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Grip is applied to produce normal surface forces that

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 54E 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 54E

Grip is applied to produce normal surface forces that compress the object being gripped. Examples include two people shaking hands, or a nurse squeezing a patient’s forearm to stop bleeding. The article “Investigation of Grip Force, Normal Force, Contact Area, Hand Size, and Handle Size for Cylindrical Handles” (Human Factors, 2008: 734–744) included the following data on grip strength (N) for a sample of 42 individuals: a. ?Construct a stem-and-leaf display based on repeating each stem value twice, and comment on interesting features. b. ?Determine the values of the fourths and the fourthspread. ?c. ?Construct a boxplot based on the five-number summary, and comment on its features. d. ?How large or small does an observation have to be to qualify as an outlier? An extreme outlier? Are there any outliers? e. ?By how much could the observation 403, currently the largest, be decreased without affecting fs?

Step-by-Step Solution:

Answer a)Step 1 of 5 1/688 2/6 3/3 4/1 5/46 6/68 7/7 8/7 9/158 10/69 11/18 12/77 13/5 14/57 15/1 16/8 17/2 18/39 19/0 20/0 21/0 22/09 23/03 24/84 25/9 29/4 32/9 40/3 Left side values represents stem Right side values represents leaf Here most of the stems having very less values and stems are more Step 2 of 5 b)Lower fourth=88 Upper fourth=207.5 Fourth spread=Upper fourth-Lower fourth =207.5-88 =119.5 c)Step 3 of 5 Step 4 of 5 d)Minimum value=16 Maximum value=403 Lower outlier bound is = lower fourth value - 1.5(fourth spread) = 88 - 1.5(119.5) = -91.25 (no "too small" outliers) There are no outlier in the low end of the distribution Upper outlier bound is = upper fourth value + 1.5(fourth spread) = 207.5 + 1.5(119.5) = 386.75 (403 is an outlier) There is one outlier in the high end of the distribution Step 5 of 5 e)The observation 403, currently the largest, can be decreased without affecting fs is 403-386.75=16.25

Step 4 of 5

Chapter 1, Problem 54E is Solved
Step 5 of 5

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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