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:
[tex]\begin{gathered} \mu\text{ = }\frac{\sum ^{}_{}x_i}{N} \\ \mu\text{ = }\frac{31+41+35+22+38+31}{6} \\ \mu\text{ = }\frac{198}{6} \\ \mu\text{ = }33 \end{gathered}[/tex]Find the summation of the square of each deviation from the mean
[tex]\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}[/tex]The standard deviation is given by the formula:
[tex]\begin{gathered} \sigma\text{ = }\sqrt[]{\frac{\sum ^{}_{}(x_i-\mu)^2}{N}} \\ \sigma\text{ = }\sqrt[]{\frac{222}{6}} \\ \sigma\text{ = }\sqrt[]{37} \\ \sigma\text{ = }6.08 \end{gathered}[/tex]The standard deviation of the population = 6.08 (rounded to 2 decimal places)
