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

Explanation:

Step1: Calculate the sample variance

The formula for sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\), where \(n\) is the number of data points, \(\bar{x}\) is the sample mean.
First, \(n = 12\), \(\bar{x}=45.8\)
\(\sum_{i = 1}^{12}(x_{i}-\bar{x})^{2}=(50 - 45.8)^{2}+(44 - 45.8)^{2}+(53 - 45.8)^{2}+(47 - 45.8)^{2}+(39 - 45.8)^{2}+(39 - 45.8)^{2}+(50 - 45.8)^{2}+(44 - 45.8)^{2}+(53 - 45.8)^{2}+(32 - 45.8)^{2}+(47 - 45.8)^{2}+(50 - 45.8)^{2}+(48 - 45.8)^{2}\)
\(=(4.2)^{2}+(-1.8)^{2}+(7.2)^{2}+(1.2)^{2}+(-6.8)^{2}+(-6.8)^{2}+(4.2)^{2}+(-1.8)^{2}+(7.2)^{2}+(-13.8)^{2}+(1.2)^{2}+(4.2)^{2}+(2.2)^{2}\)
\(=17.64+3.24 + 51.84+1.44+46.24+46.24+17.64+3.24+51.84+190.44+1.44+17.64+4.84\)
\(=453.6\)
\(s^{2}=\frac{453.6}{12 - 1}=\frac{453.6}{11}\approx41.24\)

Step2: Calculate the sample standard deviation

The formula for sample standard deviation \(s=\sqrt{s^{2}}\)
Since \(s^{2}\approx41.24\), then \(s=\sqrt{41.24}\approx6.4\)

Answer:

The sample standard deviation is \(s = 6.4\) thousand dollars.