Question

sample annual salaries (in thousands of dollars) for employees at a company are listed: 35, 52, 58, 54, 36, 36, 35, 62, 58, 31, 54, 35, 48. (a) find the sample mean and sample standard deviation. (b) each employee in the sample is given a 5% 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)? (round to one decimal place as needed.) (b) the sample mean is \\( \bar{x} = 47.2 \\) thousand dollars. (round to one decimal place as needed.) the sample standard deviation is \\( s = 10.8 \\) thousand dollars. (round to one decimal place as needed.) (c) the sample mean is \\( \bar{x} = 3.7 \\) 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: Recall the property of standard deviation

When each data point \( x_i \) is divided by a constant \( c \), the new standard deviation \( s' \) is related to the original standard deviation \( s \) by the formula \( s'=\frac{s}{c} \). Here, we are dividing each original salary by 12, so \( c = 12 \).

Step2: Identify the original standard deviation

From part (b), the original sample standard deviation \( s=10.8 \) (in thousands of dollars).

Step3: Calculate the new standard deviation

Using the formula \( s'=\frac{s}{12} \), substitute \( s = 10.8 \):
\( s'=\frac{10.8}{12}=0.9 \)

Answer:

\( 0.9 \)