Question

sample annual salaries (in thousands of dollars) for employees at a company are listed. 37 32 49 63 30 30 37 32 49 26 63 37 46 (a) find the sample mean and sample standard deviation. (b) each employee in the sample is given a 6% raise. find the sample mean and sample standard deviation for the revised data set (c) to calculate the monthly salary, divide each original salary by 12. 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 \\( \overline { x } = 40.8 \\) thousand dollars. (round to one decimal place as needed.) the sample standard deviation is \\( s = 12.2 \\) thousand dollars. (round to one decimal place as needed.) (b) the sample mean is \\( \overline { x } = 43.3 \\) thousand dollars. (round to one decimal place as needed.) the sample standard deviation is \\( s = 13 \\) thousand dollars. (round to one decimal place as needed.) (c) the sample mean is \\( \overline { x } = \\) thousand dollars. (round to one decimal place as needed.)

Explanation:

Step1: Recall the property of mean and standard deviation when data is scaled

If \(y = \frac{x}{k}\) (where \(x\) is the original data and \(y\) is the new - data, \(k\) is a non - zero constant), the mean of \(y\) is \(\bar{y}=\frac{\bar{x}}{k}\) and the standard deviation of \(y\) is \(s_y=\frac{s_x}{k}\). Here \(k = 12\), \(\bar{x}=40.8\) (from part (a)) and \(s_x = 12.2\) (from part (a)).

Step2: Calculate the new mean

We know that the formula for the new mean \(\bar{y}\) when each data point \(x_i\) is divided by \(12\) (i.e., \(y_i=\frac{x_i}{12}\)) is \(\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{\sum_{i = 1}^{n}\frac{x_i}{12}}{n}=\frac{1}{12}\frac{\sum_{i = 1}^{n}x_i}{n}\).
Since \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=40.8\), then \(\bar{y}=\frac{40.8}{12}\).

$$ \bar{y}=\frac{40.8}{12}=3.4 $$

Step3: Calculate the new standard deviation

The formula for the standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\). If \(y_i=\frac{x_i}{12}\), then \(s_y=\sqrt{\frac{\sum_{i = 1}^{n}(y_i-\bar{y})^2}{n - 1}}=\sqrt{\frac{\sum_{i = 1}^{n}(\frac{x_i}{12}-\frac{\bar{x}}{12})^2}{n - 1}}=\frac{1}{12}\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\).
Since \(s_x=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}} = 12.2\), then \(s_y=\frac{12.2}{12}\).

$$ s_y=\frac{12.2}{12}\approx1.0 $$

Answer:

The sample mean is \(3.4\) thousand dollars and the sample standard deviation is \(1.0\) thousand dollars.