In Exercises 5–20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-1, where we found measures of center. Here, we find measures of variation.) Then answer the given questions.

Foot Lengths Listed below are foot lengths in inches of randomly selected Army women measured in the 1988 Anthropometric Survey (ANSUR). Are the statistics representative of the current population of all Army women?

10.4 9.3 9.1 9.3 10.0 9.4 8.6 9.8 9.9 9.1 9.1

The foot lengths of 11 Army women measured in the year 1988 are provided.

Therange of a dataset, which has the same units as the variable, is calculated using the given formula:

$\text{Range}=\text{Highest}\text{Value}-\text{Lowest}\text{Value}$

Thesample variance,which has the square of the units of the original variable, is expressed as follows:

$s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1}$

Here,

$x_i$denotes the value of the variable;

$\overline{x}$denotes the average of the values.

Thesample standard deviation, which has the same units as the variable, is expressed as follows:

$s=\sqrt{s^2}$

The range is computed as follows:

$$\begin{gathered} \text{Range}=\text{Highest}\text{Value}-\text{Lowest}\text{Value} \\ =10.4-8.6 \\ =1.80\text{inches} \end{gathered}$$

Thus, the range of foot lengths is equal to 1.80 inches.

The sample mean is computed as follows:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{10.4+9.3+....+9.1}{11} \\ =9.45\text{inches} \end{gathered}$$

Thus, the sample mean is equal to 9.45 inches.

Substitute the data values in the variance formula:

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(10.4-9.45\right)}^2+{\left(9.3-9.45\right)}^2+...+{\left(9.1-9.45\right)}^2}{11-1} \\ =0.27{\left(\text{inches}\right)}^2 \end{gathered}$$

Thus, the variance of foot lengths is equal to 0.27 inches square.

The standard deviation is calculated as follows:

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{0.27} \\ =0.52\text{inches} \end{gathered}$$

Thus, the standard deviation of foot lengths is equal to 0.52 inches.

Since the foot lengths wererecorded in conducted by ANSUR in 1988, the values willnot hold good for the current population of Army women.

Thus, the statistics are not representative of the current population of Army women.