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.

Football Player Numbers Listed below are the jersey numbers of 11 players randomly selected from the roster of the Seattle Seahawks when they won Super Bowl XLVIII. What do the results tell us?

89 91 55 7 20 99 25 81 19 82 60

The jersey numbers of a sample of 11 players are given.

The number of observations $\left(n\right)$ is 11.

Different statistical values are calculated to identify the amount of dispersion/variation present in the data. Some of them are as follows:

Range: It tells the overall expanse of the data. It is calculated as the difference between the lowest values and the highest value.

Variance: It is the average value of the square of the deviations of the individual values from the mean $\left(\overline{x}\right)$. The formula of the sample variance is

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

Standard deviation: It is defined as the square root of the computed variance.

The range is calculated as

.$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =99-7 \\ =92.0 \end{gathered}$$

Therefore, the range of the jersey numbers is equal to 92.0.

The mean of the sample is calculated as

$$\begin{gathered} \overline{x}=\frac{{\sum }_{1=1}^n{x}_i}{n} \\ =\frac{89+91+...+60}{11} \\ =\frac{628}{11} \\ \approx 57.1 \end{gathered}$$

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Thus, the mean of the sample is 57.1.

The sample variance $\left(s^2\right)$ is calculated as

$$\begin{gathered} s^2=\frac{{\sum }_{1=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(89-57.1\right)}^2+{\left(91-57.1\right)}^2+...+{\left(60-57.1\right)}^2}{11-1} \\ =\frac{11494.909}{10} \\ \approx 1149.5 \end{gathered}$$

Therefore, the variance of the jersey numbers is equal to 1149.5.

The sample standard deviation $\left(s\right)$is equal to

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{1149.5} \\ \approx 33.9 \end{gathered}$$

Therefore, the standard deviation of the jersey numbers is equal to 33.9.

As the jersey numbers are a form of categorical data, any mathematical measure for this data has no significance. Thus, the values of range, standard deviation, and variance have no meaning.