Question

The data shows the total number of employee medical leave days taken for on-the-job accidents in the first six months of the year 19,15, 22, 16, 23, 19. Find the standard deviationThe standard deviation is about(Type an integer or decimal rounded to the nearest hundredth as needed.)

Answer 1

SOLUTION

The given data set is

$19,15,22,16,23,19$

The mean of the data is

$\begin{gathered} \bar{x}=\frac{19+15+22+16+23+19}{6} \\ \bar{x}=\frac{114}{6} \\ \bar{x}=19 \end{gathered}$

The standard deviation is calculated using

$\sigma=\sqrt{\frac{1}{N}\sum(x-\bar{x})^2}$

Substituting the data gives

$\begin{gathered} \sigma=\sqrt{\frac{(19-19)^2+(15-19)^2+(22-19)^2+(16-19)^2+(23-19)^2+(19-19)^2}{6}} \\ \sigma=8.333 \end{gathered}$

Therefore the standard deviation is 8.333