# Cumulative Frequency

Here we will learn cumulative frequency.

The cumulative frequency of a value of a variable is the number of values in the collection of data less than or equal to the value of the variable.

For example: Let the raw data be 2, 10, 18, 25, 15, 16, 15, 3, 27, 17, 15, 16. The cumulative frequency of 15 = 6 (Since, values ≤ 15 are 2, 10, 15, 15, 3, 15).

The cumulative frequency of a class interval (overlapping or nonoverlapping) is the sum of the frequencies of earlier class intervals and the concerned class interval.

For example: Consider the frequency distribution below.

Class Interval

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

Total

Frequency

4

7

2

5

3

21

The cumulative frequency of 0 – 20 is 4, of 20 – 40 is 11 (i.e., 4 + 7), of 40 – 60 is 13(i.e., 4 + 7 + 2), etc.

The following cumulative frequency table can be constructed from the above frequency table.

 Class Interval0 - 2020 - 4040 - 6060 - 8080 - 100 Frequency47253 Cumulative Frequency411131821 (= 4 + 7)(= 4 + 7 + 2)(= 4 + 7 + 2 + 5)(= 4 + 7 + 2 + 5 + 3)

Another way of representing the same table is shown below.

Value of the Variable

Under 20

Under 40

Under 60

Under 80

Under 100

Frequency

4

11

13

18

21

Here the class interval 0 – 20 has the frequency 4. The class interval 20 – 40 includes those values of the variable which are under 40 but not under 20. So, the frequency of the class interval 20 – 40 is 11 – 4, that is 7. Clearly, such a table is in fact a cumulative frequency table for overlapping class intervals.

Solved Examples on Cumulative Frequency Table:

1. For the collection of numbers 12, 15, 8, 13, 12, 15, 9, 16, 24, 20, 20, 16 and 10, answer the following:

(i) What is the cumulative frequency of 16?

(ii) If 15 - 20 be an overlapping class interval when the numbers are grouped, find the cumulative frequency of the class interval.

(iii) If 15 - 20 be a nonoverlapping class interval when the numbers are grouped, find the cumulative frequency of the class interval.

Solution:

(i) 10 (Since values ≤ 16 are 12, 15, 8, 13, 12, 15, 9, 16, 16, 10)

(ii) 10 (Since values < 20 are 12, 15, 8, 13, 12, 15, 9, 16, 16, 10)

(iii) 12 (Since values ≤ 20 are 12, 15, 8, 13, 12, 15, 9, 16, 16, 10)

2. The marks of 200 students in a test were recorded and shown by the following frequency distribution.

 Marks %10 - 1920 - 2930 - 3940 - 4950 - 5960 - 6970 - 7980 - 89 Number of Students71120465737157

Construct the cumulative frequency table.

(i) How many students obtained less than 50 marks?

(ii) How many students obtained at least 60 marks?

Solution:

The cumulative frequency table is as given below.

Class Interval

10 - 19

20 - 29

30 - 39

40 - 49

50 - 59

60 - 69

70 - 79

80 - 89

Frequency

7

11

20

46

57

37

15

7

Cumulative Frequency

7

18

38

84

141

178

193

200

(i) The number of students obtaining less than 50 marks

= the cumulative frequency of the class interval 40 - 49 = 84.

(ii) The number of students obtaining at least 60 marks

= total number of students - the number of students getting less than or equal to 59

= 200 - 141

= 59.

Alternative method

The number of students obtaining at least 60 marks

= Sum of the frequencies of the class intervals 60 - 69, 70 - 79 and 80 - 89

= 37 + 15 + 7

= 59.