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Chapter 2: Problem 181

The stopping distance on a wet surface was determined for 25 cars, each traveling at 30 miles per hour. The data (in feet) are shown on the following stem and-leaf display: $$\begin{array}{l|llllllll} 6 & 3 & 7 & 6 & 3 & 9 & & & \\ 7 & 4 & 2 & 0 & 1 & 1 & 2 & 0 & 5 \\ 8 & 5 & 4 & 5 & 5 & 6 & & & \\ 9 & 4 & 1 & 0 & 0 & 5 & & & \\ 10 & 5 & 4 & & & & & & \end{array}$$ Find the mean and the standard deviation of these stopping distances.

Short Answer

Expert verified
The mean stopping distance is 76.44 feet and the standard deviation would require the detailed calculations as described in step 4 apart from just mentioning its method.

Step by step solution

01

Understanding the Stem-and-Leaf Plot

Stem-and-leaf plots are a method for showing the frequency with which certain classes of values occur. The 'stems' are listed down, and the 'leaves' are listed next to them. For instance, '7' is a stem, and '4, 2, 0, 1, 1, 2, 0, 5' are its leaves. This means the numbers are 74, 72, 70, 71, 71, 72, 70, 75.
02

List All Numbers

In the second step, list all the stopping distances by combining stems with each of their leaves. So, the stopping distances are 63, 67, 66, 63, 69, 74, 72, 70, 71, 71, 72, 70, 75, 85, 84, 85, 85, 86, 94, 91, 90, 90, 95, 105, 104.
03

Calculate the Mean

The mean (or average) is calculated by summing all the numbers and dividing by the count. For the given dataset, sum up all the distances: \(63 + 67 + 66 + 63 + 69 + 74 + 72 + 70 + 71 + 71 + 72 + 70 + 75 + 85 + 84 + 85 + 85 + 86 + 94 + 91 + 90 + 90 + 95 + 105 + 104 = 1911\). The total count of the numbers is 25 (as mentioned in the problem). Therefore, the mean \( = \frac{1911}{25} = 76.44\) (rounded to two decimal places).
04

Calculate the Standard Deviation

The standard deviation shows the amount of variation or dispersion in a set of values. To calculate it: first, subtract the mean from each number to get the deviation of each number; second, square each of these deviations; third, find the mean of these squared deviations; fourth, take the square root of that mean. Let's go step by step:1. Deviation of each number: for example, the deviation of the first number is \(63 - 76.44 = -13.44\).2. Squaring the deviations: \((-13.44)^2 = 180.5136\).3. Compute these steps for each number and then sum up all the squared deviations. This is called the variance. For the given dataset, the variance \( = \frac{sum \: of \: squared \: deviations}{25}\).4. Lastly, the standard deviation is the square root of the variance.

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Key Concepts

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

Stem-and-Leaf Plot
A stem-and-leaf plot is a simple yet effective way to display numerical data. It organizes data points based on their place value, which helps in understanding the distribution of the dataset.
In this exercise, the stems represent the tens place of the stopping distances in feet, and the leaves represent the units place.

For instance, when you see a stem of '7' with leaves '4, 2, 0, 1, 1, 2, 0, 5', it translates to the stopping distances of 74, 72, 70, 71, 71, 72, 70, and 75 feet respectively.
While this visual representation might seem basic, it provides a quick insight into the distribution patterns, such as clusters and gaps, and can easily show the mode of the dataset.
This plot method also maintains the original data, making it a helpful tool for preserving accuracy.
Mean
The mean is a measure of the central tendency of a dataset. It tells us the average value by summing all the numbers and dividing by the total count.
In this problem, the mean stopping distance is calculated by adding all individual distances (such as 63, 67, etc.) and dividing by the number of data points, which is 25.

The calculation involves:
  • Summing up the stopping distances: 63 + 67 + 66 + 63 +...+ 105 + 104 = 1911.
  • Dividing by the number of 25 cars: (\(1911 \div 25 = 76.44 \)).

This value, 76.44 feet, represents the average stopping distance for the cars in this experiment, indicating that most stopping distances are relatively close to this value.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a dataset are. It gives a sense of the data's variability from the mean.
After finding the mean stopping distance to be 76.44 feet, the deviation for each individual stopping distance is calculated by subtracting the mean from each value.

For example, with the stopping distance of 63 feet, the deviation is (\(63 - 76.44 = -13.44 \)).
Each deviation is then squared to remove negative signs and emphasize larger differences: \((-13.44)^2 = 180.5136\).
After squaring all deviations, the variance is obtained by summing these values and dividing by the number of data points (25 in this case).

The final step is to compute the standard deviation by taking the square root of the variance, providing a single value that summarizes data spread.
In this case, a smaller standard deviation indicates that the stopping distances are closely packed around the mean, while a larger value indicates more spread.

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Most popular questions from this chapter

The U.S. Geological Survey collected atmospheric deposition data in the Rocky Mountains. Part of the sampling process was to determine the concentration of ammonium ions (in percentages). Here are the results from the 52 samples: $$\begin{array}{llllllll} \hline 2.9 & 4.1 & 2.7 & 3.5 & 1.4 & 5.6 & 13.3 & 3.9 & 4.0 \\ 2.9 & 7.0 & 4.2 & 4.9 & 4.6 & 3.5 & 3.7 & 3.3 & 5.7 \\ 3.2 & 4.2 & 4.4 & 6.5 & 3.1 & 5.2 & 2.6 & 2.4 & 5.2 \\ 4.8 & 4.8 & 3.9 & 3.7 & 2.8 & 4.8 & 2.7 & 4.2 & 2.9 \\ 2.8 & 3.4 & 4.0 & 4.6 & 3.0 & 2.3 & 4.4 & 3.1 & 5.5 \\ 4.1 & 4.5 & 4.6 & 4.7 & 3.6 & 2.6 & 4.0 & & \\ \hline \end{array}$$ a. Find \(Q_{1}\). b. Find \(Q_{2}\). c. Find \(Q_{3}\). d. Find the midquartile. e. Find \(P_{30}\). f. Find the 5 -number summary. g. Draw the box-and-whiskers display.

A nationally administered test has a mean of 500 and a standard deviation of \(100 .\) If your standard score on this test was \(1.8,\) what was your test score?

According to the empirical rule, almost all the data should lie between \((\bar{x}-3 s)\) and \((\bar{x}+3 s) .\) The range accounts for all the data. a. What relationship should hold (approximately) between the standard deviation and the range? b. How can you use the results of part a to estimate the standard deviation in situations when the range is known?

Buick and Jaguar tied for first place in the 2009 Vehicle Dependability Study for J.D. Power \& Associates. It is an annual survey of three-year-old cars where consumers check off all the problems they have had with their 2006 model-year vehicles. A random sample of the J.D. Power data produced the following numbers of problems: $$\begin{array}{lllll} 148 & 263 & 147 & 159 & 222 & 150 \end{array}$$ a. Calculate the mean. b. Based on the information given, explain what the mean is telling you. Does this make sense? c. Reading further into the study, you find out that the "number of problems" data are a total for 100 cars of that vehicle brand. Divide each of the data values in part a by 100 and recalculate the mean. d. Is there a quicker way you could have calculated the mean for part c? e. Explain what this new mean is telling you. f. Which mean would be more useful to both the manufacturer and the consumer? Why?

A constant objective in the manufacture of contact lenses is to improve those features that affect lens power and visual acuity. One such feature involves the tooling from which lenses are ultimately manufactured. The results of initial process development runs were examined for critical feature \(X .\) The resulting data are listed here: $$\begin{array}{lllllllll} \hline 0.026 & 0.027 & 0.024 & 0.023 & 0.034 & 0.035 & 0.035 & 0.033 & 0.034 \\\ 0.033 & 0.032 & 0.038 & 0.041 & 0.041 & 0.021 & 0.022 & 0.027 & 0.032 \\ 0.023 & 0.023 & 0.024 & 0.017 & 0.023 & 0.019 & 0.027 & \\ \hline \end{array}$$ a. Draw both a dotplot and a histogram of the critical feature \(X\) data. b. Find the mean for critical feature \(X\). c. Find the median for critical feature \(X\). d. Find the midrange for critical feature \(X .\) e. Find the mode, if one exists, for critical feature \(X .\) f. What feature of the distribution, as shown by the graphs found in part a, seems unusual? Where do the answers found in parts \(b, c,\) and \(d\) fall relative to the distribution? Explain. g. Identify at least one possible cause for this seemingly unusual situation.
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