Question

sample annual salaries (in thousands of dollars) for employees at a company are listed.
35 52 58 54 36 36 35 52 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 revised data set.
(c) to calculate the monthly salary, divide each original salary by 12. find the sample me 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} = 44.9 \\) thousand dollars.
(round to one decimal place as needed.)
the sample standard deviation is \\( s = 10.3 \\) thousand dollars.
(round to one decimal place as needed.)

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

Explanation:

Step1: Recall the property of mean under scaling

If each data point \( x_i \) is multiplied by a constant \( k \), the new mean \( \bar{x}' \) is \( k\times\bar{x} \), where \( \bar{x} \) is the original mean. Here, a 5% raise means each salary is multiplied by \( 1 + 0.05=1.05 \).

Step2: Calculate the new mean

The original mean \( \bar{x} = 44.9 \) (in thousands of dollars). So the new mean \( \bar{x}'=1.05\times44.9 \).
\[
1.05\times44.9 = (1 + 0.05)\times44.9=44.9+0.05\times44.9 = 44.9 + 2.245=47.145\approx47.1
\]

Answer:

47.1