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प्रश्न
If the median of the following data is 32.5, find the missing frequencies.
| Class interval: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | Total |
| Frequency: | f1 | 5 | 9 | 12 | f2 | 3 | 2 | 40 |
उत्तर
| Class interval | Frequency | Cumulative frequency |
| 0 - 10 | f1 | f1 |
| 10 - 20 | 5 | 5 + f1 |
| 20 - 30 | 9 | 14 + f1 |
| 30 - 40 | 12 | 26 + f1 |
| 40 - 50 | f2 | 26 + f1 + f2 |
| 50 - 60 | 3 | 29 + f1 + f2 |
| 60 - 70 | 2 | 31 + f1 + f2 |
| N = 40 |
Given
Median = 32.5
The median class = 30 - 40
l = 30, h = 40 - 30 = 10, f = 12 and F = 14 + f1
Median `=l+(N/2-F)/fxxh`
`rArr32.5=30+(20-(14+f1))/12xx10`
`rArr32.5-30=(20-14-f1)/12xx10`
`rArr2.5=(6-f1)/12xx10`
`rArr2.5=(6-f1)/6xx5`
⇒ 2.5 x 6 = (6 - f1) x 5
⇒ 15 = (6 - f1) x 5
⇒ 15/5 = 6 - f1
⇒ 3 = 6 - f1
⇒ f1 = 6 - 3
⇒ f1 = 3
Given sum of frequencies = 40
⇒ f1 + 5 + 9 + 12 + f2 + 3 + 2 = 40
⇒ 3 + 5 + 9 + 12 + f2 + 3 + 2 = 40
⇒ 34 + f2 = 40
⇒ f2 = 40 - 34
⇒ f2 = 6
∴ f1 = 3 and f2 = 6
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| Age (in years) | No. of Patients |
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| 20-30 | 42 |
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| Weight (in kg) | 40−45 | 45−50 | 50−55 | 55−60 | 60−65 | 65−70 | 70−75 |
| Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
If the median of the distribution given below is 28.5, find the values of x and y.
| Class interval | Frequency |
| 0 - 10 | 5 |
| 10 - 20 | x |
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| Total | 60 |
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| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of Students |
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