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प्रश्न
Find the mean of the following distribution by step deviation method:
| Class Interval | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 10 | 6 | 8 | 12 | 5 | 9 |
उत्तर
| Class interval |
Frequency (f) |
Class mark (x) |
d = `(x - A)/h` (A = 55) | fd |
| 20-30 | 10 | 25 | -3 | -30 |
| 30-40 | 6 | 35 | -2 | -12 |
| 40-50 | 8 | 45 | -1 | -8 |
| 50-60 | 12 | A = 55 | 0 | 0 |
| 60-70 | 5 | 65 | 1 | 5 |
| 70-80 | 9 | 75 | 2 | 18 |
| Total | 50 | -27 |
Here A = 55, h = 10
Mean = `A + (sumfd)/(sumf) xx h`
`= 55 + (-27)/50 xx 10`
= 55 - 5.4
= 49.6
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| No.of students | 5 | 10 | 11 | 20 | 27 | 38 | 40 | 29 | 14 | 6 |
Using a graph paper, draw an Ogive for the above distribution. Use your Ogive to estimate:
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An incomplete frequency distribution is given below
| Variate | Frequency |
| 10 – 20 | 12 |
| 20 – 30 | 30 |
| 30 – 40 | ? |
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| 60 – 70 | 25 |
| 70 – 80 | 18 |
| Total | 229 |
Median value is 46, the missing frequency is:
