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TRIMESTER – 5 MID-TERM QUESTION PAPER QUANTATIVE TECHNIQUES TIME ALLOWED : 2 hours Maximum Marks: 50 Attempt all questions : (It is an open book test. Please show complete working notes) Q-1) Construct a sub-divided bar chart for the 3 types of expenditures in dollars of a 10 marks family of four for the years 2002, 2003, 2004 and 2005 as given below: Year 2002 2003 2004 2005 Expenditure Food 3000 3500 4000 5000 Education 2000 3000 3500 5000 Other 3000 4000 5000 6000 Total 8000 10500 12500 16000 OR Construct “Less than ogive” and “More than ogive” 10 marks Class Interval Frequency Cumulative Frequency Less than Cumulative Frequency More than 15 Upton 25 25 “ 35 35 “ 45 45 “ 55 55 “ 65 65 “ 75 5 3 7 5 3 7 5 8 15 20 23 30 30 25 22 15 10 7 OR Write the difference between Bar diagram and Histogram Ans: Bar Diagram : It is very commonly used and is better for representation of qualitative data Bars are simply vertical lines, where the lengths of the bars are proportional to there corresponding numerical values. The width of the bar is un important but all bars should have the same width so as not to confuse the reader of the diagram. Additionally the bar should be equally spaced. With revenue represented by the vertical axis and the year represented by the horizontal axis. Eg.: The bar diagram can be constructed as follows : 120 100 Revenue 80 60 40 1980 1981 1982 Year In statistics, a histogram is a graphical display of tabulated frequencies, shown as bars. It shows what proportion of cases fall into each of several categories: it is a form of data binning. The categories are usually specified as non-overlapping intervals of some variable. The categories (bars) must be adjacent. The intervals are generally of the same size. Histogram : Eg.: In other words a histogram represents a frequency distribution by means of rectangles whose widths represent class intervals and whose areas are proportional to the corresponding frequencies. They only place the bars together to make it easier to compare data. Q-2) Calculate the Karl Pearson’s Coefficient of Skewness for the following Class Interval f 5-10 10-15 15-20 20-25 25-30 30-35 35-40 5 10 15 20 15 10 5 Ans: Calculation of coefficient of Skewness : Values 5—10 10--15 15--20 20--25 25--30 30--35 35--40 Mid Point X 7.5 12.5 17.5 22.5 27.5 32.5 37.5 f d d/5 fd1 fd2 cf 5 10 15 20 15 10 5 N=80 -15 -10 -5 0 5 10 15 -3 -2 -1 0 1 2 3 -15 -20 -15 0 15 20 15 Efd1=0 45 40 15 0 15 40 45 200 5 15 30 50 65 75 80 15 marks Karl Person’s Cofficient of Skewness = 3(Mean-Median)/Std. Deviation Mean = A + E fd’/N x C = 22.5 + 0/80 x 5 = 22.5 Median is the size of N/2 or 80/2 or 40th Item Thus median lies in class 20-25 By interpolation : Median = l1 + N/2-cf/f x c = 20 + 40-30/20 x 5 = 22.5 Std. Deviation of z = 1.6 x 5 = 8 Cofficient of Skewness = 3(Mean- Median)/Std. Deviation = 3(22.5-22.5)/8 =0 Coefficient of Skewness is 0, the distribuition is symmetrical. OR Calculate the i) Quartile Deviation iii) Coefficient of Skewness 15 marks Class Interval f cf 5-10 10-15 15-20 20-25 25-30 30-35 35-40 5 10 15 20 15 10 5 5 15 30 50 65 75 80 Q-3) Do Any Two A OR B OR C 20 marks A). If a new drug has been found to be effective 40% of the time, then what is the probability that in a random sample of 4 patients, it will be effective on 2 of them? Find “Effective” as success and “Non-effective”. 10 marks Ans A) p= .4(since the drug is effective 40% of the time) q = (l-p) = (l-.4) = .6 x= 20 n= 4 Then : P(x) = (n/x) (p)2 (q)2 = (4/2) (.4)2 (.6)2 = 4!/2!2!(.4)2 (.6)2 = 6 X.16 X.36 = 0.3456. The mean and the standard deviation for the distribution are given as follows : μ = np Sigma =( npq)2 B).Explain any with Example in brief a) Addition Rule OR b) Multiplication Rule OR c) Conditional Probability 10 marks Ans :B) Addition Rule : When two events are mutually exclusive, then the probability that one or the other event will occur, is the sum of their separate probabilities For example: if we roll a single die then probability that it will come up with face 5 or 6 where event A is face 5 and event B is face 6 is given by : P[A or B] =P[A] + P[B] P[5 or 6] = P[5] + P[6] =1/6+1/6 =2/6 =1/3 P[A or B] is written as P[AUB] which is also known as Prob. [a union B] b) Multiplication Rule : If event A and B are independent events then the probability that they will both occur is the product of their seprate probabilities. Events A and B are independent if and only if , P[AB] = P[A] x P[B] Example : if we toss a coin twice the probability that the first results in head and second toss results in a tail is given by : P[H,T] = P[H] x P[T] =½x½=¼ If events A and B are not independent meaning that the probability of occurrence of one events is dependent or conditional upon the occurrence or non occurrence of the other event then the probability that they will both occur is given by : P [AB] = P[A]x P[B] P[AB] = P[A] x P[B/A] Where P[B/A] means the probability of event B on the condition of the out come of event A. C.) The probability that a student passes a test in Statistics is 2/3, and the probability that he passes both a test in statistics and a test in mathematics is 14/45. The probability that he passes the test in at least one test is 4/5. What is the probability that he passes the test in Mathematics? 10 marks Q-4) A babysitter has 5 children under her supervision with average age of 6 years. But individually, the age of each child be as follows: X1 = 2 X2 = 4 X3 = 6 Calculate Population Mean and Standard Deviation. X4 = 8 X5 = 10 5 marks OR Explain a) Null Hypothesis OR b) Alternative Hypothesis 5 marks Ans : a) Null Hypothesis : formally describes some aspect of the statistical behaviour of a set of data; this description is treated as valid unless the actual behaviour of the data contradicts this assumption. Thus, the null hypothesis is contrasted against another hypothesis. Statistical hypothesis testing is used to make a decision about whether the data contradicts the null hypothesis: this is called significance testing. A null hypothesis is never proven by such methods, as the absence of evidence against the null hypothesis does not establish it. In other words, one may either reject, or not reject the null hypothesis; one cannot accept it. Failing to reject it gives no strong reason to change decisions predicated on its truth, but it also allows for the possibility of obtaining further data and then re-examining the same hypothesis. Example: one may want to compare the test scores of two random samples of men and women, and ask whether or not one group (population) has a mean score (which really is) different from the other. A null hypothesis would be that the mean score of the male population was the same as the mean score of the female population: H0 : μ1 = μ2 where: H0 = the null hypothesis μ1 = the mean of population 1, and μ2 = the mean of population 2.