Question

2-10. Suppose that everyone in a company receives a pay raise of $200 per month. How does that affect the mean monthly pay for that organization? How does it affect the stan- dard deviation of monthly pay?

Answer 1

2.10) Let $x_1,x_2,...,x_n$ be the monthly pay received by n employees of the company.

The mean monthly pay for the organization is

$\bar{x}=\frac{\sum_{i=1}^n x_i}{n}$

The standard deviation of monthly pay is

$s=\sqrt\frac{\sum_{i=1}^n (x_i-\bar{x})^2}{n-1}$

Everyone in the company receives a pay raise of $200 per month

Now the new values are $x_1+200,x_2+200,...,x_n+200$

The new mean monthly pay for the organization is

$\begin{align*} \bar{x}_n&=\frac{\sum_{i=1}^n (x_i+200)}{n}\\ &=\frac{\sum_{i=1}^n x_i+\sum_{i=1}^n200}{n}\\ &=\frac{\sum_{i=1}^n x_i}{n}+\frac{n\times 200}{n}\\ &=\bar{x}+200 \end{align*}$

ans:the mean monthly pay of the organization increases by $200

The new standard deviation of monthly pay is

$\begin{align*} s_n&=\sqrt\frac{\sum_{i=1}^n (x_i+200-\bar{x}_n)^2}{n-1}\\ &=\sqrt\frac{\sum_{i=1}^n (x_i+200-(\bar{x}+200))^2}{n-1}\\ &=\sqrt\frac{\sum_{i=1}^n (x_i-\bar{x})^2}{n-1}\\ &=s \end{align*}$

ans:the standard deviation of monthly pay does not change

2.13)Let $y_1,y_2,...,y_n$ be the annual pay received by n employees of the company.

The mean annual pay for the organization is

$\bar{y}=\frac{\sum_{i=1}^n y_i}{n}$

The standard deviation of annual pay is

$s=\sqrt\frac{\sum_{i=1}^n (y_i-\bar{y})^2}{n-1}$

Everyone receives an annual pay raise of 5%. The new annual pay is $1.05\times y_1,1.05\times y_2,...,1.05\times y_n$

The new mean annual pay for the organization is

$\bar{y}_n=\frac{\sum_{i=1}^n 1.05\times y_i}{n}=\frac{1.05\times\sum_{i=1}^n y_i}{n}=1.05\bar{y}$

ans: The new mean annual pay for the organization raises by 1.05 times, that is raises by 5%.

The new standard deviation of annual pay is

$\begin{align*} s_n&=\sqrt\frac{\sum_{i=1}^n (1.05\times y_i-\bar{y}_n)^2}{n-1}\\ &=\sqrt\frac{\sum_{i=1}^n (1.05\times y_i-1.05\times \bar{y})^2}{n-1}\\ &=\sqrt\frac{\sum_{i=1}^n 1.05^2(y_i-\bar{y})^2}{n-1}\\ &=\sqrt\frac{1.05^2\sum_{i=1}^n (y_i-\bar{y})^2}{n-1}\\ &=1.05\sqrt\frac{\sum_{i=1}^n (y_i-\bar{y})^2}{n-1}\\ &=10.5s \end{align*}$

ans: The new standard deviation of annual pay for the organization raises by 1.05 times, that is raises by 5%.