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Question
The following distribution gives the number of accidents met by 160 workers in a factory during a month.
| No. of accidents(x) | 0 | 1 | 2 | 3 | 4 |
| No. of workers (f) | 70 | 52 | 34 | 3 | 1 |
Find the average number of accidents per worker.
Solution
Let the assumed mean (A) = 2
| No. of Accidents (x1) | No. of workers (f1) |
u1 = x1 - A = x1 - 2 |
f1u1 |
| 0 | 70 | -2 | -140 |
| 1 | 52 | -1 | -52 |
| 2 | 34 | 0 | 0 |
| 3 | 3 | 1 | 3 |
| 4 | 1 | 2 | 2 |
| N = 160 | `sumf_1"u"_1=-187` |
Average no of accidents per day workers `=A+(sumf_1"u"_1)/N`
`=2+(-187)/160`
`=2-187/160`
`=(320-187)/160`
`=133/160`
= 0.83
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