Chapter 2.2, Problem 11E

(a)

To determine

To graph:

A histogram for the given data.

Explanation of Solution

Given information:

Hourly Wage at First Job (in Dollars)
Class	Frequency
$7.50\text{-}8.49$	12
$8.50\text{-}9.49$	50
$9.50\text{-}10.49$	48
$10.50\text{-}11.49$	45
$11.50\text{-}12.49$	34

Formula used:

Class mid-point $=\frac{\text{upper}\text{limit}\text{of}\text{class}+\text{lower}\text{limit}\text{of}\text{class}}{2}$

$p$$=\frac{\text{lower}\text{limit}\text{of}\text{succeding}\text{class}-\text{upper}\text{limit}\text{of}\text{the}\text{class}}{2}$

For adjustment of class boundaries subtract $p$ from lower limit and $p$ with upper limit of each class.

Calculation:

Compute the class mid-point of each class.

For the class $7.50\text{-}8.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{7.50+8.49}{2}\\ & =& 7.995\end{array}$

For the class $8.50\text{-}9.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{8.50+9.49}{2}\\ & =& 8.995\end{array}$

For the class $9.50\text{-}10.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{9.50+10.49}{2}\\ & =& 9.995\end{array}$

For the class $10.50\text{-}11.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{10.50+11.49}{2}\\ & =& 10.995\end{array}$

For the class $11.50\text{-}12.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{11.50+12.49}{2}\\ & =& 11.995\end{array}$

Calculate the adjustment factor.

$\begin{array}{rcl}p& =& \frac{{\displaystyle 8.50-8.49}}{{\displaystyle 2}}\\ & =& \frac{{\displaystyle .01}}{{\displaystyle 2}}\\ & =& 0.005\end{array}$

Construct the table with adjusted class boundary and class mid-point.

Age at time of First Marriage (in Years)
Class boundary	Mid-point	Frequency
$7.495\text{-}8.495$	7.995	12
$8.495\text{-}9.495$	8.995	50
$9.495\text{-}10.495$	9.995	48
$10.495\text{-}11.495$	10.995	45
$11.495\text{-}12.495$	11.995	34

Table 1

Graph:

Construct the histogram corresponding to the table 1.

Figure 1

Interpretation:

Figure 1 represents the histogram for the given data.

Statistics Concept Introduction

A histogram is a bar graph of a frequency distribution of quantitative data.

Put classes along $x\text{-axis}$ and frequency along $y\text{-axis}$. Mark class mid- point of every class along $x\text{-axis}$. The width of each bar represents the width of each class. Width of each class should be same and each bar should touch each other.

(b)

To determine

To calculate:

The relative frequency for each class.

Answer to Problem 11E

Solution:

Required relative frequency table is,

Class	Relative Frequency
$7.50\text{-}8.49$	$6\%$
$8.50\text{-}9.49$	$26\%$
$9.50\text{-}10.49$	$25\%$
$10.50\text{-}11.49$	$25\%$
$11.50\text{-}12.49$	$18\%$

Explanation of Solution

Given information:

Hourly Wage at First Job (in Dollars)
Class	Frequency
$7.50\text{-}8.49$	12
$8.50\text{-}9.49$	50
$9.50\text{-}10.49$	48
$10.50\text{-}11.49$	45
$11.50\text{-}12.49$	34

Formula used:

Relative Frequency

$=\frac{f}{n}$

$\begin{array}{l}n=\text{sum}\text{of}\text{frequencies}\\ f=\text{frequency}\text{for}\text{the}\text{class}\end{array}$

Calculation:

Compute $n:$.

$\begin{array}{rll}n& =& 12+50+48+45+34\\ & =& 189\end{array}$

Compute relative frequencies for each class in the following table.

Class	Frequency	Relative Frequency
$7.50\text{-}8.49$	12	$\frac{f}{n}=\frac{12}{189}\approx 0.06=6\%$
$8.50\text{-}9.49$	50	$\frac{f}{n}=\frac{50}{189}\approx 0.26=26\%$
$9.50\text{-}10.49$	48	$\frac{f}{n}=\frac{48}{189}\approx 0.25=25\%$
$10.50\text{-}11.49$	45	$\frac{f}{n}=\frac{45}{189}\approx 0.24=24\%$
$11.50\text{-}12.49$	34	$\frac{f}{n}=\frac{34}{189}\approx 0.18=18\%$

Table 2

Conclusion:

Thus, the required relative frequency table is,

Class	Relative Frequency
$7.50\text{-}8.49$	$6\%$
$8.50\text{-}9.49$	$26\%$
$9.50\text{-}10.49$	$25\%$
$10.50\text{-}11.49$	$24\%$
$11.50\text{-}12.49$	$18\%$

(c)

To determine

To graph:

A relative frequency histogram for the given data.

Explanation of Solution

Given information:

Hourly Wage at First Job (in Dollars)
Class	Relative Frequency
$7.50\text{-}8.49$	$6\%$
$8.50\text{-}9.49$	$26\%$
$9.50\text{-}10.49$	$25\%$
$10.50\text{-}11.49$	$24\%$
$11.50\text{-}12.49$	$18\%$

Formula used:

Class mid-point $=\frac{\text{upper}\text{limit}\text{of}\text{class}+\text{lower}\text{limit}\text{of}\text{class}}{2}$

$p$$=\frac{\text{lower}\text{limit}\text{of}\text{succeding}\text{class}-\text{upper}\text{limit}\text{of}\text{the}\text{class}}{2}$

For adjustment of class boundaries subtract $p$ from lower limit and $p$ with upper limit of each class.

Calculation:

Compute the class mid-point for each class.

For the class $7.50\text{-}8.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{7.50+8.49}{2}\\ & =& 7.995\end{array}$

For the class $8.50\text{-}9.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{8.50+9.49}{2}\\ & =& 8.995\end{array}$

For the class $9.50\text{-}10.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{9.50+10.49}{2}\\ & =& 9.995\end{array}$

For the class $10.50\text{-}11.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{10.50+11.49}{2}\\ & =& 10.995\end{array}$

For the class $11.50\text{-}12.49$,

$\begin{array}{rcl}\text{Class mid‐point}& =& \frac{\text{upper\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}+\text{lower\hspace{0.17em}limit\hspace{0.17em}of\hspace{0.17em}class}}{2}\\ & =& \frac{11.50+12.49}{2}\\ & =& 11.995\end{array}$

Calculate the adjustment factor.

$\begin{array}{rcl}p& =& \frac{{\displaystyle 8.50-8.49}}{{\displaystyle 2}}\\ & =& \frac{{\displaystyle .01}}{{\displaystyle 2}}\\ & =& 0.005\end{array}$

Construct the table with adjusted class boundary and class mid-point.

Age at time of First Marriage (in Years)
Class boundary	Mid-point	Relative Frequency
$7.495\text{-}8.495$	7.995	$6\%$
$8.495\text{-}9.495$	8.995	$26\%$
$9.495\text{-}10.495$	9.995	$25\%$
$10.495\text{-}11.495$	10.995	$24\%$
$11.495\text{-}12.495$	11.995	$18\%$

Table 3

Graph:

Construct the histogram corresponding to the table 3.

Figure 2

Interpretation:

Figure 2 represents the relative frequency histogram for the given data.

Statistics Concept Introduction

A histogram is a bar graph of a frequency distribution of quantitative data.

Put classes along $x\text{-axis}$ and relative frequency along $y\text{-axis}$. Mark class mid- point of every class along $x\text{-axis}$. The width of each bar represents the width of each class. Width of each class should be same and each bar should touch each other.