In Exercises 5–8, identify the class width, class midpoints, and class boundaries for the given frequency distribution. Also identify the number of individuals included in the summary. The frequency distributions are based on real data from Appendix B.

Age (yr) of Best Actor when Oscar Was Won	Frequency
20-29	1
30-39	28
40-49	36
50-59	15
60-69	6
70-79	1

Data are given on the ages of the best actors when they won the Oscar.

The class width of an interval is computed by subtracting two consecutive lower class limits.

Here, the lower class limit of the first class interval is equal to 20, and the lower class limit of the second class interval is equal to 30.

Thus, the class width is as follows.

$$\begin{gathered} \text{Class}\text{width}=30-20 \\ =10 \end{gathered}$$

The midpoint of a class interval is calculated using the following formula:

$\text{Midpoint}=\frac{\text{lower}\text{limit}+\text{upper}\text{limit}}{2}$

Thus, the midpoints of the class intervals are computed as shown below.

Thus, the midpoints are 24.5, 34.5, 44.5, 54.5, 64.5, and 74.5.

The value of the gap between each successive interval divided by 2 is subtracted from the lower limit and added to the upper limit of a class interval to obtain the class boundaries.

The gap is calculated as shown below.

$$\begin{gathered} \text{Gap}=2^{nd}\text{lower}\text{class}\text{limit}-1^{st}\text{upper}\text{class}\text{limit} \\ =30-29 \\ =1 \end{gathered}$$

The value equal to $\frac{1}{2}=0.5$is subtracted from the lower class limits and added to the upper-class limits of each interval. Thus, the class boundaries are obtained as shown below.

The total number of individuals included in the given frequency distribution is equal to the sum of the frequencies of all classes. The total frequency is computed below.

$$\begin{gathered} \text{Total}\text{frequency}=1+28+36+15+6+1 \\ =87 \end{gathered}$$

Therefore, the total number of individuals included in the given frequency distribution is 87.