251y0815 6/11/08ECO251 QBA1FIRST EXAMJune 11, 2008 Name: ___KEY_________________ Student Number: _________________________Class Hour: _____________________ Remember – Neatness, or at least legibility, counts. In most non-multiple-choice questions an answer needs a calculation or short explanation to count. Show your work! The exam is normed on 50 points, so that any grade above 48 is an A+ and grades wrap around.Part I. (7 points) The following numbers are to be considered a sample of prices of gasoline taken at five gas stations.1x 4.14 4.19 4.25 4.01 3.99Compute the following: Show your work!a) The Median (1)b) The Standard Deviation (3)c) The 3rd quintile (2)d) The Coefficient of variation (1)Extra credit beyond this point!e) The harmonic mean (1.5)f) The root-mean square (1.5)g) The geometric mean (3 ways!) (3)Solution: My calculations are below. xlog is a logarithm to the base 10 and xln is a natural logarithm.Row x orderinx 2x x1 xlog xln 1 4.14 3.99 17.1396 0.241546 0.617000 1.42070 2 4.19 4.01 17.5561 0.238663 0.622214 1.43270 3 4.25 4.14 18.0625 0.235294 0.628389 1.44692 4 4.01 4.19 16.0801 0.249377 0.603144 1.38879 5 3.99 4.25 15.9201 0.250627 0.600973 1.38379 20.58 84.7584 1.21551 3.07172 7.07290a) The Median (1): the median is the middle number when the data is in order – 4.14b) The Standard Deviation (3): We have 58.20x, 7584.842x,116.4558.20nxx, 01278.0405112.04116.457584.8412222nxnxs. So that113049.001278.0 s.If you chose to annoy me by using the definitional formula, you should have gotten the following.Row x xx 2xx 1 4.14 0.024 0.000576 2 4.19 0.074 0.005476 3 4.25 0.134 0.017956 4 4.01 -0.106 0.011236 5 3.99 -0.126 0.015876 20.58 0.000 0.05112You would have 58.20x,116.4558.20nxx, 05112.02xx 1251y0815 6/11/08 01278.0405112.0122nxxs. So that 113049.001278.0 sI had better repeat the table from the last page.Row x orderinx 2x x1 xlog xln 1 4.14 3.99 17.1396 0.241546 0.617000 1.42070 2 4.19 4.01 17.5561 0.238663 0.622214 1.43270 3 4.25 4.14 18.0625 0.235294 0.628389 1.44692 4 4.01 4.19 16.0801 0.249377 0.603144 1.38879 5 3.99 4.25 15.9201 0.250627 0.600973 1.38379 20.58 84.7584 1.21551 3.07172 7.07290c) The 3rd quintile (2): The third quintile has 60.53of the data below it. The data must be in order. banplocation .60.3660.1 . 17.414.419.460.014.460.034340.1xxxxxpd) The Coefficient of variation (1): 0247.116.4113049.0xsC or 2.47%Extra credit beyond this point!e) The harmonic mean (1.5): The formula table says xnxh111 or99.3101.4125.4119.4114.41511hx 243102.021551.151. So.1135.4243102.01111xnxhOf course some of you decided that009718.058.2015199.301.425.419.414.4151?99.3101.4125.4119.4114.4151111xnxhThis is, of course, an easier way to do the problem. It is also wrong and unreasonable (since it is not between 3.99 and 4.25), and you will get an A for the course if you can prove to me that it is not wrong! And please don’t try any math if you get on “Are you smarter than a fifth-grader.”f) The root-mean square (1.5): The formula table says 22211xnxorxnxrmsrms95168.1657584.842rmsx. So 11724.495168.16 rmsxg) The geometric mean (3): The formula table says nnngxxxxxx1321. So 11476.461428.117961428.117999.301.425.419.414.42.055ngxxThe formula table also says )ln(1ln xnxg, but I said in class that this could be either natural logs or logs to the base 10. If we use logarithms to the base 10 we get 507172.3)log(1log xnxg 614344.0 and 11476.410614344.0gx. If we use 2251y0815 6/11/08natural logarithms we get )ln(1ln xnxg 41458.1507290.7 and11476.441458.1exg3251y0815 6/11/08Part II. (18 points) According to Anderson, Sweeny and Williams a bank found the following as a sample of 30 waiting times (in seconds) for service. Row Time Frequency 1 60-119.99 6 2 120-179.99 10 3 180-239.99 8 4 240-299.99 4 5 300-359.99 2a. Calculate the Cumulative Frequency (1)b. Calculate the Mean (1)c. Calculate the Median (2)d. Calculate the Mode (1)e. Calculate the Variance (3)f. Calculate the Standard Deviation (2)g. Calculate the Interquartile Range (3) h. Calculate a Statistic showing Skewness and interpret it (3)i. Make a frequency polygon of the data (Neatness Counts!)(2) Solution: Note that unreasonable answers are answers where the mean, median, mode, first quartile and third quartile do not fall between 60 and 360. If we use the computational method, we get the following. xis the midpoint of the class.Row Class f F x fx 2fx 3fx 1 60-119.99 6 6 90 540 48600 4374000 2 120-179.99 10 16 150 1500 225000 33750000 3 180-239.99 8 24 210 1680 352800 74088000 4 240-299.99 4 28 270 1080 291600 78732000 5 300-359.99 2 30 330 660 217800 71874000 30 5460 1135800 262818000If we use the definitional method, we get the following. xis the midpoint of the class. I usually tell peoplethat they are wasting their time if they use the definitional method. Because of the large numbers here that may not be true.Row Class f F x fx xx xxf 2xxf 3xxf 1 60-119.99 6 6 90 540 -92 -552 50784 -4672128 2 120-179.99 10 16 150 1500 -32 -320 10240 -327680 3 180-239.99 8 24 210 1680 28 224 6272 175616 4 240-299.99 4 28 270 1080 88 352 30976 2725888 5 300-359.99 2 30 330 60 148 296 43808 6483584 30 5460 0 142080 4385280If you used the computational method, you would have gotten 30fn and 5460fx, so that the mean is 0000.182305460nfxx. You
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