/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 A study carried out to investiga... [FREE SOLUTION] | 91Ó°ÊÓ

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A study carried out to investigate the distribution of total braking time (reaction time plus acceleratorto-brake movement time, in msec) during real driving conditions at \(60 \mathrm{~km} / \mathrm{h}\) 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 \(=500\) sd \(=96\) minimum \(=220\) maximum \(=925\) 5 th percentile \(=400 \quad 10\) th percentile \(=430\) 90 th percentile \(=64095\) th percentile \(=720\) What can you conclude about the shape of a histogram of this data? Explain your reasoning.

Short Answer

Expert verified
The histogram is likely right-skewed, with a longer tail to the right.

Step by step solution

01

Understand Central Tendency

The mean of the braking times is 535 msec, the median is 500 msec, and the mode is also 500 msec. When the mean is greater than the median, it suggests that the data may be right-skewed, as extreme values on the right can pull the mean upwards.
02

Analyze Percentile Information

The 5th percentile is 400 and the 10th percentile is 430, while the 90th percentile is 640 and the 95th percentile is 720. The spread between successive lower percentiles (30 for 5th to 10th) is less than the spread between upper percentiles (80 from 90th to 95th), which indicates the possibility of more data stretched out on the right side, confirming possible right skew.
03

Review Spread and Extremes

The standard deviation is relatively large at 96 msec, and the range of the data from minimum (220) to maximum (925) is quite wide. A greater range and standard deviation, especially with the high maximum compared to the mean, can indicate a skew, often to the right if the higher values are extreme.
04

Consider Histogram Shape Conclusion

Given all these observations — where the mean is greater than the median, the upper percentile spread is larger, and the data range is wide with higher extremes — the histogram is likely right-skewed, meaning a longer tail to the right.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Tendency
In the context of analyzing data, central tendency provides a summary of the dataset's average value. It includes measures like the mean, median, and mode. In our braking time study, the mean is 535 msec, while both the median and mode are 500 msec. When interpreting these values, the relative position of the mean compared to the median can reveal a lot about the distribution's shape.

If the mean is greater than the median, as in this case, it often indicates a right-skewed distribution. This happens because the mean is affected by all values in the dataset, including potential high outliers that pull it upwards. In contrast, the median and mode are more resistant to such high values.

Therefore, the information that the mean exceeds the median suggests the presence of outliers or a tail extending toward higher values, corroborating the suspicion of a right-skewed distribution.
Percentiles in Data
Percentiles are particularly useful in understanding the spread and skewness of data. A percentile indicates the value below which a given percentage of observations fall. For instance, the 5th and 10th percentiles in the braking time data are 400 and 430 msec, respectively.

Analyzing these percentiles, we notice that the spread between the lower percentiles is smaller (30 msec) than the spread between the higher percentiles, like between the 90th (640) and 95th (720) percentiles where the difference is 80 msec. Larger gaps at higher percentiles typically imply more variation or stretching of the data towards the higher end.

This further supports the identification of a right-skewed distribution. More data are concentrated in lower ranges, with fewer but more spread out higher values, contributing to the longer tail on the right side of the histogram.
Histogram Interpretation
Interpreting histograms is key in visualizing and understanding data distributions. A histogram visualizes the frequency of data points within certain ranges, called bins. When analyzing the shape of the histogram for our braking times, acknowledging all gathered data is crucial.

Given the statistical insights, such as the right-skewness indicated by our central tendency and percentiles, the histogram would likely have the bulk of the data on the left side with a tail trailing off to the right. In real terms, this could mean many drivers had braking times around 500 msec, while fewer experienced much longer times.

Hence, a visual representation of the histogram would show more frequent occurrences of values nearer the median and mode, with decreasing frequencies extending to the right.
Standard Deviation Significance
Standard deviation is a measure of data spread or dispersion relative to its mean and is crucial in understanding data variability. In our braking time study, the standard deviation is 96 msec, which indicates considerable variability among the data points.

A larger standard deviation signifies that data points are more spread out from the mean, which matches our findings about the data's range being significant from 220 msec to 925 msec. Under normal circumstances, wide range and high standard deviation support the notion of a dataset with substantial diversity in response times, further affirming the right-skew model if those long-braking times are infrequent yet far above the mean.

Thus, the standard deviation helps to substantiate claims about skewness by highlighting variability factors, which are vital in assessing the extent of the data's tail and overall symmetry.

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Most popular questions from this chapter

a. For what value of \(c\) is the quantity \(\sum\left(x_{i}-c\right)^{2}\) minimized? [Hint: Take the derivative with respect to \(c\), set equal to 0 , and solve.] b. Using the result of part (a), which of the two quantities \(\sum\left(x_{i}-\bar{x}\right)^{2}\) and \(\sum\left(x_{i}-\mu\right)^{2}\) will be smaller than the other (assuming that \(\bar{x} \neq \mu) ?\)

a. Let \(a\) and \(b\) be constants and let \(y_{i}=a x_{i}+b\) for \(i=1,2, \ldots, n\). What are the relationships between \(\bar{x}\) and \(\bar{y}\) and between \(s_{x}^{2}\) and \(s_{y}^{2}\) ? b. The Australian army studied the effect of high temperatures and humidity on human body temperature (Neural Network Training on Human Body Core Temperature Data, Technical Report DSTO TN-0241, Combatant Protection Nutrition Branch, Aeronautical and Maritime Research Laboratory). They found that, at \(30^{\circ} \mathrm{C}\) and \(60 \%\) relative humidity, the sample average body temperature for nine soldiers was \(38.21^{\circ} \mathrm{C}\), with standard deviation \(.318^{\circ} \mathrm{C}\). What are the sample average and the standard deviation in \({ }^{\circ} \mathrm{F}\) ?

In a study of warp breakage during the weaving of fabric (Technometrics, 1982: 63), 100 specimens of yarn were tested. The number of cycles of strain to breakage was determined for each yarn specimen, resulting in the following data: \(\begin{array}{rrrrrrrrrr}86 & 146 & 251 & 653 & 98 & 249 & 400 & 292 & 131 & 169 \\ 175 & 176 & 76 & 264 & 15 & 364 & 195 & 262 & 88 & 264 \\ 157 & 220 & 42 & 321 & 180 & 198 & 38 & 20 & 61 & 121 \\ 282 & 224 & 149 & 180 & 325 & 250 & 196 & 90 & 229 & 166 \\ 38 & 337 & 65 & 151 & 341 & 40 & 40 & 135 & 597 & 246 \\ 211 & 180 & 93 & 315 & 353 & 571 & 124 & 279 & 81 & 186 \\ 497 & 182 & 423 & 185 & 229 & 400 & 338 & 290 & 398 & 71 \\ 246 & 185 & 188 & 568 & 55 & 55 & 61 & 244 & 20 & 284 \\ 393 & 396 & 203 & 829 & 239 & 236 & 286 & 194 & 277 & 143 \\ 198 & 264 & 105 & 203 & 124 & 137 & 135 & 350 & 193 & 188\end{array}\) a. Construct a relative frequency histogram based on the class intervals \(0-100,100-200, \ldots\), and comment on features of the distribution. b. Construct a histogram based on the following class intervals: \(0-50,50-100,100-150\), \(150-200, \quad 200-300, \quad 300-400, \quad 400-500\), \(500-600,600-900 .\) c. If weaving specifications require a breaking strength of at least 100 cycles, what proportion of the yam specimens in this sample would be considered satisfactory?

At the beginning of the 2007 baseball season each American League team had nine starting position players (this includes the designated hitter but not the pitcher). Here are the salaries for the New York Yankees and the Cleveland Indians in thousands of dollars: \(\begin{array}{llllll}\text { Yankees: } & 12000 & 600 & 491 & 22709 & 21600 \\\ & 13000 & 13000 & 15000 & 23429 & \\ \text { Indians: } & 3200 & 3750 & 396 & 383 & 1000 \\ & 3750 & 917 & 3000 & 4050 & \end{array}\) Construct a comparative boxplot and comment on interesting features. Compare the salaries of the two teams. The Indians won more games than the Yankees in the regular season and defeated the Yankees in the playoffs.

Let \(\bar{x}_{n}\) and \(s_{n}^{2}\) denote the sample mean and variance for the sample \(x_{1}, \ldots, x_{n}\) and let \(\bar{x}_{n+1}\) and \(s_{n+1}^{2}\) denote these quantities when an additional observation \(x_{n+1}\) is added to the sample. a. Show how \(\bar{x}_{n+1}\) can be computed from \(\bar{x}_{n}\) and \(x_{n+1}\). b. Show that $$ n s_{n+1}^{2}=(n-1) s_{n}^{2}+\frac{n}{n+1}\left(x_{n+1}-\bar{x}_{n}\right)^{2} $$ so that \(s_{n+1}^{2}\) can be computed from \(x_{n+1}, \bar{x}_{n}\), and \(s_{n^{-}}^{2}\). c. Suppose that a sample of 15 strands of drapery yarn has resulted in a sample mean thread elongation of \(12.58 \mathrm{~mm}\) and a sample standard deviation of \(.512 \mathrm{~mm}\). A 16 th strand results in an elongation value of \(11.8\). What are the values of the sample mean and sample standard deviation for all 16 elongation observations?

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