Question

sample annual salaries (in thousands of dollars) for employees at a company are listed.
50 44 53 47 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 } = \square \\) thousand dollars.
(round to one decimal place as needed.)

Explanation:

Step1: Recall the property of mean when each data is increased

If each data value \(x_i\) in a data - set is increased by a constant \(c\), the new mean \(\bar{y}\) of the data - set \(y_i=x_i + c\) is given by \(\bar{y}=\bar{x}+c\). Here \(c = 2\) (since the raise is \(2000\) and the original data is in thousands of dollars). The original mean \(\bar{x}=45.8\).
\(\bar{y}=45.8 + 2\)

Step2: Recall the property of standard deviation when each data is increased

If each data value \(x_i\) in a data - set is increased by a constant \(c\), the standard deviation \(s_y\) of the data - set \(y_i=x_i + c\) is the same as the standard deviation \(s_x\) of the original data - set. The formula for the sample standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\). When \(y_i=x_i + c\), \(\bar{y}=\bar{x}+c\), and \(y_i-\bar{y}=(x_i + c)-(\bar{x}+c)=x_i-\bar{x}\). So \(s_y=\sqrt{\frac{\sum_{i = 1}^{n}(y_i-\bar{y})^2}{n - 1}}=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}=s_x\)

Answer:

The sample mean is \(47.8\) thousand dollars and the sample standard deviation is \(6.1\) thousand dollars.