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 = 6.1\\) thousand dollars.
(round to one decimal place as needed.)

(b) the sample mean is \\(\bar{x} = 48.2\\) thousand dollars.
(round to one decimal place as needed.)

the sample standard deviation is \\(s = 6.0\\) thousand dollars.
(round to one decimal place as needed.)

(c) the sample mean is \\(\bar{x} = 40.2\\) 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 original sample statistics

Using the Mean Calculation and Sample Standard Deviation knowledge points
The sample data (in thousands of dollars) is:

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

The sample size is \(n = 13\).
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 \(\sum (x - \bar{x})^2\) is:

$$ \sum (x - \bar{x})^2 \approx 440.3077 $$

The sample variance is:

$$ s^2 = \frac{440.3077}{12} \approx 36.6923 $$

The sample standard deviation is:

$$ s = \sqrt{36.6923} \approx 6.057 \approx 6.1 $$

Analyze effect of adding a constant

Using the Linear Transformation of Data knowledge point
Adding a constant \(c = 3\) to each data point:

$$ \bar{x}_{\text{new}} = \bar{x} + 3 = 45.2 + 3 = 48.2 $$

$$ s_{\text{new}} = s = 6.1 $$

Analyze effect of subtracting a constant

Using the Linear Transformation of Data knowledge point
Subtracting a constant \(c = 5\) from each data point:

$$ \bar{x}_{\text{new}} = \bar{x} - 5 = 45.2 - 5 = 40.2 $$

$$ s_{\text{new}} = s = 6.1 $$

Answer:

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