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प्रश्न
If the mean of the following frequency distribution is 18, find the missing frequency.
| Class interval | 11 – 13 | 13 – 15 | 15 – 17 | 17 – 19 | 19 – 21 | 21 – 23 | 23 – 25 |
| Frequency | 3 | 6 | 9 | 13 | f | 5 | 4 |
उत्तर
| Class | Mid point `(bb(x_i))` |
Frequency `(bb(f_i))` |
`bb(f_ix_i)` |
| 11 – 13 | 12 | 3 | 36 |
| 13 – 15 | 14 | 6 | 84 |
| 15 – 17 | 16 | 9 | 144 |
| 17 – 19 | 18 | 13 | 234 |
| 19 – 21 | 20 | f | 20f |
| 21 – 23 | 22 | 5 | 110 |
| 23 – 25 | 24 | 4 | 96 |
| `sumf_i = 40 + f` | `sumf_ix_i = 704 + 20f` |
Mean = `(sumf_ix_i)/(sumf_i)`
∴ 18 = `(704 + 20f)/(40 + f)` ...[∵ Given, mean = 18]
⇒ 18(40 + f) = 704 + 20f
⇒ 720 + 18f = 704 + 20f
⇒ 2f = 16
⇒ f = 8
So, missing frequency f is 8.
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| 300 - 350 | 30 |
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| 400 - 450 | 16 |
| 450 - 500 | 7 |
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| Frequency | 9 | 12 | 15 | 10 | 14 |
Find the mean of the following data, using direct method:
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11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
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The mean of the following frequency data is 42, Find the missing frequencies x and y if the sum of frequencies is 100
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| 10 | 5 |
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