Answer
Classes with greatest frequency: 0 –7, 8–15; frequency = 8
Classes with least frequency: 16 –23, 24 –31, 32–39 ; frequency = 3
Work Step by Step
1. The number of classes (5)
2. The minimum data entry is 0 and the maximum data entry is 39, so the range is $39 - 0 = 39$. Class width = range/number of classes: $39/5 = 7.8 ≈ 8 $.
3. The minimum data of 0 entry is a convenient lower limit for the first class. Find the lower limits of the remaining 4 classes, add the class width of 8 to the lower limit of each previous class. So, the lower limits of the other classes are 0 + 8 = 8, 8 + 8 = 16, and so on. The upper limit of the first class is 7, which is one less than the lower limit of the second class. The upper limits of the other classes are 7 + 8 = 15, 15 + 8 = 23, and so on
4. Tallying is the done for each data entry in the appropriate class.
5. Midpoint = (Lower class limit) + (Upper class limit) /2, So, midpoint of first class $(0+7)/2 = 3.5 $ and so on.
6. To find the relative frequency of a class, divide the frequency ,f by the sample size n. Relative frequency for the first Class frequency Sample size $8/25 = 0.32$ and so on.
7. The cumulative frequency of a class is the sum of the frequencies of that class and all previous classes. The cumulative frequency of the last class is equal to the sample size n.
8. The frequency distribution is shown below
