Example Construction of Frequency Distribution

Construct a frequency distribution with suitable class interval size of marks obtained by 50 students of a class 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 approximate

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: For finding the class boundaries, we take half of the difference between lower class limit of the 2nd class and upper class limit of the 1st class\frac{{20 - 19}}{2} = \frac{1}{2} = 0.5. This value is subtracted from lower class limit and added in 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

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