Question

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 Actress When Oscar Was Won	Frequency
20-29	29
30-39	34
40-49	14
50-59	3
60-69	5
70-79	1
80-89	1

Short answer

The class width for each interval is equal to 10 actresses.

The midpoints are 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, and 84.5.

The class boundaries are 19.5, 29.5, 39.5, 49.5, 59.5, 69.5, 79.5, and 89.5.

The total number of individuals included is 87.

Step-by-step solution

Given information

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

Class width

The class width of an interval is computed by subtracting a lower class limit from the succeeding lower class limit.

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.

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

As it can be observed that all class intervals have equal width, the class width for each interval is equal to 1.

Class midpoint

The midpoint of a class interval has the following expression:

$\text{Midpoint}=\frac{\text{Lower}\text{limit}+\text{Upper}\text{limit}}{2}$

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

Class interval	Midpoint
20-29	$$\begin{gathered} \text{Midpoint}=\frac{20+29}{2} \\ =24.5 \end{gathered}$$
30-39	$$\begin{gathered} l\text{Midpoint}=\frac{30+39}{2} \\ =34.5 \end{gathered}$$
40-49	$$\begin{gathered} \text{Midpoint}=\frac{40+49}{2} \\ =44.5 \end{gathered}$$
50-59	$$\begin{gathered} \text{Midpoint}=\frac{50+59}{2} \\ =54.5 \end{gathered}$$
60-69	$$\begin{gathered} \text{Midpoint}=\frac{60+69}{2} \\ =64.5 \end{gathered}$$
70-79	$$\begin{gathered} \text{Midpoint}=\frac{70+79}{2} \\ =74.5 \end{gathered}$$
80-89	$$\begin{gathered} \text{Midpoint}=\frac{80+89}{2} \\ =84.5 \end{gathered}$$

Thus, the midpoints are 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, and 84.5.

Class boundaries

To compute the class boundaries, 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.

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.

Class interval	Lower class boundaries	Upper class boundaries
20-29	$20-0.5=19.5$	$29+0.5=29.5$
30-39	30 - 0.5 = 29.5	39 + 0.5 = 39.5
40-49	40 - 0.5 = 39.5	49 + 0.5 = 49.5
50-59	50 - 0.5 = 49.5	59 + 0.5 = 59.5
60-69	60 - 0.5 = 59.5	69 + 0.5 = 69.5
70-79	70 - 0.5 = 69.5	79 + 0.5 = 79.5
80-89	80 - 0.5 = 79.5	89 + 0.5 = 89.5

Therefore, the class boundaries are 19.5, 29.5, 39.5, 49.5, 59.5, 69.5, 79.5, and 89.5.

Total number of individuals

The sum of all the frequencies gives the total number of individuals included in the given frequency distribution.

$$\begin{gathered} \text{Total}\text{frequency}=29+34+14+3+5+1+1 \\ =87 \end{gathered}$$

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