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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

Short Answer

Expert verified

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

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

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

The total number of individuals included is equal to 87.

Step by step solution

01

Given information

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

02

Class width

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.

Classwidth=30-20=10

As it can be observed that all class intervals have equal width, the class width for each interval is equal to 10.
03

Class midpoint

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

Midpoint=lowerlimit+upperlimit2

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

Class interval

Midpoint

20-29

Midpoint=20+292=24.5

30-39

Midpoint=30+392=34.5

40-49

Midpoint=40+492=44.5

50-59

Midpoint=50+592=54.5

60-69

Midpoint=60+692=64.5

70-79

Midpoint=70+792=74.5

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

04

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 to obtain the class boundaries.

The gap is calculated as shown below.

Gap=2ndlowerclasslimit-1stupperclasslimit=30-29=1

The value equal to 12=0.5is 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

Therefore, the class boundaries are 19.5, 29.5, 39.5, 49.5, 59.5, 69.5, and 79.5.
05

Total number of individuals

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.

Totalfrequency=1+28+36+15+6+1=87

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

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