Chapter 14.2, Problem 21E

(a)

To determine

To find: The sum of the wages in each class.

Answer to Problem 21E

The sum of wages in each class.

Explanation of Solution

Given:

The given table is as:

Weekly Wages	Frequency
$\begin{array}{l}\$130\text{-}\$140\\ \$140\text{-}\$150\\ \$150\text{-}\$160\\ \$160\text{-}\$170\\ \$170\text{-}\$180\\ \$180\text{-}\$190\\ \$190\text{-}\$200\end{array}$	$\begin{array}{l}11\\ 24\\ 30\\ 10\\ 13\\ 8\\ 4\end{array}$

Calculation:

The sum is calculated as:

Multiply the average weekly wage by the frequency to obtain the sum of all wages for each class.

  $\begin{array}{c}\text{Sum of wages}=\text{Average weekly wage×Frequency}\\ 135\times 11=1485\\ 145\times 24=3480\\ 155\times 30=4650\end{array}$

Further simplified as:

  $\begin{array}{c}165\times 10=1650\\ 175\times 13=2275\\ 185\times 8=1480\\ 195\times 4=780\end{array}$

The sum of wages in each class.

(b)

To determine

To find: The sum of all of the wages in the frequency distribution.

Answer to Problem 21E

The sum of all of the wages in the frequency distribution is $15800$ .

Explanation of Solution

Sum the wages for each class to determine the total amount paid out in wages weekly. $\begin{array}{c}\text{Total wages}=1485+3480+4650+1650+2275+1480+780\\ =15800\end{array}$

The sum of all of the wages in the frequency distribution is $15800$ .

(c)

To determine

To find: The number of employees in the frequency distribution.

Answer to Problem 21E

The number of employees in the frequency distributionis $100$ .

Explanation of Solution

Sum the number of workers in each class together to determine the total number of workers:

  $\begin{array}{c}\text{Total workers}=11+24+30+10+13+8+4\\ =100\end{array}$

The number of employees in the frequency distribution is $100$ .

(d)

To determine

To find: The mean weekly wage in the frequency distribution.

Answer to Problem 21E

The mean weekly wage in the frequency distribution is $158$ .

Explanation of Solution

The mean is calculated as:

  $\begin{array}{c}\overline{X}=\frac{\text{Total amount paid out weekly wages}}{\text{number of employees to determine the average wage}}\\ =\frac{15800}{100}\\ =158\end{array}$

The mean weekly wage in the frequency distribution is $158$ .

(e)

To determine

To find: The median class of the frequency distribution.

Answer to Problem 21E

The median class of the frequency distributionis $150\text{-}160$ .

Explanation of Solution

Since there are one hundred employees, if the data is arranged sequentially according to weekly wages, the median would be average of the 50 and 51st value,both of which fall into the $150\text{-}160$ category.

The median class of the frequency distribution is $150\text{-}160$ .

(f)

To determine

To find: The median weekly wage in the frequency distribution.

Answer to Problem 21E

The median class of the frequency distribution is $150\text{-}160$ .

Explanation of Solution

Since there are one hundred employees, if the data is arranged sequentially according to weekly wages, the median would be average of the 50 and 51st value, both of which fall into the $150\text{-}160$ category. The average of this category would then be taken to determine the median value.

The median class of the frequency distribution is $150\text{-}160$ .

So the median weekly wage in the frequency distribution is:

  $\begin{array}{c}{M}_d=\frac{150+160}{2}\\ =\frac{310}{2}\\ =155\end{array}$

The median weekly wage in the frequency distribution is $155$ .

(g)

To determine

To explain: The reason of mean and median are good measures of central tendency in this situation.

Answer to Problem 21E

The reason is given below.

Explanation of Solution

Both the mean and median are good measures of central tendency in this situation because of their similarity in value. Both can be taken as a valid measure of central tendency.

The explanation is given above.