solution;
a) From the given information,
N = 53,min=42 and max = 129
K = 1+3.322log₁₀53
= 1+3.322(1.724)
= 1+5.728
= 6.728
Number of classes is 7.
b) find the width of the class interval first
width = max – min /classes
= 129-42/7
= 87/7 =12.42
Take the next largest number 13 as class width. Since the minimum value is 42
The lower limit iof the first class could be 40.
So,
The upper limit of the last class is,
40+(13 x 7) = 131
As the upper limit of the last class covers the entire range of values,
It is better to take the lower limit of the first class as 40
The lower limit of the first calss could be 40.
