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Question
Given that the mean of the following frequency distribution is 30, find the missing frequency ‘f’.
| Class Interval | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 |
| Frequency | 4 | 6 | 10 | f | 6 | 4 |
Solution
| Class interval |
Frequency (f) |
x | d = x - A | t = `"d"/"i"` | ft |
| 0 - 10 | 4 | 5 | - 30 | - 3 | - 12 |
| 10 - 20 | 6 | 15 | - 20 | - 2 | - 12 |
| 20 - 30 | 10 | 25 | - 10 | - 1 | - 10 |
| 30 - 40 | f | 35 = A | 0 | 0 | 0 |
| 40 - 50 | 6 | 45 | 10 | 1 | 6 |
| 50 - 60 | 4 | 55 | 20 | 2 | 8 |
Step deviation method,
∑f = 30 + f, A = 35, Class width (i) = 10
and ∑ft = - 34 + 14 = - 20
∴ Mean = A + `(sum "ft")/(sum "f") xx "i"`
`=> 30 = 35 + (- 20)/(30 + "f") xx 10`
`=> - 5 = (- 200)/(30 + "f")`
`=> 30 + "f" = 200/5`
⇒ 30 + f = 40
⇒ f = 40 - 30
⇒ f = 10
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