Question

sample annual salaries (in thousands of dollars) for employees at a company are listed. 35 49 44 63 28 28 35 49 44 30 63 35 50 (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 = 42.5 thousand dollars. (round to one decimal place as needed.) (b) the sample mean is x = thousand dollars. (round to one decimal place as needed.) (c) the sample standard deviation is s = 12 thousand dollars. (round to one decimal place as needed.)

Explanation:

Step1: Calculate sample mean

The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points.
We have the data set: $35,49,44,63,28,28,35,49,44,30,63,35,50$.
$n = 13$
$\sum_{i=1}^{13}x_{i}=35 + 49+44 + 63+28+28+35+49+44+30+63+35+50=552$
$\bar{x}=\frac{552}{13}\approx42.5$ (rounded to one decimal place)

Step2: Calculate sample standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$
First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$:
$(35 - 42.5)^{2}=(-7.5)^{2}=56.25$
$(49 - 42.5)^{2}=(6.5)^{2}=42.25$
$(44 - 42.5)^{2}=(1.5)^{2}=2.25$
$(63 - 42.5)^{2}=(20.5)^{2}=420.25$
$(28 - 42.5)^{2}=(-14.5)^{2}=210.25$
$(28 - 42.5)^{2}=(-14.5)^{2}=210.25$
$(35 - 42.5)^{2}=(-7.5)^{2}=56.25$
$(49 - 42.5)^{2}=(6.5)^{2}=42.25$
$(44 - 42.5)^{2}=(1.5)^{2}=2.25$
$(30 - 42.5)^{2}=(-12.5)^{2}=156.25$
$(63 - 42.5)^{2}=(20.5)^{2}=420.25$
$(35 - 42.5)^{2}=(-7.5)^{2}=56.25$
$(50 - 42.5)^{2}=(7.5)^{2}=56.25$

$\sum_{i = 1}^{13}(x_{i}-\bar{x})^{2}=56.25+42.25 + 2.25+420.25+210.25+210.25+56.25+42.25+2.25+156.25+420.25+56.25+56.25 = 1781$

$s=\sqrt{\frac{1781}{12}}\approx12.2$ (rounded to one decimal place)

Step3: Adjust for monthly salary

If we divide each original salary by 12, the new mean $\bar{y}=\frac{\bar{x}}{12}=\frac{42.5}{12}\approx3.5$ (rounded to one decimal place)
The new standard - deviation $t=\frac{s}{12}=\frac{12.2}{12}\approx1.0$ (rounded to one decimal place)

Step4: Adjust for 6% raise

If we have a 6% raise, we multiply each value by $1 + 0.06=1.06$.
The new mean after 6% raise for the monthly data: $\bar{z}=3.5\times1.06 = 3.7$ (rounded to one decimal place)
The new standard - deviation after 6% raise for the monthly data: $u = 1.0\times1.06=1.1$ (rounded to one decimal place)

Answer:

(a) $42.5$
(b) $12.2$
(c) Mean: $3.7$, Standard deviation: $1.1$
(d) When we divide by 12, the mean and standard deviation are also divided by 12. When we apply a 6% raise (multiply by 1.06), the mean and standard deviation are multiplied by 1.06. This is because the mean and standard deviation are linear functions of the data in the sense that if $y_{i}=a + bx_{i}$ (dividing by 12 is $y_{i}=\frac{1}{12}x_{i}$ and 6% raise is $y_{i}=1.06x_{i}$), the new mean $\bar{y}=a + b\bar{x}$ and new standard deviation $s_{y}=|b|s_{x}$ where $\bar{x}$ and $s_{x}$ are the original mean and standard deviation.