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. Herewe find measures of variation.) Then answer the given questions.

Caffeine in Soft Drinks Listed below are measured amounts of caffeine (mg per 12oz of drink) obtained in one can from each of 20 brands (7-UP, A&W Root Beer, Cherry Coke, . . ., Tab). Are the statistics representative of the population of all cans of the same 20 brands consumed by Americans?

0 0 34 34 34 45 41 51 55 36 47 41 0 0 53 54 38 0 41 47

The amounts of caffeine present in one can for a sample of 20 cans of 20 different brands are given.

The sample size (n) is 20.

The amount of spread in data is calculated by computing the measures of dispersion. The three measures of dispersion that are used the most are as follows:

The rangeis obtained by subtracting the greatest and the smallest values.

The formula of varianceis given as

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

Here,

x represents the values in the sample, and

$\overline{x}$is the sample mean.

The standard deviation is calculated by taking the square root of the value of variance.

The range is calculated as

$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =55-0 \\ =55.0 \end{gathered}$$

Therefore, the sample range is equal to 55.0 mg.

The sample mean is calculated as

$$\begin{gathered} \overline{x}=\frac{{\sum }_{1=1}^n{x}_i}{n} \\ =\frac{0+0+34+...+47}{20} \\ =\frac{651}{20} \\ \approx 32.6 \end{gathered}$$

Thus, the sample mean is 32.6 mg.

The variance of the sample is calculated as

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(0-32.6\right)}^2+{\left(0-32.6\right)}^2+{\left(34-32.6\right)}^2+...+{\left(47-32.6\right)}^2}{20-1} \\ =\frac{7854.95}{19} \\ =413.4 \end{gathered}$$

.

Therefore, the sample variance is equal to 413.4${\text{mg}}^2$.

The sample standard deviation is calculated as

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{413.4} \\ \approx 20.3 \end{gathered}$$

Therefore, the sample standard deviation is equal to 20.3 mg.

Some brands are consumed far more frequently in the United States in comparison to other brands, but the sample consists of each of the 20 brands that Americans use, which indicates that the 20 brands are all equally weighted in the computations.

So, it can be observed that the sample is not representative of the entire population of brands of drinks. Thus, the statistics are not considered representative of all the cans of 20 brands that Americans popularly use.