Example Construction of Frequency Distribution

Construct a frequency distribution with the suitable class interval size of marks obtained by 50 students of a class, which are given below:

23, 50, 38, 42, 63, 75, 12, 33, 26, 39, 35, 47, 43, 52, 56, 59, 64, 77, 15, 21, 51, 54, 72, 68, 36, 65, 52, 60, 27, 34, 47, 48, 55, 58, 59, 62, 51, 48, 50, 41, 57, 65, 54, 43, 56, 44, 30, 46, 67, 53

 

Solution:

Arrange the marks in ascending order as

12, 15, 21, 23, 26, 27, 30, 33, 34, 35, 36, 38, 39, 41, 42, 43, 43, 44, 46, 47, 47, 48, 48, 50, 50, 51, 51, 52, 52, 53, 54, 54, 55, 56, 56, 57, 58, 59, 59, 60, 62, 63, 64, 65, 65, 67, 68, 72, 75, 77

Minimum Value = 12   Maximum = 77

Range = Maximum Value – Minimum Value = 77 - 12 = 65

Number of Classes = 1 + 3.322\log N

Number of Classes =  1 + 3.322\log 50

Number of Classes = 1 + 3.322(1.69897)

Number of Classes = 1 + 5.64 = 6.64 or 7, approximately.

Class Interval Size (h) = \frac{{Range}}{{No.{\text{ }}of{\text{ }}Classes}} = \frac{{65}}{7} = 9.3 or 10

 

Marks
Class Limits
C.L
Number of
Students
f
Class
Boundary
C.B
Class
Marks
x
10 - 19
2
9.5 - 19.5
\frac{{10 + 19}}{2} = 14.5
20 - 29
4
19.5 - 29.5
\frac{{20 + 29}}{2} = 24.5
30 - 39
7
29.5 - 39.5
\frac{{30 + 39}}{2} = 34.5
40 - 49
10
39.5 - 49.5
\frac{{40 + 49}}{2} = 44.5
50 - 59
16
49.5 - 59.5
\frac{{50 + 59}}{2} = 54.5
60 - 69
8
59.5 - 69.5
\frac{{60 + 69}}{2} = 64.5
70 - 79
3
69.5 - 79.5
\frac{{70 + 79}}{2} = 74.5
50

Note: To find the class boundaries, we take half of the difference between the lower class limit of the 2nd class and the upper class limit of the 1st class\frac{{20 - 19}}{2} = \frac{1}{2} = 0.5. This value is subtracted from the lower class limit and is added to the upper class limit to get the required class boundaries.

 

Frequency Distribution by Exclusive Method

Class Boundray C.B
Frequency
f
10 - 19
2
20 - 29
4
30 - 39
7
40 - 49
10
50 - 59
16
60 - 69
8
70 - 79
3
 
50