Question

all 5 members of a family work. their hourly wages (in dollars) are the following. 21, 18, 13, 22, 26. assuming that these wages constitute an entire population, find the standard deviation of the population. round your answer to two decimal places. (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.
$n = 5$, $x_1=21$, $x_2 = 18$, $x_3=13$, $x_4=22$, $x_5=26$.
$\mu=\frac{21 + 18+13+22+26}{5}=\frac{100}{5}=20$.

Step2: Calculate the squared differences

$(x_1-\mu)^2=(21 - 20)^2=1$;
$(x_2-\mu)^2=(18 - 20)^2=4$;
$(x_3-\mu)^2=(13 - 20)^2 = 49$;
$(x_4-\mu)^2=(22 - 20)^2=4$;
$(x_5-\mu)^2=(26 - 20)^2=36$.

Step3: Calculate the variance

The population variance $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^2}{n}$.
$\sigma^{2}=\frac{1+4 + 49+4+36}{5}=\frac{94}{5}=18.8$.

Step4: Calculate the standard deviation

The population standard deviation $\sigma=\sqrt{\sigma^{2}}$.
$\sigma=\sqrt{18.8}\approx4.34$.

Answer:

$4.34$