Question

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

Short answer

The value of the mean is equal to 35.17 years.

The value of the standard deviation is equal to 19.64 years.

The value of the variance is equal to 385.69 years.

Step-by-step solution

Given information

The distribution of the ages of moviegoers is provided.

Compute the midpoints

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:

Class Interval	Midpoint
2-11	\(\begin{array}{r}{\rm{Midpoint}} = \frac{{2 + 11}}{2}\\ = 6.5\end{array}\)
12-17	\(\begin{array}{c}{\rm{Midpoint}} = \frac{{12 + 17}}{2}\\ = 14.5\end{array}\)
18-24	\(\begin{array}{c}{\rm{Midpoint}} = \frac{{18 + 24}}{2}\\ = 21\end{array}\)
25-39	\(\begin{array}{c}{\rm{Midpoint}} = \frac{{25 + 39}}{2}\\ = 32\end{array}\)
40-49	\(\begin{array}{c}{\rm{Midpoint}} = \frac{{40 + 49}}{2}\\ = 44.5\end{array}\)
50-59	\(\begin{array}{c}{\rm{Midpoint}} = \frac{{50 + 59}}{2}\\ = 54.5\end{array}\)
60-80	\(\begin{array}{c}{\rm{Midpoint}} = \frac{{60 + 80}}{2}\\ = 70\end{array}\)

Calculate the mean

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

Class Interval	f	Midpoint(x)	fx
2-11	7	6.5	45.5
12-17	15	14.5	217.5
18-24	19	21	399
25-39	19	32	608
40-49	15	44.5	667.5
50-59	11	54.5	599.5
60-80	14	70	980
	\(\sum f = N = 100\)	\(\sum {x = 243} \)	\(\sum {fx = 3517} \)

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.

Calculate the standard deviation

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

Class Interval	f	Midpoint(x)	fx	\({x^2}\)	\(f{x^2}\)
2-11	7	6.5	45.5	42.25	295.75
12-17	15	14.5	217.5	210.25	3153.75
18-24	19	21	399	441	8379
25-39	19	32	608	1024	19456
40-49	15	44.5	667.5	1980.25	29703.75
50-59	11	54.5	599.5	2970.25	32672.75
60-80	14	70	980	4900	68600
	\(\sum f = N = 100\)	\(\sum {x = 243} \)	\(\sum {fx = 3517} \)	\(\sum {{x^2}} = 11568\)	\(\sum {f{x^2}} = 162261\)

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.

Calculate the variance

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.