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

## Problem 59E Chapter 1

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Probability and Statistics for Engineering and the Sciences | 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...

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