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प्रश्न
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f1 and f2.
| Class Interval | 0 − 20 | 20 − 40 | 40 − 60 | 60 − 80 | 80 − 100 | 100 − 120 |
| Frequency | 5 | f1 | 10 | f2 | 7 | 8 |
उत्तर
Arithmetic Mean `(barx)` = 62.8
Sum of all the frequencies `(sumf_"i")` = 50
Let the missing frequencies be f1 and f2
| Class interval | frequency fi |
mid value xi |
fixi |
| 0 − 20 | 5 | 10 | 0 |
| 20 − 40 | f1 | 30 | 30 f1 |
| 40 − 60 | 10 | 50 | 500 |
| 60 − 80 | f2 | 70 | 70 f2 |
| 80 − 100 | 7 | 90 | 630 |
| 100 − 120 | 8 | 110 | 880 |
| `sumf_"i" = 30 + f_1 + f_2` | `sumf_"i"x_"i" = 2060 + 30f_1 + 70f_2` |
`30 + f_1 + f_2` = 50
`f_1 + f_2` = 20 ...(1)
Mean = `(sumf_"i"x_"i")/(sumf_"i")`
62.8 = `(2060 + 30f_1 + 70f_2)/50`
3140 = `2060 + 30f_1 + 70f_2`
∴ `30f_1 + 70f_2` = 3140 − 2060
`30f_1 + 70f_2` = 1080
(÷ by 10) ⇒ `3f_1 + 7f_2` = 108 ...(2)
(1) × 3 ⇒ `3f_1 + 3f_2` = 60 ...(3)
(2) × 3 ⇒ `3f_1 + 7f_2` = 108 ...(4)
(−) (−) (−)
(3) − (4) ⇒ `−4f_2` = − 48
`f_2 = 48/4 ⇒ f_2` = 12
Substitute of the value of f2 in (1)
f1 + 12 = 20
⇒ f1 = 20 – 12 = 8
The Missing frequency is 8 and 12.
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