Here we will learn how to find the mean of classified data (continuous and discontinuous).
If the class marks of the class intervals be m_{1}, m_{2}, m_{3}, m_{4}, ……, m_{n} and the frequencies of the corresponding classes be f_{1}, f_{2}, f_{3}, f_{4}, …….., f_{n} then the mean of the distribution is given by
Mean = A or (\(\overline{x}\)) = \(\frac{m_{1}f_{1} + m_{2}f_{2} + m_{3}f_{3} + m_{4}f_{4} + .... + m_{n}f_{n}}{f_{1} + f_{2} + f_{3} + f_{4} + .... + f_{n}}\)
Symbolically, A = \(\frac{\sum m_{i}f_{i}}{\sum f_{i}}\)
This is the direct method of finding the mean of classified data.
Solved Examples on Mean of Classified Data (Continuous and Discontinuous)
1. Find the mean of the following frequency distribution.
Class Interval
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
Frequency
4
11
8
7
10
5
Solution:
Here, the calculations are done in the table given below.
Class Interval |
Class Mark (m_{i}) |
Frequency (f_{i}) |
m_{i}f_{i} |
0 - 10 |
5 |
4 |
20 |
10 - 20 |
15 |
11 |
165 |
20 - 30 |
25 |
8 |
200 |
30 - 40 |
35 |
7 |
245 |
40 - 50 |
45 |
10 |
450 |
50 - 60 |
55 |
5 |
275 |
\(\sum f_{i}\) = 45 |
\(\sum m_{i}f_{i}\) = 1355 |
Therefore, mean A = \(\frac{\sum m_{i}f_{i}}{\sum f_{i}}\)
= \(\frac{1355}{45}\)
= 30\(\frac{1}{9}\)
2. Find the mean of the following frequency distribution.
Class Interval
11 - 20
21 - 30
31 - 40
41 - 50
51 - 60
Frequency
12
10
15
16
20
Solution:
After making the class intervals overlapping, we do the following calculations.
Class Interval |
Class Mark (m_{i}) |
Frequency (f_{i}) |
m_{i}f_{i} |
10.5 - 20.5 |
15.5 |
12 |
186.0 |
20.5 - 30.5 |
25.5 |
10 |
255.0 |
30.5 - 40.5 |
35.5 |
15 |
532.5 |
40.5 - 50.5 |
45.5 |
16 |
728.0 |
50.5 - 60.5 |
55.5 |
20 |
1110.0 |
\(\sum f_{i}\) = 73 |
\(\sum m_{i}f_{i}\) = 2811.5 |
Therefore, mean A = \(\frac{\sum m_{i}f_{i}}{\sum f_{i}}\)
= \(\frac{2811.5}{73}\)
= 38.51 (Approx.).
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