Discrete Frequency Distribution
The process of preparing this kind of distribution is very simple. We just have to count the number of times a particular value is repeated which is termed as the frequency of that class. In order to facilitate the counting prepare a column of tallies. In another column, place all the possible values of variable from the lowest to the highest. This put a bar (vertical line) which opposite the particular value to which it elates. To facilitate the counting, blocks of 5 bars are prepared and some space is left in between every block. We finally count the number of bars and get the frequency.
The process shall be clear from the following examples:
Illustration:
In a survey of 35 families in a village, the number of children per family was recorded and the following data obtained
|
1
|
0
|
2
|
3
|
4
|
5
|
6
|
|
7
|
2
|
3
|
4
|
0
|
2
|
5
|
|
8
|
4
|
5
|
12
|
6
|
3
|
2
|
|
7
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6
|
5
|
3
|
3
|
7
|
8
|
|
9
|
7
|
9
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4
|
5
|
4
|
3
|
Represent the data in the form of a discrete frequency distribution.
Solution:
Frequency distribution of the number of children
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No. of children
|
tallies
|
frequency
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|
0
|
||
|
2
|
|
1
|
|
|
1
|
|
2
|
||||
|
4
|
|
3
|
|||| |
|
6
|
|
4
|
||||
|
5
|
|
5
|
||||
|
5
|
|
6
|
|||
|
3
|
|
7
|
||||
|
4
|
|
8
|
||
|
2
|
|
9
|
||
|
2
|
|
10
|
-
|
0
|
|
11
|
-
|
0
|
|
12
|
|
|
1
|
|
|
|
Total 35
|
It is clear from the table that the number of children varied from 0 to 12. There were 2 families with no child, 5 families with 4 children and only one family with 12 children.
A simple formula to obtain the estimate of appropriate class interval, i.e. i is:
i = [(L - S)/k]
Where, L = largest item
S = smallest item
k = the number of classes
For example, if the salary of 100 employees in a commercial undertaking varied between $500 and 45500 and we want to form 10 classes, then the class interval would be:
i = [(l - S)/k]
L = 5500, S = 500, k = 10
i = [(5500 - 500)/10] = 5000/10 = 500