for Rs. 20 per drum. The variable cost is Rs. 55 per drum. During the last three years, monthly demand was recorded and showed the following frequencies: Monthly demand (No. of drums): 2,000 3,000 6,000 8 16 12 Frequency (No. of months) : (a) Prepare an appropriate pay-off table. (b) Advise the production management on the number of drums that should be produced next month. (xvi) A stockist of a particular commodity makes a profit of Rs. 30 on each sale made within the same week of purchase; otherwise he incurs a loss of Rs. 30 on each item. No. of items sold within the same week : 5 6 7 8 9 10 11 Frequency 0 9 12 24 9 6 0 : (a) Find out the optimum number of items the stockist should buy every week in order to maximize the profit. (b) Calculate the expected value of perfect information. (xvii) A physician purchases a particular vaccine on Monday of each week. The vaccine must be used within the week following, otherwise it becomes worthless. The vaccine costs Rs. 20 per dose and the physician charges Rs. 60 per dose. In the past 50 weeks, the physician has administered the vaccine in the following quantities: Doses per week : 20 25 40 60 No. of weeks 5 15 25 5 : (a) Draw up a pay-off matrix. (b) Obtain a regret matrix. (c) Determine the optimum number of doses the physician should buy. (d) The maximum amount the physician would be willing to pay per week for perfect information about the number of doses expected to be demanded in a week. Ans. 8. Egg, Deluxe, Deluxe, Deluxe, 9. S3, S1, S1, S1, 10. Obtain option, No action, Buy Land, obtain option, buy land or obtain option 11. A3 A1 A2 A2 A2 or A3, 12. Property, Exp. Return = 6% 13. (i) (ii), 14. Type I ER = 340, 15. 3000 drums 16. 8 units, EP = 210, EVPI = 25.50, 17. 40, EMV = 1210, EVPI = 1210. ❑❑❑ 204