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The article “Can We Really Walk Straight” (Amer. J. of

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 73E 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 73E

The article “Can We Really Walk Straight?” (Amer. J. of Physical Anthropology, 1992: 19–27) reported on an experiment in which each of 20 healthy men was asked to walk as straight as possible to a target 60 m away at normal speed. Consider the following observations on cadence (number of strides per second): Use the methods developed in this chapter to summarize the data; include an interpretation or discussion wherever appropriate. [Note: The author of the article used a rather sophisticated statistical analysis to conclude that people cannot walk in a straight line and suggested several explanations this.]

Step-by-Step Solution:

Answer : Step 1 : Consider the following observations on cadence (number of strides per second): 0.95 0.78 0.85 0.93 0.92 0.93 0.95 1.05 0.93 0.93 0.86 1.06 1 1.06 0.92 0.96 0.85 0.81 0.81 0.96 Sum 18.51 Begin by computing mean and standard deviation of the given data. Use the formula for the mean ,noting that there are 20 observation. So we have to find mean. The formula of the mean is xi x = n Substitute the value i = 18.51 and n=20. x = 18.51 20 x = 0.9255 Therefore mean is 0.9255 Now we have to find the standard deviation. 2 x (x x) x (x x)2 0.95 0.00060025 0.78 0.02117025 0.85 0.00570025 0.93 2.025E-05 0.92 3.025E-05 0.93 2.025E-05 0.95 0.00060025 1.05 0.01550025 0.93 2.025E-05 0.93 2.025E-05 0.86 0.00429025 1.06 0.01809025 1 0.00555025 1.06 0.01809025 0.92 3.025E-05 0.96 0.00119025 0.85 0.00570025 0.81 0.01334025 0.81 0.01334025 0.96 0.00119025 2 (x x) = 0.124495 The formula of the standard deviation is. (x x) s = i n1 Substitute the valueix , n and x. s = 0.124495 201 s = 0.119495 s =0.006552368 s = 0.08096.8938 Therefore standard deviation is 606.8938. Now we have to find stem and leaf. 0.95 0.78 0.85 0.93 0.92 0.93 0.95 1.05 0.93 0.93 0.86 1.06 1 1.06 0.92 0.96 0.85 0.81 0.81 0.96 Then stem and leaf is given below. Stem Leaf 0.7 8 0.8 1 1 5 5 6 0.9 2 2 3 3 3 3 5 5 6 6 1 0 5 6 6 Consider the data. We arranged the data smallest to highest. Using excel we get the output. 0.78 0.81 0.81 0.85 0.85 0.86 0.92 0.92 0.93 0.93 0.93 0.93 0.95 0.95 0.96 0.96 1 1.05 1.06 1.06 Minimum value = 0.78 Maximum value = 1.06 Median = 0.93 Lower fourth = 0.8575 Upper fourth = 0.96 By using excel we get. Lower fourth is 0.855 and Upper fourth is 0.96 We have drawn a scatter plot. The data appears to be a bit skewed toward smaller values (negatively skewed). There are no outliers. The mean and the median are close in value.

Step 2 of 1

Chapter 1, Problem 73E is Solved
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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