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Chapter 1: Q81SE (page 50)

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 (鈥淎 Field Study on Braking Responses During Driving,鈥� Ergonomics, 1995: 1903鈥�1910):

mean= 535 median=500 mode=500

sd=96 minimum=220 maximum=925

5th percentile=400 10th percentile=430

90th percentile =640 95th percentile= 720

What can you conclude about the shape of a histogram of this data? Explain your reasoning.

Short Answer

Expert verified

The distribution is positively skewed.

Step by step solution

01

Given information

The summary information on thedistribution of total braking time (reaction time plus accelerator-to-brake movement time, in ms) is provided.

02

Determine the shape of histogram

The mean of the provided data is 535.

The median of the provided data is 500.

The mode of the provided data is 500.

It can be observed that the mean(535) is greater than the median value (500).Therefore, the distribution is positively skewed.

Thus, the distribution is positively skewed.

The skewness can also be observed from the values below 5th percentile and above 95th percentile.

Clearly,

\(\begin{aligned}{\rm{Median}} - {10^{th}}{\rm{percentile}} &= 500 - 430\\ &= 70\\{90^{th}}{\rm{percentile}} - {\rm{Median}} &= 640 - 500\\ &= 140\end{aligned}\)

This implies right tail is more extended as compared to left.

Also, the distribution appears to be have only one mode.

Thus, it can be concluded that the histogram would be unimodal and positively skewed.

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122.2 124.2 124.3 125.6 126.3 126.5 126.5 127.2 127.3

127.5 127.9 128.6 128.8 129.0 129.2 129.4 129.6 130.2

130.4 130.8 131.3 131.4 131.4 131.5 131.6 131.6 131.8

131.8 132.3 132.4 132.4 132.5 132.5 132.5 132.5 132.6

132.7 132.9 133.0 133.1 133.1 133.1 133.1 133.2 133.2

133.2 133.3 133.3 133.5 133.5 133.5 133.8 133.9 134.0

134.0 134.0 134.0 134.1 134.2 134.3 134.4 134.4 134.6

134.7 134.7 134.7 134.8 134.8 134.8 134.9 134.9 135.2

135.2 135.2 135.3 135.3 135.4 135.5 135.5 135.6 135.6

135.7 135.8 135.8 135.8 135.8 135.8 135.9 135.9 135.9

135.9 136.0 136.0 136.1 136.2 136.2 136.3 136.4 136.4

136.6 136.8 136.9 136.9 137.0 137.1 137.2 137.6 137.6

137.8 137.8 137.8 137.9 137.9 138.2 138.2 138.3 138.3

138.4 138.4 138.4 138.5 138.5 138.6 138.7 138.7 139.0

139.1 139.5 139.6 139.8 139.8 140.0 140.0 140.7 140.7

140.9 140.9 141.2 141.4 141.5 141.6 142.9 143.4 143.5

143.6 143.8 143.8 143.9 144.1 144.5 144.5 147.7 147.7

a. Construct a stem-and-leaf display of the data by first deleting (truncating) the tenths digit and then repeating each stem value five times (once for leaves 1 and 2, a second time for leaves 3 and 4, etc.). Why is it relatively easy to identify a representative strength value?

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The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives(flight hours/\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)) to reach a given crack size in fastener holes intended for use in military aircraft (鈥淪tatistical Crack Propagation in Fastener Holes Under Spectrum Loading,鈥� J. Aircraft,1983: 1028鈥�1032):

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1.011 1.064 1.109 1.132 1.140 1.153 1.253 1.394

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