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प्रश्न
Find the mean of the following frequency distribution by the short cut method :
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 9 | 12 | 15 | 10 | 14 |
उत्तर
| Class Interval | xi | fi | A = 25 d = x - A |
`f_i d` |
| 0-10 | 5 | 9 | -20 | -180 |
| 10-20 | 15 | 12 | -10 | -120 |
| 20-30 | A = 25 | 15 | 0 | 0 |
| 30-40 | 35 | 10 | 10 | 100 |
| 40-50 | 45 | 14 | 20 | 280 |
| Total | 60 | 80 |
`barx = A + (Σf_i d)/(Σf_i)`
`barx = 25 + 80/60`
`barx = 25 + 1.33`
`barx = 26.33`
`therefore` Mean = 26.33
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संबंधित प्रश्न
1) Using step–deviation method, calculate the mean marks of the following distribution.
2) State the modal class.
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| Frequency | 5 | 20 | 10 | 10 | 9 | 6 | 12 | 8 |
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| Age-years | 16 – 18 | 18 – 20 | 20 – 22 | 22 – 24 | 24 – 26 |
| No. of students | 2 | 7 | 21 | 17 | 3 |
Using the information given in the adjoining histogram, calculate the mean.
The marks obtained by 120 students in a mathematics test is given below:
| Marks | No. of students |
| 0 – 10 | 5 |
| 10 – 20 | 9 |
| 20 – 30 | 16 |
| 30 – 40 | 22 |
| 40 – 50 | 26 |
| 50 – 60 | 18 |
| 60 – 70 | 11 |
| 70 – 80 | 6 |
| 80 – 90 | 4 |
| 90 – 100 | 3 |
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- the lower quartile.
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| class | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
| Frequency | 7 | 5 | 10 | 12 | 6 |
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| Class | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 |
| Frequency | 7 | 10 | 14 | 17 | 15 | 11 | 6 |
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233, 173, 189, 208, 194, 204, 194, 185, 200 and 220.
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