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The distribution of the ages of moviegoers is provided.

Consider the last class interval 鈥�60 and older鈥� 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.