Question

the ages (in years) of the 6 employees at a particular computer store are the following. 40, 28, 32, 32, 36, 24 send data to calculator send data to excel assuming that these ages constitute an entire population, find the standard deviation of the population. (if necessary, consult a list of formulas.)

Explanation:

Step1: Calculate the mean

The mean $\mu$ of a population is $\mu=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points.
$n = 6$, and $\sum_{i=1}^{6}x_{i}=40 + 28+32+32+36+24=192$.
$\mu=\frac{192}{6}=32$.

Step2: Calculate the squared differences

For each data - point $x_{i}$, calculate $(x_{i}-\mu)^2$.
$(40 - 32)^2=64$, $(28 - 32)^2 = 16$, $(32 - 32)^2=0$, $(32 - 32)^2 = 0$, $(36 - 32)^2=16$, $(24 - 32)^2 = 64$.

Step3: Calculate the variance

The population variance $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^2}{n}$.
$\sum_{i = 1}^{6}(x_{i}-\mu)^2=64 + 16+0+0+16+64 = 160$.
$\sigma^{2}=\frac{160}{6}=\frac{80}{3}\approx26.67$.

Step4: Calculate the standard deviation

The population standard deviation $\sigma=\sqrt{\sigma^{2}}$.
$\sigma=\sqrt{\frac{80}{3}}\approx5.16$.

Answer:

$5.16$