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Blood cocaine concentration (mg/L) was determined both for

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 59E Chapter 1

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 59E

Blood cocaine concentration (mg/L) was determined both for a sample of individuals who had died from cocaineinduced excited delirium (ED) and for a sample of those who had died from a cocaine overdose without excited delirium; survival time for people in both groups was at most 6 hours. The accompanying data was read from a comparative boxplot in the article “Fatal Excited Delirium Following Cocaine Use” (?J. of Forensic Sciences, ?1997: 25–1). a.? ?Determine the medians, fourths, and fourth spreads for the two samples. b?.? re there any outliers in either sample? Any extreme outliers? c.Construct a comparative boxplot, and use it as a basis for comparing and contrasting the ED and non-ED samples

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Problem 59E Answer: Step1: We have Blood cocaine concentration (mg/L) was determined both for a sample of individuals who had died from cocaine induced excited delirium (ED) and for a sample of those who had died from a cocaine overdose without excited delirium; survival time for people in both groups was at most 6 hours. The accompanying data was read from a comparative boxplot in the article “Fatal Excited Delirium Following Cocaine Use” (J. of Forensic Sciences, 1997: 25–1). ED 0 0 0 0 0.1 0.1 0.1 0.1 0.2 0.2 0.3 0.3 0.3 0.4 0.5 0.7 0.8 1 1.5 2.7 2.8 3.5 4 8.9 9.2 11.7 21 Non ED 0 0 0 0 0 0.1 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.5 0.5 0.6 0.8 0.9 1 1.2 1.4 1.5 1.7 2 3.2 3.5 4.1 4.3 4.8 5 5.6 5.9 6 6.4 7.9 8.3 8.7 9.1 9.6 9.9 11 11.5 12.2 12.7 14 16.6 17.8 We need to find, a. Determine the medians, fourths, and fourth spreads for the two samples. b. Are there any outliers in either sample Any extreme outliers c.Construct a comparative boxplot, and use it as a basis for comparing and contrasting the ED and non-ED samples Step2: a). 1).Consider, ED 0 0 0 0 0.1 0.1 0.1 0.1 0.2 0.2 0.3 0.3 0.3 0.4 0.5 0.7 0.8 1 1.5 2.7 2.8 3.5 4 8.9 9.2 11.7 21 To find the lower fourth and upper fourth, we need to find median of given data and it is the middle value in a given data. Now, •Median = 0.4 (If there is an odd number of observations, the median is the middle value.) •Consider lower half data that is {0,0,0,....,0.3} lower fourth = 0.1 (If there is an odd number of observations, the median is the middle value.) •Consider upper half data that is {0.5,0.7,0.8,....,21} lower fourth = 2.8 (If there is an odd number of observations, the median is the middle value.) Then, The fourth spread is defined as fs = upper fourth lower fourth = 2.8 - 0.1 = 2.7. 2).Consider, Non ED 0 0 0 0 0 0.1 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.5 0.5 0.6 0.8 0.9 1 1.2 1.4 1.5 1.7 2 3.2 3.5 4.1 4.3 4.8 5 5.6 5.9 6 6.4 7.9 8.3 8.7 9.1 9.6 9.9 11 11.5 12.2 12.7 14 16.6 17.8 To find the lower fourth and upper fourth, we need to find median of given data and it is the middle value in a given data. Now, 12th value +13th value •Median = 2 (If there is an even number of observations, the median is the average of the two middle values.) = 1.521.7 = 3.2/2 = 1.6 •Consider lower half data that is {0,0,0,....,1.4} lower fourth = 25th value +26th v(If there is an even number of observations, the median is 2 the average of the two middle values.) = 0.2+0.3 2 = 0.5/2 = 0.25 •Consider upper half data that is {2,3.2,3.5,....,17.8} lower fourth = 48th value +49th va(If there is an even number of observations, the median 2 is the average of the two middle values.) = 7.9+8.3 2 = 16.2/2 = 8.1 Then, The fourth spread is defined as fs = upper fourth lower fourth = 8.1 - 0.25 = 7.85 Step3: b). 1).Consider, ED: mild outliers are less than outlier = lower fourth - 1.5(fourth spread) = 0.1 – 1.5(2.7) = –3.95 greater than outlier = upper fourth + 1.5(fourth spread) = 2.8 + 1.5(2.7) = 6.85 Extreme outliers are less than outlier = lower fourth - 3(fourth spread) = 0.1 – 3(2.7) = –8.0 greater than outlier = upper fourth + 3(fourth spread) = 2.8 + 3(2.7) = 10.9. So, the two largest observations (11.7, 21.0) are extreme outliers and the next two largest values (8.9, 9.2) are mild outliers. There are no outliers at the lower end of the data. 2).Consider, Non-ED: mild outliers are less than outlier = lower fourth - 1.5(fourth spread) = 0.3 – 1.5(7.85) = –11.475 greater than outlier = upper fourth + 1.5(fourth spread) = 8.1 + 1.5(7.85) = 19.875. Note that there are no mild outliers in the data, hence there cannot be any extreme outliers, either. Step4: c). A comparative boxplot appears below. The outliers in the ED data are clearly visible. There is noticeable positive skewness in both samples; the Non-ED sample has more variability than the Ed sample; the typical values of the ED sample tend to be smaller than those for the Non-ED sample.

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Chapter 1, Problem 59E is Solved
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Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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Blood cocaine concentration (mg/L) was determined both for