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 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 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 x = 3.4 thousand dollars.
(round to one decimal place as needed.)
the sample standard deviation is s = 1.0 thousand dollars.
(round to one decimal place as needed.)
(d) what can you conclude from the results of (a), (b), and (c)?
a. when each entry is multiplied by a constant k, the sample mean and the sample standard deviation remain unaffected.
b. when each entry is multiplied by a constant k, the new sample mean is k·x and the sample standard deviation remains unaffected.
c. when each entry is multiplied by a constant k, the new sample standard deviation is k·s and the sample mean remains unaffected.
d. when each entry is multiplied by a constant k, the new sample mean is k·x and the new sample standard deviation is k·s.

Explanation:

Step1: Analyze part (b)

In part (b), each salary is multiplied by \(k = 1+0.06=1.06\). The original mean \(\bar{x}=40.8\), and the new mean is \(1.06\times40.8 = 43.3\) (rounded to one decimal). The original standard deviation \(s = 12.2\), and the new standard deviation is \(1.06\times12.2=13.0\) (rounded to one decimal).

Step2: Analyze part (c)

In part (c), each salary is multiplied by \(k=\frac{1}{12}\). The original mean \(\bar{x}=40.8\), and the new mean is \(\frac{40.8}{12}=3.4\). The original standard deviation \(s = 12.2\), and the new standard deviation is \(\frac{12.2}{12}\approx1.0\)

Answer:

D. When each entry is multiplied by a constant \(k\), the new sample mean is \(k\cdot\bar{x}\) and the new sample standard deviation is \(k\cdot s\)