Question

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.

California Smokers In the California Health Interview Survey, randomly selected adults are interviewed. One of the questions asks how many cigarettes are smoked per day, and results are listed below for 50 randomly selected respondents. How well do the results reflect the smoking behavior of California adults?

9 10 10 20 40 50 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Short answer

The range of the number of cigarettes smoked per day is 50.0.

The variance of the number of cigarettes smoked per day is 89.7.

The standard deviation of the number of cigarettes smoked per day is 9.5.

As people may wrongly report the number of cigarettes they smoke per day in the survey, the results of the measures of variation do not present the actual picture of the smoking behavior of California adults.

Step-by-step solution

Given information

The number of cigarettes smoked per day by 50 randomly selected adults of California is listed.

Computation of range

Therange depicts the difference between the maximum value and the minimum value reported in the sample.

$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =50-0 \\ =50.0 \end{gathered}$$

Therefore, the range of the data is 50.0.

Computation of variance and standard deviation

For computing the variance and the standard deviation, the sample mean is required.

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{9+10+....+0}{50} \\ =2.8 \end{gathered}$$

Thus, the sample mean is 2.8.

Thevariancedepicts the spread of the data from a central value, which is the arithmetic mean of the data.

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(9-2.8\right)}^2+{\left(10-2.8\right)}^2+...+{\left(0-2.8\right)}^2}{50-1} \\ =89.7 \end{gathered}$$

Therefore, the variance of the data is 89.7.

Thestandard deviation also shows the spread of the data. It is the square root value of variance.

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{89.7} \\ =9.5 \end{gathered}$$

Therefore, the standard deviation of the data is 9.5.

Interpretation

There is a high chance ofmisreportingof the number of cigarettes smoked per day by the respondent. Thus, these values of measures of variation are not based on the actual number of cigarettes smoked, and theycannot reflect the true smoking behavior of Californian adults.