: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.

Speed Dating In a study of speed dating conducted at Columbia University, female subjects were asked to rate the attractiveness of their male dates, and a sample of the results is listed below (1 = not attractive; 10 = extremely attractive). Can the results be used to describe the variation among attractiveness ratings for the population of adult males?

5 8 3 8 6 10 3 7 9 8 5 5 6 8 8 7 3 5 5 6 8 7 8 8 8 7

The attractiveness ratings of the male counterparts given by a group of 26 females are provided.

Therangeis one of the quantities used to measure the dispersion of data. Its units are equal to the units of the sample observations. It is calculated as shown below:

.$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =10-3 \\ =7.0 \end{gathered}$$

Therefore, the range of the sample is 7.0.

The sample mean needs to be computed to calculate the sample variance and the sample standard deviation.

The sample mean is computed as follows:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{5+8+....+7}{26} \\ =6.6 \end{gathered}$$

Thus, the sample mean is equal to 16.6.

Thevarianceis another quantity that is used to measure the dispersion of data. Its units are the square of the units of the sample observations. It is calculated as shown below:

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(5-6.6\right)}^2+{\left(8-6.6\right)}^2+...+{\left(7-6.6\right)}^2}{26-1} \\ =3.5 \end{gathered}$$

Therefore, the variance of the sample is 3.5.

Thestandard deviationis computed by calculating the square root of the variance term. Its units are the same as the units of the sample observations.

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{3.5} \\ =1.9 \end{gathered}$$

Therefore, the standard deviation of the sample is 1.9.

The data is measured on anordinal scale, with ratings of attractiveness from 1 to 10. The mean value is not defined for ordinal data. Thus, the dispersion from the mean hasno meaning for the given data.

The values do not depict the amount of variation in attractiveness ratings for the population of all adult males.