Chapter 4.4, Problem 40E

a.

To determine

Draw a histogram for the travel time distribution.

Answer to Problem 40E

The histogram for the travel time distribution is given below:

Explanation of Solution

Calculation:

The data represent the relative frequency for travel time to work for a large sample of adults who did not work at home.

Software procedure:

Step-by-step procedure to obtain the width using MINITAB software:

• Choose Calc > Calculator.
• Enter the column of width under Store result in variable.
• Enter the formula ‘Upper’-‘Lower’ under Expression.
• Click OK.

The range value is stored in the column “width”.

Step-by-step procedure to obtain the density using MINITAB software:

• Choose Calc > Calculator.
• Enter the column of Frequency under Store result in variable.
• Enter the formula ‘Frequency’/‘width’ under Expression.
• Click OK.

The density is stored in the column “Frequency”.

Step by step procedure to draw the relative frequency histogram using MINITAB software:

• Select Graph > Bar chart.
• In Bars represent select Values from a table.
• In One column of values select Simple.
• Enter Frequency in Graph variables.
• Enter Travel Time (Minutes) in categorical variable.
• Select OK.
• Click on the axis of the graph and in Gap between clusters enter 0.

Thus, the histogram for travel time is obtained.

b.

To determine

Delineate the features, such as center, shape, and variability of the histogram from Part (a).

Explanation of Solution

From the histogram, it is observed that the distribution of travel time is slightly positively skewed with a single maximum frequency density, that is, a single mode. Most of the values of travel time lie between 5 and 45. The center of the distribution is approximately 20.

c.

To determine

Explain whether it is appropriate to use the empirical rule to make statements about the travel time distribution.

Answer to Problem 40E

No, it is not appropriate to use empirical rule to make statements about the travel time distribution.

Explanation of Solution

By observing the histogram, the travel time data are skewed right. However, the empirical rule is applicable for data that are distributed normally, which is bell-shaped and symmetric.

Therefore, in this context, it is not appropriate to use the empirical rule to make the statements about the travel time distribution.

d.

To determine

Elucidate the reason why the travel time distribution could not be well approximated by a normal curve.

Explanation of Solution

It is given that the approximate mean and standard deviation for the travel time distribution are 27 minutes and 24 minutes, respectively.

The observation of the travel time should not be negative. In general, the travel time begins from 0.

If the distribution of the travel time is approximately normal, then the percentage of the travel time that fall below 0 is obtained as given below:

$\begin{array}{c}P\left(X<0\right)=P\left(z<\frac{0-27}{24}\right)\\ \approx P\left(z<-1.12\right)\end{array}$

Use the standard normal probabilities (cumulative z-curve area) table to find the z-value:

Procedure:

For z at –1.12:

• Locate –1.1 in the left column of the table.
• Obtain the value in the corresponding row below 0.02.

That is, $P\left(z<-1.12\right)=0.1314$.

Therefore, the percentage of the travel time that fall below 0 is 13.14% that is not possible.

Hence, the distribution of the travel time cannot be approximated by the normal curve.

e.

To determine

Find the percentage of travel time between 0 and 75 minutes using Chebyshev’s rule.

Obtain the percentage of travel time between 0 and 47 minutes using Chebyshev’s rule.

Answer to Problem 40E

The percentage of travel time between 0 and 75 minutes is at least 75%.

The percentage of travel time between 0 and 47 minutes is at least 21%.

Explanation of Solution

Calculation:

The approximate mean and standard deviation for the travel time distribution are 27 minutes and 24 minutes, respectively.

The values at 2 standard deviations away from the mean is obtained as follows:

$\begin{array}{c}\left\{\begin{array}{l}\text{The value at 2 standard deviation }\\ \text{away from the mean}\end{array}\right\}=\text{mean}\pm 2\text{standard devition}\\ =27\pm 2\left(24\right)\\ =27\pm 48\\ =\left(27-48,27+48\right)\\ =\left(-21\text{ minutes},75\text{ minutes}\right)\end{array}$

The observation of the travel time should not be negative. Therefore, the travel time that begins at –21 minutes lies in 2 standard deviation below the mean that is not possible. Travel time 75 minutes lies 2 standard deviations above from the mean.

Chebyshev’s rule:

For any number k, where $k\ge 1$, the percentage of the observation within k the standard deviations of the mean is at least $100\left(1-\frac{1}{k^2}\right)\%$.

The z-score gives the number of standard deviations that an observation of 75 minutes is away from the mean. Here, the z-scores for 0 minutes and 75 minutes are –2 and 2, respectively. Thus, these observations of time are 2 standard deviations from the mean that is $k=2$.

(i).

The percentage of travel time between 0 and 75 minutes is obtained as given below:

Use Chebyshev’s rule as follows:

$\begin{array}{c}100\left(1-\frac{1}{k^2}\right)\%=100\left(1-\frac{1}{2^2}\right)\%\\ =100\left(1-0.25\right)\%\\ =100\left(0.75\right)\%\\ =75\%\end{array}$

In this context, at least 0.75 proportion of observations lie within 2 standard deviations of the mean.

Therefore, the percentage of travel time between 0 and 75 minutes is 75%.

(ii).

A travel time of 47 minutes is $0.833\left(=\frac{47-27}{24}\right)$ standard deviations above the mean.

A travel time of 0 minutes is $1.125\left(=\frac{27}{24}\right)$ standard deviations below the mean.

Since 1.125 is larger than 0.833, these observations of time are 1.125 standard deviations from the mean that is $k=1.125$.

The percentage of the travel times between 0 and 47 minutes is obtained as given below:

Use Chebyshev’s rule as follows:

$\begin{array}{c}100\left(1-\frac{1}{k^2}\right)\%=100\left(1-\frac{1}{{1.125}^2}\right)\%\\ =100\left(1-0.79\right)\%\\ \approx 100\left(0.21\right)\%\\ =21\%\end{array}$

Thus, the percentage of the travel time between 0 and 47 minutes is 21%.

f.

To determine

Explain the way why the statements in Part (e) agree with the actual percentage for the travel time distribution.

Explanation of Solution

Calculation:

The actual percentage of travel time between 0 and 75 minutes is obtained as given below:

$\begin{array}{c}\left(\begin{array}{l}\text{Percentage between }\\ \text{0 and 75}\end{array}\right)=100\left(1-\left[\begin{array}{l}\text{interval between }\\ \text{90 or more}\end{array}\right]-\left\{\frac{\begin{array}{l}\text{interval between }\\ \text{60 to < 90}\end{array}}{2}\right\}\right)\%\\ =100\left(1-0.02-\frac{0.05}{2}\right)\%\\ =100\left(1-0.02-0.025\right)\%\\ =100\left(0.955\right)\%\\ =95.5\%\end{array}$

Therefore, the actual percentage of travel time between 0 and 75 minutes is approximately 95.5% and differs a lot from the percentage obtained using Chebyshev’s rule.

The actual percentage of travel times between 0 and 47 minutes is obtained as given below:

$\begin{array}{c}\left(\begin{array}{l}\text{Percentage between }\\ \text{0 and 47}\end{array}\right)=100\left(\begin{array}{l}1-\left[\begin{array}{l}\text{interval between }\\ \text{90 or more}\end{array}\right]-\left\{\begin{array}{l}\text{interval between }\\ \text{60 to < 90}\end{array}\right\}-\\ \left\{\frac{\begin{array}{l}\text{interval between }\\ \text{45 to < 60}\end{array}}{2}\right\}\end{array}\right)\%\\ =100\left(1-0.02-0.05-\frac{0.06}{2}\right)\%\\ =100\left(1-0.02-0.05-0.03\right)\%\\ =100\left(0.90\right)\%\\ =90\%\end{array}$

Therefore, the actual percentage of travel time between 0 and 47 minutes is approximately 90% and differs a lot from the percentage obtained using Chebyshev’s rule.