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Solved: A study carried out to investigate the

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 82E 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 82E

A study carried out to investigate the distribution of total braking time (reaction time plus accelerator-to-brake movement time, in ms) during real driving conditions at 60 km/hr gave the following summary information on the distribution of times (“A Field Study on Braking Responses During Driving,” Ergonomics, 1995: 1903–1910): What can you conclude about the shape of a histogram of this data? Explain your reasoning.

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Step 1 of 3

Problem 82E Answer: Step1: We have A study carried out to investigate the distribution of total braking time (reaction time plus accelerator-to-brake movement time, in ms) during real driving conditions at 60 km/hr gave the following summary information on the distribution of times (“A Field Study on Braking Responses During Driving,” Ergonomics, 1995: 1903–1910): Mean = 535, Median = 500, Mode = 500 Standard deviation = 96, Minimum = 220, Maximum = 925 5th Percentile = 400, 10th percentile = 430 90th percentile = 640, 95th percentile = 720 We need to conclude about the shape of a histogram of this data Explain your reasoning. Step2: Here, we can see the mean is somewhat larger than the median and mode. That is mean = 535 > median = 500 Also, the maximum and higher (90th and 95th) percentiles are farther from the center (that is mean, median, or mode) than are the minimum and lower (5th and 10th) percentiles. That is 95th percentile - median = 720 - 500 = 220 Median - 5th percentile = 500 - 400 = 100 This tells me the histogram is skewed to the right. We can in fact plot a density histogram with the available information. Make the breakpoints for the class intervals equal to the given quantiles, divide the percentage in the class by the class interval width. For instance, there are 5% of the date between the minimum and the 5th percentile, so the height of the bars over the interval (220,400) is 0.05/(400-220). Similarly, there are 40% between the 10th and 50th percentiles. Thus, the height of the bar over (430,500) is 0.4/(500-430). A sketch of the histogram is below. It looks skewed right, sort of.

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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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