Ages of MoviegoersThe table below shows the distribution of the ages of moviegoers(based on data from the Motion Picture Association of America). Use the data to estimate themean, standard deviation, and variance of ages of moviegoers.Hint:For the open-ended categoryof “60 and older,” assume that the category is actually 60–80.

Age	2-11	12-17	18-24	25-39	40-49	50-59	60 and older
Percent	7	15	19	19	15	11	14

The distribution of the ages of moviegoers is provided.

Consider the last class interval “60 and older” as 60-80.

The midpoint of a class interval has the following expression:

\({\rm{Midpoint}} = \frac{{{\rm{Lower}}\;{\rm{limit}} + {\rm{Upper}}\;{\rm{limit}}}}{2}\)

Thus, the midpoints of the class intervals are computed as shown:

The following table shows the calculations required to compute the mean value:

Themean value is computed below:

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {f{x_i}} }}{N}\\ = \frac{{3517}}{{100}}\\ = 35.17\end{array}\)

The mean value is equal to 35.17 years.

The following table shows the computations necessary for calculating the standard deviation:

The value of the standard deviation is computed below:

\(\begin{array}{c}\sigma = \sqrt {\frac{{\sum {f{x^2}} }}{N} - {{\left( {\frac{{\sum {fx} }}{N}} \right)}^2}} \\ = \sqrt {\frac{{162261}}{{100}} - {{\left( {\frac{{3517}}{{100}}} \right)}^2}} \\ = 19.64\end{array}\)

Therefore, the standard deviation is equal to 19.64 years.

The variance is computed as follows:

\(\begin{array}{c}{\sigma ^2} = \frac{{\sum {f{x^2}} }}{N} - {\left( {\frac{{\sum {fx} }}{N}} \right)^2}\\ = \frac{{162261}}{{100}} - {\left( {\frac{{3517}}{{100}}} \right)^2}\\ = 385.69\end{array}\)

The value of the variance is equal to 385.69 years squared.