Question

sample annual salaries (in thousands of dollars) for employees at a company are listed.
49 42 46 54 40 40 49 42 46 32 54 49 45

(a) find the sample mean and sample standard deviation.
(b) each employee in the sample is given a $3000 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 $5000 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)?

(a) the sample mean is \\(\bar{x} = 45.2\\) thousand dollars.
(round to one decimal place as needed.)

the sample standard deviation is \\(s = \\) thousand dollars.
(round to one decimal place as needed.)

Explanation:

Calculate sample statistics

Using the Mean Calculation and Sample Standard Deviation knowledge points
The sample data set consists of \(n = 13\) values:

$$ \{49, 42, 46, 54, 40, 40, 49, 42, 46, 32, 54, 49, 45\} $$

The sum of the values is:

$$ \sum x = 49 + 42 + 46 + 54 + 40 + 40 + 49 + 42 + 46 + 32 + 54 + 49 + 45 = 588 $$

The sample mean is:

$$ \bar{x} = \frac{588}{13} \approx 45.23 \approx 45.2 $$

The sum of squared differences from the mean is:

$$ \sum (x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n} $$

$$ \sum x^2 = 49^2 + 42^2 + 46^2 + 54^2 + 40^2 + 40^2 + 49^2 + 42^2 + 46^2 + 32^2 + 54^2 + 49^2 + 45^2 = 27048 $$

$$ \sum (x - \bar{x})^2 = 27048 - \frac{588^2}{13} = 27048 - \frac{345744}{13} \approx 27048 - 26595.6923 = 452.3077 $$

The sample standard deviation is:

$$ s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}} = \sqrt{\frac{452.3077}{12}} \approx \sqrt{37.6923} \approx 6.139 \approx 6.1 $$

Analyze linear transformations

Using the Linear Transformation of Data knowledge point
Adding a constant \(c\) to each data value increases the mean by \(c\) but leaves the standard deviation unchanged:

$$ \bar{x}_{\text{new}} = \bar{x} + c, \quad s_{\text{new}} = s $$

Subtracting a constant \(c\) from each data value decreases the mean by \(c\) but leaves the standard deviation unchanged:

$$ \bar{x}_{\text{new}} = \bar{x} - c, \quad s_{\text{new}} = s $$

Answer:

The sample standard deviation is s = <blank>\(6.1\)</blank> thousand dollars.