Question

sample annual salaries (in thousands of dollars) for employees at a company are listed.
50 44 53 47 39 39 39 50 44 53 32 47 50 48
(a) find the sample mean and sample standard deviation.
(b) each employee in the sample is given a $2000 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 the sample standard deviation for the revised data set
(d) what can you conclude from the results of (a), (b), and (c)?
(a) the sample mean is \\( \bar { x } = 45.8 \\) thousand dollars.
(round to one decimal place as needed.)
the sample standard deviation is \\( s = 6.1 \\) thousand dollars.
(round to one decimal place as needed.)
(b) the sample mean is \\( \bar { x } = 47.8 \\) thousand dollars.
(round to one decimal place as needed.)
the sample standard deviation is \\( s = 6.1 \\) thousand dollars.
(round to one decimal place as needed.)
(c) the sample mean \\( \bar { x } = \square \\) thousand dollars
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for the mean

The formula for the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). When each data - point \(x_{i}\) is changed to \(y_{i}=x_{i}-2\) (since \(2000\) dollars is \(2\) thousand dollars), the new mean \(\bar{y}\) is given by \(\bar{y}=\frac{\sum_{i = 1}^{n}(x_{i}- 2)}{n}\).
Using the property of summation \(\sum_{i = 1}^{n}(x_{i}-2)=\sum_{i = 1}^{n}x_{i}-2n\). Then \(\bar{y}=\frac{\sum_{i = 1}^{n}x_{i}-2n}{n}=\frac{\sum_{i = 1}^{n}x_{i}}{n}-2\).
We know from part (a) that \(\bar{x} = 45.8\).

Step2: Calculate the new mean

Substitute \(\bar{x}\) into the formula for \(\bar{y}\). So \(\bar{y}=\bar{x}-2\).
Since \(\bar{x}=45.8\), then \(\bar{y}=45.8 - 2=43.8\).

Step3: Recall the formula for the standard deviation

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\). When \(y_{i}=x_{i}-2\), the new standard deviation \(s_{y}=\sqrt{\frac{\sum_{i = 1}^{n}[(x_{i}-2)-(\bar{x}-2)]^{2}}{n - 1}}\).
Simplify the expression inside the square - root: \((x_{i}-2)-(\bar{x}-2)=x_{i}-\bar{x}\). So \(s_{y}=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\).

Answer:

The sample mean is \(43.8\) thousand dollars and the sample standard deviation is \(6.1\) thousand dollars.