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प्रश्न
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.
| Class | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
| Frequency | 5 | f1 | 10 | f2 | 7 | 8 |
उत्तर
| Class interval | Mid value(x1) | Frequency(f1) | f1x1 |
| 0 - 20 | 10 | 5 | 50 |
| 20 - 40 | 30 | f1 | 30f1 |
| 40 - 60 | 50 | 10 | 500 |
| 60 - 80 | 70 | f2 | 70f2 |
| 80 - 100 | 90 | 7 | 630 |
| 100 - 120 | 110 | 8 | 880 |
| N = 50 | `sumf_1x_1=30f_1+70f_2+2060` |
Given
Sum of frequency = 50
⇒ 5 + f1 + 10 + f2 + 7 + 8 = 50
⇒ f1 + f2 = 50 - 5 - 10 - 7 - 8
⇒ f1 + f2 = 20
⇒ 3f1 + 3f2 = 60 .........(1)[multiply both side by '3']
And Mean = 62.8
`rArr(sumf_1x_1)/N=62.8`
`rArr(30f_1+70f_2+2060)/50=62.8`
⇒ 30f1 + 70f2 + 2060 = 62.8 x 50
⇒ 30f1 + 70f2 + 2060 = 3140
⇒ 30f1 + 70f2 = 3140 - 2060
⇒ 30f1 + 70f2 = 1080
⇒ 3f1 + 7f2 = 108 ............(2)[Both sides divided by 10]
Subtract equation (1) from equation (2)
⇒ 3f1 + 7f2 - (3f1 + 3f2) = 108 - 60
⇒ 3f1 + 7f2 - 3f1 - 3f2 = 48
⇒ 4f2 = 48
⇒ f2 = 48/4 = 12
Put value of f2 in equation (1)
⇒ 3f1 + 3f2 = 60
⇒ 3f1 + 3(12) = 60
⇒ 3f1 + 36 = 60
⇒ 3f1 = 60 - 36
⇒ 3f1 = 24
⇒ f1 = 24/3 = 8
∴ f1 = 8 and f2 = 12
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