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प्रश्न
1) Using step–deviation method, calculate the mean marks of the following distribution.
2) State the modal class.
| Class Interval | 50 - 55 | 55 - 60 | 60 - 65 | 65 - 70 | 70 - 75 | 75 - 80 | 80 - 85 | 85 – 90 |
| Frequency | 5 | 20 | 10 | 10 | 9 | 6 | 12 | 8 |
उत्तर
1)
| Class interval |
Frequency (f) |
x | d = x – A= x – 67.5 |
t = d/i i = 5 |
f xt |
| 50 – 55 | 5 | 52.5 | -15 | -3 | -15 |
| 55 – 60 | 20 | 57.5 | -10 | -2 | -40 |
| 60 – 65 | 10 | 62.5 | -5 | -1 | -10 |
| 65 – 70 | 10 | 67.5 | 0 | 0 | 0 |
| 70 – 75 | 9 | 72.5 | 5 | 1 | 9 |
| 75 – 80 | 6 | 77.5 | 10 | 2 | 12 |
| 80 – 85 | 12 | 82.5 | 15 | 3 | 36 |
| 85 – 90 | 8 | 87.5 | 20 | 4 | 32 |
| Total | 80 | 24 |
Assumed mean (A) = 67.5
Class size, i = 5
Mean = `A + i (sum ft)/(sum f)``
`= 67.5 + 5 xx 24/80`
`= 67.5 + 1.5`
= 69
Thus, the mean of the given data is 69.
2) Modal class is the class corresponding to the greatest frequency.
So, the modal class is 55 – 60.
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