Chapter 2.1, Problem 8E

(a)

To determine

To find:

The class width for the given frequency distribution.

Answer to Problem 8E

Solution:

The class width of the given frequency distribution is $0.99$ which is approximately equal to $1.0$.

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
$10.50-11.49$	92
$11.50-12.49$	78
$12.50-13.49$	68
$13.50-14.49$	45
$14.50-15.49$	34

The upper limit of the distribution is given as $15.49$ and the lower limit is given as $10.50$. The number of classes is 5.

Formula used:

The formula to calculate the class width is,

$\text{Class}\text{width}=\frac{\text{Upper}\text{limit}-\text{Lower}\text{limit}}{\text{Number}\text{of}\text{class}}$

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute $15.49$ for upper limit and $10.50$ for lower limit and 5 for the number of classes in the formula,

$\begin{array}{rll}\text{Class}\text{width}& =& \frac{15.49-10.50}{5}\\ & =& \frac{4.99}{5}\\ & =& 0.998\\ & \approx & 1.0\end{array}$

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

Answer to Problem 8E

Solution:

The class boundary for each class of the given frequency distribution is,

Class	Frequency	Class boundaries
$10.50-11.49$	92	$10.495-11.495$
$11.50-12.49$	78	$11.495-12.495$
$12.50-13.49$	68	$12.495-13.495$
$13.50-14.49$	45	$13.495-14.495$
$14.50-15.49$	34	$14.495-15.495$

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
$10.50-11.49$	92
$11.50-12.49$	78
$12.50-13.49$	68
$13.50-14.49$	45
$14.50-15.49$	34

Formula used:

The formula to calculate the class boundary is,

$\text{Class}\text{boundary}=\frac{\text{Upper}\text{limit}\text{of}\text{one}\text{class}+\text{Lower}\text{limit}\text{of}\text{next}\text{class}}{\text{2}}$

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since, the largest number is 11.49 and 11.50 is the smallest number in the next class, then, substitute $11.49$ for upper limit and $11.50$ for lower limit in the formula,

$\begin{array}{rll}\text{Class}\text{boundary}& =& \frac{11.49+11.50}{2}\\ & =& \frac{22.99}{2}\\ & =& 11.495\end{array}$

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
$10.50-11.49$	92	$10.495-11.495$
$11.50-12.49$	78	$11.495-12.495$
$12.50-13.49$	68	$12.495-13.495$
$13.50-14.49$	45	$13.495-14.495$
$14.50-15.49$	34	$14.495-15.495$

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

Answer to Problem 8E

Solution:

The midpoint of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint
$10.50-11.49$	92	$10.495-11.495$	$10.995$
$11.50-12.49$	78	$11.495-12.495$	$11.995$
$12.50-13.49$	68	$12.495-13.495$	$12.995$
$13.50-14.49$	45	$13.495-14.495$	$13.995$
$14.50-15.49$	34	$14.495-15.495$	$14.995$

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
$10.50-11.49$	92
$11.50-12.49$	78
$12.50-13.49$	68
$13.50-14.49$	45
$14.50-15.49$	34

Formula used:

The formula to calculate the midpoint of any class is,

$\text{Midpoint}=\frac{\text{Upper}\text{limit}+\text{Lower}\text{limit}}{\text{2}}$

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute $11.49$ for upper limit and $10.50$ for lower limit in the formula,

$\begin{array}{rll}\text{Midpoint}& =& \frac{11.49+10.50}{2}\\ & =& \frac{21.99}{2}\\ & =& 10.995\end{array}$

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
$10.50-11.49$	92	$10.495-11.495$	$10.995$
$11.50-12.49$	78	$11.495-12.495$	$11.995$
$12.50-13.49$	68	$12.495-13.495$	$12.995$
$13.50-14.49$	45	$13.495-14.495$	$13.995$
$14.50-15.49$	34	$14.495-15.495$	$14.995$

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

Answer to Problem 8E

Solution:

The relative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency
$10.50-11.49$	92	$10.495-11.495$	$10.995$	$\frac{92}{317}=29\%$
$11.50-12.49$	78	$11.495-12.495$	$11.995$	$\frac{78}{317}=25\%$
$12.50-13.49$	68	$12.495-13.495$	$12.995$	$\frac{68}{317}=21\%$
$13.50-14.49$	45	$13.495-14.495$	$13.995$	$\frac{45}{317}=14\%$
$14.50-15.49$	34	$14.495-15.495$	$14.995$	$\frac{34}{317}=11\%$

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
$10.50-11.49$	92
$11.50-12.49$	78
$12.50-13.49$	68
$13.50-14.49$	45
$14.50-15.49$	34

Formula used:

The formula to calculate the relative frequency of any class is,

$\text{Relative}\text{frequency}=\frac{f}{N}$

Here $N$ is the sum of the frequencies.

Calculation:

The sum of the frequencies is,

$\begin{array}{rll}N& =& 92+78+68+45+34\\ & =& 317\end{array}$

For the first class,

Substitute 92 for $f$ and 317 for $N$ in the formula,

$\begin{array}{rll}\text{Relative frequency}& =& \frac{92}{317}\\ & =& 0.29\\ & =& 29\%\end{array}$

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
$10.50-11.49$	92	$10.495-11.495$	$10.995$	$\frac{92}{317}=29\%$
$11.50-12.49$	78	$11.495-12.495$	$11.995$	$\frac{78}{317}=25\%$
$12.50-13.49$	68	$12.495-13.495$	$12.995$	$\frac{68}{317}=21\%$
$13.50-14.49$	45	$13.495-14.495$	$13.995$	$\frac{45}{317}=14\%$
$14.50-15.49$	34	$14.495-15.495$	$14.995$	$\frac{34}{317}=11\%$

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

Answer to Problem 8E

Solution:

The cumulative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
$10.50-11.49$	92	$10.495-11.495$	$10.995$	$\frac{92}{317}=29\%$	92
$11.50-12.49$	78	$11.495-12.495$	$11.995$	$\frac{78}{317}=25\%$	$92+78=170$
$12.50-13.49$	68	$12.495-13.495$	$12.995$	$\frac{68}{317}=21\%$	$170+68=238$
$13.50-14.49$	45	$13.495-14.495$	$13.995$	$\frac{45}{317}=14\%$	$238+45=283$
$14.50-15.49$	34	$14.495-15.495$	$14.995$	$\frac{34}{317}=11\%$	$283+34=317$

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
$10.50-11.49$	92
$11.50-12.49$	78
$12.50-13.49$	68
$13.50-14.49$	45
$14.50-15.49$	34

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
$10.50-11.49$	92	$10.495-11.495$	$10.995$	$\frac{92}{317}=29\%$	92
$11.50-12.49$	78	$11.495-12.495$	$11.995$	$\frac{78}{317}=25\%$	$92+78=170$
$12.50-13.49$	68	$12.495-13.495$	$12.995$	$\frac{68}{317}=21\%$	$170+68=238$
$13.50-14.49$	45	$13.495-14.495$	$13.995$	$\frac{45}{317}=14\%$	$238+45=283$
$14.50-15.49$	34	$14.495-15.495$	$14.995$	$\frac{34}{317}=11\%$	$283+34=317$