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. CLABE Statistics Homework assignment - Problem sheet 1 . 1. A supervisor of a plant kept records of the time (in seconds) that employees needed to complete a particular task. The data are summarized as follows: Time 30 ` 40 40 ` 50 50 ` 60 60 ` 80 80 ` 100 100 ` 150 10 15 20 30 24 20 Frequency (a) Graph the data with a histogram. (b) Discuss possible errors occurring if the frequency is used instead of the density. SOLUTION 0.008 0.003 Density 0.013 0.017 (a) The histogram is the following: 30 40 50 60 80 100 150 Time (b) In this case it is not possible to use frequencies in place of densities because the interval classes have dierent sizes. The wrong histogram is plotted below. 1 0.084 0.126 Frequency 0.168 0.252 Wrong histogram 30 40 50 60 80 100 150 Time 2. You want to buy a new house and, for this reason, last week you visited 10 dierent ats which are on sale. All the ats are located in the same area and are similar for dimension and other basic characteristics. The prices (X , in thousands of euros) of the ats you visited are given below 176, 153, 215, 185, 168, 197, 159, 162, 181, 160 (a) Calculate the mean and the standard deviation of the data. (b) Provide the ve number summary of the data. (c) Represent the data in a histogram with the following values as delimiters for the interval classes: 150; 170; 190; 220. (d) You know that of a and b Y = aX + b in the case where a and b are two positive ȳ = 489 and sY = 48.59. where constants. Calculate the values SOLUTION (a) The mean is x̄ = 175.6, the sample variance is s2X = 377.82 and the standard deviation sX = 19.44. (b) The data, in increasing order, are 153 159 160 162 168 176 181 185 197 215 and therefore the ve number summary is given by min=153 Q1 = (159 + 160)/2 = 159.5 Me = Q2 = (168 + 176)/2 = 172 2 (1) Q3 = (185 + 197)/2 = 191 Max=215. 0.0067 Density 0.0150 0.0250 (c) The histogram is given below 150 170 190 220 price 48.59 = |a|19.44 and, therefore, |a| = 2.5. It is said in the a is positive, so that a = 2.5. Furthermore, ȳ = ax̄ + b, that is, 489 = 2.5 × 175.6 + b and b = 489 − 2.5 × 175.6 = 50. (d) Since sY = |a|sX , that is text of the problem that 3. A sample of students at a local high school were asked what their plans were after graduation. Possible responses were college (C), military (M), work (W), and other (O). The following data were collected. M C C W O C W W W C C C M W O O M W O O M W W C C C C C C C (a) Create a frequency table for the data. (b) Create a bar chart for the data. Use relative frequency for the vertical axis. (c) What proportion of students in the sample plan to enter college after graduation? (d) What proportion of students in the sample plan to either work or enter the military after graduation? SOLUTION (a) The frequency table for the data is as follows 3 Plans after graduation category freq. rel.freq C 13 0.43 M 4 0.13 O 5 0.17 W 8 0.27 30 1.00 0.3 0.2 0.0 0.1 relative frequency 0.4 0.5 (b) The bar chart for the data is as follows C M O W Plans after graduation (c) The proportion of students in the sample who plan to enter college after graduation is equal to 0.43. (d) The proportion of students in the sample who plan to either work or enter the military after graduation is equal to 0.4. 4. The following data set contains the number of calories for chicken sandwiches at Burger King. 670, 920, 1160, 340, 570, 750, 930, 770, 970, 260, 370, 310, 800, 450, 520, 630 4 (a) Find the range of this data set. (b) Find the ve-number summary. (c) Find the interquartile range. (d) Construct a boxplot. SOLUTION (a) The range of the given data set is R = 1160 − 260 = 900; (b) The ve-number summary is: min=260, 800+920 2 = 860, Q1 = 370+450 2 = 410, Me = 630+670 2 = 650, Q3 = Max=1160; (c) The interquartile range is IQR= 860 − 410 = 450. 400 600 800 1000 (d) The boxplot of the data is given below Calories 5. A researcher in an alcoholism treatment center, interested in summarizing the length of stay in the center for rst-time patients, randomly selects 14 records of individuals institutionalized within the previous two years. The length of stay in the center, in days, are as follows 7, 6, 6, 3, 8, 5, 4, 5 7, 5, 4, 5, 6, 6, 7. (a) Create a frequency table for the data; (b) provide a suitable graphical representation of the data; (c) calculate the sample mean and variance of the data. SOLUTION (a) The frequency table of the data is given below. Length of stay category freq. rel.freq cum. freq. 3 1 0.071 0.071 4 2 0.142 0.214 5 3 0.214 0.428 6 4 0.286 0.714 7 3 0.214 0.928 8 1 0.071 1.00 14 1.00 0.21 0.14 0.07 Relative freq. 0.29 (b) These data can be suitably represented by means of a barplot for discrete variables 2 3 4 5 6 7 8 9 Length of stay (c) If we denote by X the variable of interest then x̄ = 5.64, s2X = 1.94 e sX = 1.39. 6. The owner of a bakery is interested in investigating the purchase behavior of his customers. For this purpose, 12 receipts issued in June are randomly extracted. selected receipts are as follows 6 The amounts in euro of the 4.05 6.60 3.45 17.70 14.25 1.95 10.20 8.10 8.70 10.80 5.55 1.95 (a) Calculate the mean and the standard deviation of the data. (b) Calculate the median and the interquartile range of the data. (c) Represent the data in a histogram with the following values as delimiters for the interval classes: 1; 2; 5; 9; 11; 20. (d) The amount of every receipt includes a tax computed as the 21% of the net value of the receipt. You decide to carry out the analysis with respect to the net amounts. Compute points (a) and (b) above with respect to this new set of data. SOLUTION (a) The mean, the variance and the standard deviation are x̄ = 7.78, s2X = 23.93 and sX = 4, 89 respectively. (b) If we write the data points in increasing ordering we obtain the sequence 1.95 1.95 3.45 4.05 5.55 6.60 8.10 8.70 10.20 10.80 14.25 17.70 and it follows that the rst, second and third quartiles are Q1 = (3.45 + 4.05)/2 = 3.75, Q2 = (6.60 + 8.10)/2 = 7.35 and Q3 = (10.20 + 10.80)/2 = 10.5 respectively. Hence, the median of the data is Me = 7.35 whereas the interquartile range is IQR = 10.5−3.75 = 6.75. (c) The required histogram is represented below, note that the density of every bin can be read from the vertical axis. 0.083 0.019 0.056 density 0.167 histogram of receipt values 0 1 2 5 9 11 20 euro (d) Let Y denote the receipt net values. Then that • ȳ = • s2Y 1 1.21 = × x̄ = 6.43; 1 2 2 sX 1.21 = 16.34; 7 Y = 1 1.21 ×X is a linear transformation of X so • sY = 1 1.21 sX = 4.04. It is also straightforward to see that multiplying every data point by a positive constant does not change the ordering of the values and, furthermore, the median and the interquartile range of Y can be computed by multiplying the corresponding quantities of X by 1 1.21 . Hence • • the median of Y is 1 1.21 the interquartile range × 7.35 = 6.07; 1 of Y is 1.21 × 6.75 = 5.58 7. Consider the following sample of ve values and corresponding weights: (a) Calculate the arithmetic mean of the (b) Calculate the weighted mean of the xi wi 4.6 8 3.2 3 5.4 6 2.6 2 5.2 5 xi xi values without weights. values. SOLUTION (a) The arithmetic mean is equal to 4.2; (b) The weighted mean is equal to 110 24 = 4.583. 8. Let X be the age of the students enrolled in an on-line macroeconomics course. The ages of a sample of 12 students are 21 22 27 36 18 19 22 23 22 28 36 33 (a) Calculate the mean, the median and the modal age. (b) Calculate the mean of the following variables: i) ii) Y = X/(40 − X); W = X/40 − X . SOLUTION 8 (a) The mean is x̄ = 25.58. The data points, in increasing ordering, are 18 19 21 22 22 22 23 27 28 33 36 36 and it is easy to see that (b) Since Y Me = 22+23 2 = 22.5 whereas the modal age is 22. X , in order to compute the mean of Y one has to ȳ = 2.91. On the other hand, therefore w̄ = x̄/40 − x̄ = −24.94. is NOT a linear transformation of apply the transformation to every data point to obtain that W is a linear transformation of X and 9