# Mean of Classified Data

Here we will learn how to find the mean of classified data (continuous and discontinuous).

If the class marks of the class intervals be m1, m2, m3, m4, ……, mn and the frequencies of the corresponding classes be f1, f2, f3, f4, …….., fn 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 (mi) Frequency (fi) mifi 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 (mi) Frequency (fi) mifi 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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