Questions: Sample annual salaries (in thousands of dollars) for employees at a company are listed. 40, 51, 47, 54, 37, 37, 40, 51, 47, 26, 54, 40, 52 (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 (bar x) = thousand dollars. (Round to one decimal place as needed.)

Solution Steps

The sample mean \( \bar{x} \) is calculated using the formula:

\[ \bar{x} = \frac{\sum_{i=1}^N x_i}{N} \]

For the given salaries, we have:

\[ \bar{x} = \frac{576}{13} = 44.3 \]

Thus, the sample mean is:

\[ \text{Sample mean: } \bar{x} = 44.3 \text{ thousand dollars} \]

The sample standard deviation \( s \) is calculated using the formula:

\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]

Calculating the variance:

\[ \sigma^2 = 70.7 \]

Then, the standard deviation is:

\[ s = \sqrt{70.7} = 8.4 \]

Thus, the sample standard deviation is:

\[ \text{Sample standard deviation: } s = 8.4 \text{ thousand dollars} \]

After a 6% raise, the new sample mean \( \bar{x}_{\text{raised}} \) is calculated as follows:

\[ \bar{x}_{\text{raised}} = \frac{\sum_{i=1}^N (x_i \cdot 1.06)}{N} = 1.06 \cdot \bar{x} \]

Calculating the new mean:

\[ \bar{x}_{\text{raised}} = 1.06 \cdot 44.3 = \frac{610.6}{13} = 47.0 \]

Thus, the sample mean after the raise is:

\[ \text{Sample mean after 6% raise: } \bar{x}_{\text{raised}} = 47.0 \text{ thousand dollars} \]

The sample standard deviation remains unchanged when all values are increased by a constant factor. Therefore, we calculate the new variance:

\[ \sigma^2_{\text{raised}} = 79.5 \]

Then, the new standard deviation is:

\[ s_{\text{raised}} = \sqrt{79.5} = 8.9 \]

Thus, the sample standard deviation after the raise is:

\[ \text{Sample standard deviation after 6% raise: } s_{\text{raised}} = 8.9 \text{ thousand dollars} \]

To find the sample mean of monthly salaries, we divide each original salary by 12:

\[ \bar{x}_{\text{monthly}} = \frac{\sum_{i=1}^N \left(\frac{x_i}{12}\right)}{N} = \frac{48.0}{13} = 3.7 \]

Thus, the sample mean of monthly salaries is:

\[ \text{Sample mean of monthly salaries: } \bar{x}_{\text{monthly}} = 3.7 \text{ thousand dollars} \]

The variance for the monthly salaries is calculated as follows:

\[ \sigma^2_{\text{monthly}} = 0.5 \]

Then, the standard deviation is:

\[ s_{\text{monthly}} = \sqrt{0.5} = 0.7 \]

Thus, the sample standard deviation of monthly salaries is:

\[ \text{Sample standard deviation of monthly salaries: } s_{\text{monthly}} = 0.7 \text{ thousand dollars} \]

Final Answer