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Question
Find the mean of the following frequency distribution table using a suitable method:
| Class | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 - 70 |
| Frequency | 25 | 40 | 42 | 33 | 10 |
Solution
Let us choose a = 45, h = 10, then `d_i = x_i – 45 and u_i =
(x_i−45)/10`
Using step-deviation method, the given data is shown as follows:
| Weight | Number of packets `(f_i)` |
Class mark `(x_i)` | `d_i = x_i` – 45 | `u_i = (x_i−45)/ 10` |
`(f_i u_i)` |
| 20 – 30 | 25 | 35 | -20 | -2 | -50 |
| 30 – 40 | 40 | 35 | -10 | -1 | -40 |
| 40 – 50 | 42 | 45 | 0 | 0 | 0 |
| 50 – 60 | 33 | 55 | 10 | 1 | 33 |
| 60 – 70 | 10 | 65 | 20 | 2 | 20 |
| Total | `Ʃ f_i` = 150 | `Ʃ f_i u_i = `-37 |
The mean of the given data is given by,
x a+ `((Ʃ_i f_i u_i)/(Ʃ_i f_i)) xx h`
`= 45 - (37/150)xx10`
=`45-37/15`
=45-2.466
= 42.534
Hence, the mean is 42.534.
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