Answer:
The standard deviation of the population = 6.08
Explanations:
The given ages of the employers are:
31, 41, 35, 22, 38, 31
Find the mean of the dataset:
![\begin{gathered} \mu\text{ = }\frac{\sum ^{}_{}x_i}{N} \\ \mu\text{ = }(31+41+35+22+38+31)/(6) \\ \mu\text{ = }(198)/(6) \\ \mu\text{ = }33 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/uyfnbeekbc7b84pzx1cfpj1azjp6bngbwo.png)
Find the summation of the square of each deviation from the mean
![\begin{gathered} \sum ^6_(i\mathop=0)(x_i-\mu)^2=(31-33)^2+(41-33)^2+(35-33)^2+(22-33)^2+(38-33)^2+(31-33)^2 \\ \sum ^6_{i\mathop{=}0}(x_i-\mu)^2=(-2)^2+(8)^2+(2)^2+(-11)^2+(5)^2+(-2)^2 \\ \sum ^6_{i\mathop{=}0}(x_i-\mu)^2=4+64+4+121+25+4 \\ \sum ^6_{i\mathop{=}0}(x_i-\mu)^2=222 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9kwquj2kyjynqh9yle6110m6qdsx8meyw5.png)
The standard deviation is given by the formula:
![\begin{gathered} \sigma\text{ = }\sqrt[]{\frac{\sum ^{}_{}(x_i-\mu)^2}{N}} \\ \sigma\text{ = }\sqrt[]{(222)/(6)} \\ \sigma\text{ = }\sqrt[]{37} \\ \sigma\text{ = }6.08 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ix0xs5alv8vyvud81aghm2oejskt2cbfcr.png)
The standard deviation of the population = 6.08 (rounded to 2 decimal places)