Chapter 2, Problem 3SR

a.

To determine

Check whether 7 classes would be used to create a frequency distribution using the $2^k$ rule.

Answer to Problem 3SR

The number of classes for the frequency distribution that consists of 73 observations is 7.

Explanation of Solution

Selection of number of classes:

The “2 to the k rule” suggests that the number of classes is the smallest value of k, where $2^k$ is greater than the number of observations.

It is given that the data set consists of 73 observations. The value of k can be obtained as follows:

$\begin{array}{l}{2}^6=64<73\\ 2^7=128>73\end{array}$

Here, $k=7$ is the smallest value for which $2^k$ is greater than the number of observations.

Therefore, the number of classes for the frequency distribution that consists of 73 observations is 7.

b.

To determine

Check whether the class interval of 30 would summarize the data in 7 classes.

Answer to Problem 3SR

The reasonable class interval is 30.

Explanation of Solution

From the given data set, the maximum and minimum values are 488 and 320, respectively.

The formula for the class interval is given as follows:

$i\ge \frac{\left(\text{Maximum value}-\text{Minimum value}\right)}{k}$

Where, i is the class interval and k is the number of classes.

Therefore, the class interval for the given data can be obtained as follows:

$\begin{array}{c}i\ge \frac{\left(\text{Maximum value}-\text{Minimum value}\right)}{k}\\ \ge \frac{488-320}{7}\\ \ge \frac{168}{7}\\ \ge 24\end{array}$

In practice, the class interval size is usually rounded up to some convenient number such as a multiple of 10 or 100. Therefore, the reasonable class interval is 30.

c.

To determine

Construct a frequency and relative frequency table for the given data.

Answer to Problem 3SR

The frequency and relative frequency table for the given data are as follows:

Class	Frequency	Relative Frequency
300-330	2	$\frac{2}{73}\approx 0.0274$
330-360	2	$\frac{2}{73}\approx 0.0274$
360-390	17	$\frac{17}{73}\approx 0.2329$
390-420	27	$\frac{27}{73}\approx 0.3697$
420-450	22	$\frac{22}{73}\approx 0.3014$
450-480	1	$\frac{1}{73}\approx 0.0137$
480-510	2	$\frac{2}{73}\approx 0.0274$
Total	73	1

Explanation of Solution

Frequency table:

The frequency table is a collection of mutually exclusive and exhaustive classes that show the number of observations in each class.

It is given that the lower limit of the first class is 300. From Part (b), the class interval is 30. The frequency table of the 73 home runs can be constructed as follows:

Class	Frequency	Relative Frequency
300-330	2	$\frac{2}{73}\approx 0.0274$
330-360	2	$\frac{2}{73}\approx 0.0274$
360-390	17	$\frac{17}{73}\approx 0.2329$
390-420	27	$\frac{27}{73}\approx 0.3697$
420-450	22	$\frac{22}{73}\approx 0.3014$
450-480	1	$\frac{1}{73}\approx 0.0137$
480-510	2	$\frac{2}{73}\approx 0.0274$
Total	73	1

d.

To determine

Find the number of home runs traveled at a distance of 360 up to 390 feet.

Answer to Problem 3SR

The number of home runs traveled at a distance of 360 up to 390 feet is 17.

Explanation of Solution

From the frequency table in Part (c), the frequency value of the class 360-390 is 17. Therefore, the number of home runs traveled at a distance of 360 up to 390 feet is 17.

e.

To determine

Calculate the percentage of the number of home runs traveled at a distance of 360 up to 390 feet.

Answer to Problem 3SR

The percentage of the number of home runs traveled at a distance of 360 up to 390 feet is 23.288%.

Explanation of Solution

From the frequency table in Part (c), the frequency value of the class 360-390 is 17. The total number of observations is 73. The value of 17 home runs traveled can be converted into percentage as follows:

$\begin{array}{c}\text{Percentage of home runs traveled}=\frac{17}{73}\times 100\\ =23.288\%\end{array}$

Therefore, percentage of the number of home runs traveled at a distance of 360 up to 390 feet is 23.288%.

f.

To determine

Calculate the percentage of the number of home runs traveled at a distance of 390 feet or more.

Answer to Problem 3SR

The percentage of the number of home runs traveled at a distance of 390 feet or more is 71.233%.

Explanation of Solution

From the frequency table in Part (c), the sum of all the frequencies of the classes 390-420 and above is given below:

$\begin{array}{c}\left(\begin{array}{c}\text{Sum of frequencies of}\\ \left(\text{390-420}\right)\text{ and above}\end{array}\right)=27+22+1+2\\ =52\end{array}$

The total number of observations is 73. The value of 52 home runs traveled can be converted into percentage as follows:

$\begin{array}{c}\text{Percentage of home runs traveled}=\frac{52}{73}\times 100\\ =71.233\%\end{array}$

Therefore, percentage of the number of home runs traveled at a distance of 390 feet or more is 71.233%.