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.

Fireighter Fatalities Listed below are the numbers of heroic firefighters who lost their lives in the United States each year while fighting forest fires. The numbers are listed in order by year, starting with the year 2000. What important feature of the data is not revealed by any of the measures of variation?

20 18 23 30 20 12 24 9 25 15 8 11 15 34

The number of firefighters who died in the US every year for 14 years (from 2000 onwards) is listed.

The difference between the greatest and the lowest values of the dataset gives therange.

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

The variance of a sample is calculated using the formula

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

Here,

$x_i$is the ith observation of the sample, and

$\overline{x}$ is the arithmetic mean of the observations.

Thestandard deviation of a sample is calculated using the formula$s=\sqrt{s^2}$ .

The range of the observations is computed as shown below.

$$\begin{gathered} \text{Range}=\text{Greatest}\text{Value}-\text{Lowest}\text{Value} \\ =34-8 \\ =26.0 \end{gathered}$$

Therefore, the range is 26.0.

The sample mean is computed as shown below.

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{20+18+....+34}{14} \\ =18.9 \end{gathered}$$

Thus, the sample mean is 18.9.

Substitute the values in the variance formula.

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(20-19.9\right)}^2+{\left(18-18.9\right)}^2+....+{\left(34-18.9\right)}^2}{14-1} \\ =60.9 \end{gathered}$$

Therefore, the variance is 60.9.

The standard deviation is computed as shown below.

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{60.9} \\ =7.8 \end{gathered}$$

Therefore, the standard deviation is 7.8.

The values are provided for 14 consecutive years.

The measures of variation describe the extent to which the data spread on average.

As the data is time-based, the measures of variation do not reveal anyparticular cycle or trend that the variable has followed over the years. This is an important feature that is missing.