Chapter 2.1, Problem 22E

(a)

To determine

To find: the result of the mean salary and median salary if the teacher receives high salary.

Answer to Problem 22E

The new median of the salary is 1.05 times of the old median of the salary and the new mean salary is 1.05 times of the old mean salary.

Explanation of Solution

Given:

Teacher receives a 5% raise instead of a flat $1000 raise.

The range of amount will be $1400 to $3000.

Calculation:

As each teacher receives a 5% raise, the new mean and median are obtained as follows:

  $\begin{array}{l}\text{Mean}\left(\text{new}\right)\text{=}\frac{{\displaystyle \sum \left(\text{salary of each teacher+salary of each teacher×0}\text{.05}\right)}}{\text{number of teacher}}\\ =\frac{{\displaystyle \sum \left(\text{salary of each teacher}\times \left(1+0.05\right)\right)}}{\text{number of teachers}}\\ =1.05\left(\frac{{\displaystyle \sum \text{salary of each teacher}}}{\text{number of teachers}}\right)\\ =1.05\times \text{Mean}\left(\text{old}\right)\end{array}$

Therefore the new mean salary is 1.05 times of the old mean salary.

We know that change of scale does not affect the median position. So the new median of the salary is the value of the old median salary raise by 5%. That means

  $\begin{array}{l}\text{Median}\left(\text{new}\right)\text{=Median}\left(\text{old}\right)\text{+0}\text{.05}\times \text{Median}\left(\text{old}\right)\\ =1.05\times \text{Median}\left(\text{old}\right)\end{array}$

Therefore the new median of the salary is 1.05 times of the old median of the salary.

Conclusion:

Therefore, the new median of the salary is 1.05 times of the old median of the salary and the new mean salary is 1.05 times of the old mean salary.

(b)

To determine

To explain:whether there will be increase inIQR and the standard deviation if there is a 5% raise.

Answer to Problem 22E

The new inter-quartile range is 1.05 times of the old inter-quartile range and the standard deviation of the new salary is 1.05 times of the old salary's standard deviation.

Explanation of Solution

Calculation:

As each teacher receives a 5% raise, the new standard deviation is obtained as follows:

  $\begin{array}{l}\text{std}\text{.dev}\text{.}\left(\text{new}\right)=\sqrt{\frac{{\displaystyle \sum {\left[\left(1.05\times \text{salary of each teacher}\right)-\left(1.05\times \text{Mean}\left(old\right)\right)\right]}^2}}{\left(n-1\right)}}\\ =\sqrt{\frac{{1.05}^2{\displaystyle \sum {\left[\left(\text{salary of each teacher}\right)-\left(\text{Mean}\left(old\right)\right)\right]}^2}}{\left(n-1\right)}}\\ =1.05\times =\sqrt{\frac{{\displaystyle \sum {\left[\left(\text{salary of each teacher}\right)-\left(\text{Mean}\left(old\right)\right)\right]}^2}}{\left(n-1\right)}}\\ =1.05\times \text{std}\text{.dev}\text{.}\left(\text{old}\right)\end{array}$

Therefore the standard deviation of the new salary is 1.05 times of the old salary's standard deviation.

The first and third quartiles are still the medians of the first and second halves of the data, the change of scale does not affect the median position.

Thus

  $\begin{array}{l}{Q}_1\left(new\right)=Q_1\left(old\right)+0.05\times Q_1\left(old\right)\\ =1.05\times Q_1\left(old\right)\\ Q_3\left(new\right)=Q_3\left(old\right)+0.05\times Q_3\left(old\right)\\ =1.05\times Q_3\left(old\right)\end{array}$

Thus the new inter quartile range is obtained as follows:

  $\begin{array}{l}\text{IQR}\left(new\right)=Q_3\left(new\right)-Q_1\left(new\right)\hfill \\ \begin{array}{l}=1.05\times Q_3\left(old\right)-1.05\times Q_1\left(old\right)\\ =1.05\times \left(Q_3\left(old\right)-Q_1\left(old\right)\right)\end{array}\hfill \\ =1.05\times \text{IQR}\left(old\right)\hfill \end{array}$

Conclusion:

Therefore the new inter-quartile range is 1.05 times of the old inter-quartile range and the standard deviation of the new salary is 1.05 times of the old salary's standard deviation.