## Related questions with answers

Sample annual salaries (in thousands of dollars) for employees at a company are listed.

$\begin{array}{lllllllllllll}40 & 35 & 49 & 53 & 38 & 39 & 40 & 37 & 49 & 34 & 38 & 43 & 47\end{array}$

(a) Find the sample mean and sample standard deviation. (b) Each employee in the sample is given a $\$ 1000$ raise. Find the sample mean and sample standard deviation for the revised data set. (c) Each employee in the sample takes a pay cut of $\$ 2000$ from their original salary. Find the sample mean and sample standard deviation for the revised data set. (d) What can you conclude from the results of (a), (b), and (c)?

Solution

VerifiedGiven:

$40, 35, 49, 53, 38, 39, 40, 37, 49, 34, 38, 43, 47$

(a) The mean is the sum of all values divided by the number of values:

$\overline{x}=\dfrac{\sum x}{n}=\dfrac{40+35+49+ 53+38+39+40+37+49+34+38+43+47}{13}=\dfrac{542}{13}\approx 41.6923$

$n$ is the number of values in the data set.

$n=13$

The variance is the sum of squared deviations from the mean divided by $n-1$:

$\begin{align*} s^2&=\dfrac{\sum (x-\overline{x})^2}{n-1} \\ &=\dfrac{(40-41.6923)^2+(35-41.6923)^2+(49-41.6923)^2+(53-41.6923)^2+(38-41.6923)^2}{\:} \\ &\dfrac{+(39-41.6923)^2+(40-41.6923)^2+(37-41.6923)^2+(49-41.6923)^2}{\:} \\ &+\dfrac{+(34-41.6923)^2+(38-41.6923)^2+(43-41.6923)^2+(47-41.6923)^2}{13-1} \\ &\approx 35.8974 \end{align*}$

The standard deviation is the square root of the variance:

$s=\sqrt{35.8974}\approx 5.9914$

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