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Solutions for Chapter 5 10. Here is the stemplot. 48] 8 50] 7 s1]0 52 | 6799 53 | 04469 54 | 2467 55 | 03578 56 | 12358 5031 59. 5845 The distribution is roughly symmetric and single-peaked except for one low observation, which may be an outlier. 12. (a) Here is the stemplot, with the outlier. 69 455 334477 025669 YAWURWHH 3 (b) For all 18 years, x = WeDsensstorage as seagrass emsamesTestssseTIeNe ES =! = 36.56. To determine the median, we must put the data in order from smallest to largest. 16 19 24 25 25 33 33 34 34 37 37 40 42 45 46 46 49 73 Since there are 18 pieces of data, the median is the mean of the 9” and 10" pieces of data. M=S8L=0-35,5 ead Ss For the 17 years other than 2001, x = 1s:2s:2#:19+33425s ses 4ees7a33447-40e37s3449e6eds EEE Tee bets = SS = 34.41. Taking out 73 as a piece of data, we have now 17 pieces of data. 16 19 24 25 25 33 33 34 34 37 37 40 42 45 46 46 49 The median is the 4=+£=9" piece of data, namely M=34. Removing the high outlier cuts the gap between the median and the mean about in half. 20. To determine the minimum, maximum, and median, we must put the 21 pieces of data in order from smallest to largest. 4.88 5.07 5.10 5.26 5.27 5.29 5.29 5.30 5.34 5.34 5.36 5.39 5.42 5.44 5.46 5.47 5.50 5.53 5.55 5.57 5.58 5.61 5.62 5.63 5.65 5.68 5.75 5.79 5.85 The minimum is 4.88 and the maximum is 5.85. The median is the %! =2=15" piece of data, namely 5.46. Since there are 14 observations to the left of the median, Q, is the mean of the 7” and 8" pieces of data, namely 22452 = 182 = 5.295. Since there are 14 observations to the right of the median, Q, is the mean of the 22" and 23" pieces of data, namely 2558? = 1125 = 5.615, Thus, the five-number summary is 4.88, 5.295, 5.46, 5.615, 5.85. The quartiles are roughly equidistant from the median (symmetry). The minimum (a possible outlier) is farther from the median than is the maximum.
