Question

The ages (in years) of the 6 employees at a particular computer store are the following.31, 41, 35, 22, 38, 31Assuming that these ages 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.)

Answer 1

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:

$\begin{gathered} \mu\text{ = }\frac{\sum ^{}_{}x_i}{N} \\ \mu\text{ = }\frac{31+41+35+22+38+31}{6} \\ \mu\text{ = }\frac{198}{6} \\ \mu\text{ = }33 \end{gathered}$

Find the summation of the square of each deviation from the mean

$\begin{gathered} \sum ^6_{i\mathop=0}(x_i-\mu)^2=(31-33)^2+(41-33)^2+(35-33)^2+(22-33)^2+(38-33)^2+(31-33)^2 \\ \sum ^6_{i\mathop{=}0}(x_i-\mu)^2=(-2)^2+(8)^2+(2)^2+(-11)^2+(5)^2+(-2)^2 \\ \sum ^6_{i\mathop{=}0}(x_i-\mu)^2=4+64+4+121+25+4 \\ \sum ^6_{i\mathop{=}0}(x_i-\mu)^2=222 \end{gathered}$

The standard deviation is given by the formula:

$\begin{gathered} \sigma\text{ = }\sqrt[]{\frac{\sum ^{}_{}(x_i-\mu)^2}{N}} \\ \sigma\text{ = }\sqrt[]{\frac{222}{6}} \\ \sigma\text{ = }\sqrt[]{37} \\ \sigma\text{ = }6.08 \end{gathered}$

The standard deviation of the population = 6.08 (rounded to 2 decimal places)