Chapter 3.1, Problem 25E

To determine

To find:

The mean hourly wage for the given data.

Answer to Problem 25E

Solution:

The mean hourly wage for the given data is $10.20.

Explanation of Solution

Given information:

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

Definition:

The weighted mean is the mean of a data set in which each data value in the set does not hold the same relative importance.

Formula used:

When upper limit of the first class is not equal to lower limit of second class.

Class mid point is calculated.

The formulae to calculate class mid point is,

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

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 $11.50\text{-}12.49$,

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

Construct the table class mid-point.

Age at time of First Marriage (in Years)
Class	Mid-point	Frequency
$7.50-8.49$	7.995	12
$8.50-9.49$	8.995	50
$9.50-10.49$	9.995	48
$10.50-11.49$	10.995	45
$11.50-12.49$	11.995	34

Table 1

Formula used:

The formula to calculate the weighted mean of set of data is,

$\begin{array}{rcl}\text{Weighted}\text{Mean}\left(\overline{x}\right)& =& \frac{w_1{x}_1+w_2{x}_2+\mathrm{...}+w_n{x}_n}{w_1+w_2+\mathrm{...}+w_n}\\ & =& \frac{{\displaystyle \sum _{i=1}^n}\left(w_i{x}_i\right)}{{\displaystyle \sum _{i=1}^n{w}_i}}\end{array}$

Where, $x_i$ is the $i^{\text{th}}$ data value and $w_i$ is the weight of the $i^{\text{th}}$ data value and $n$ is the total number of collection.

Calculation:

Substitute. $7.995,8.995,9.995,10.995,11.995$ for $x_1,x_2,x_3,x_4,x_5$ and $12,50,48,45,34$ for $w_1,w_2,w_3,w_4,w_5$ respectively in the formula.

$\begin{array}{rll}\text{Weighted}\text{Mean}\left(\overline{x}\right)& =& \frac{w_1{x}_1+w_2{x}_2+\mathrm{...}+w_n{x}_n}{w_1+w_2+\mathrm{...}+w_n}\\ & =& \frac{\left(7.995\times 12\right)+\left(8.995\times 50\right)+\left(9.995\times 48\right)+\left(10.995\times 45\right)+\left(11.995\times 34\right)}{12+45+50+48+34}\\ & =& \frac{95.94+449.75+479.76+494.775+407.83}{189}\\ & =& 10.20\end{array}$

The mean hourly wage for the given data is $10.20.