91影视

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

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.

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

$$\begin{gathered} \stackrel{炉}{x}=\frac{{鈭�}_{i=1}^n{x}_i}{n} \\ =\frac{9+10+....+0}{50} \\ =2.8 \end{gathered}$$

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.

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

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.

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.