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

Explanation:

Step1: Count the number of data points

First, we count how many data points there are. Let's list the data: 35, 52, 58, 54, 36, 36, 35, 52, 58, 31, 54, 35, 48. Let's count them: that's 13 data points.

Step2: Calculate the sum of the data

Now we sum all the values:
\[

$$\begin{align*} &35 + 52 + 58 + 54 + 36 + 36 + 35 + 52 + 58 + 31 + 54 + 35 + 48\\ =& (35\times3) + (52\times2) + (58\times2) + (54\times2) + (36\times2) + 31 + 48\\ =& 105 + 104 + 116 + 108 + 72 + 31 + 48\\ =& 105+104 = 209\\ &209+116 = 325\\ &325+108 = 433\\ &433+72 = 505\\ &505+31 = 536\\ &536+48 = 584 \end{align*}$$

\]

Step3: Calculate the sample mean

The sample mean \(\bar{x}\) is the sum of the data divided by the number of data points \(n\). So:
\[
\bar{x} = \frac{\text{Sum}}{n} = \frac{584}{13} \approx 44.9
\]

Answer:

44.9