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प्रश्न
In an annual examination, marks (out of 90) obtained by students of Class X in mathematics are given below:
| Marks Obtained |
0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
| Number of students |
2 | 4 | 5 | 20 | 9 | 10 |
Find the mean marks.
उत्तर
Let us choose a = 52.5, h = 15, then `d_i = x_i – 52.5 and u_i =( x_i−52.5 )/ 15`
Using step-deviation method, the given data is shown as follows:
| Weight | Number of students `(f_i)` |
Class mark `(x_i)` | `d_i = x_i – `37.5 | `u_i = (x_i−52.5)/ 15` |
`(f_i u_i)` |
| 0 – 15 | 2 | 7.5 | -45 | -3 | -6 |
| 15 – 30 | 4 | 22.5 | -30 | -2 | -8 |
| 30 – 45 | 5 | 37.5 | -15 | -1 | -5 |
| 45 – 60 | 20 | 52.5 | 0 | 0 | 0 |
| 60 – 75 | 9 | 67.5 | 15 | 1 | 9 |
| 75 – 90 | 10 | 82.5 | 30 | 2 | 20 |
| Total | `Ʃ f_i` = 50 | `Ʃ f_i u_i` = 10 |
The mean of the given data is given by,
x = a + `((Ʃ _ i f_i u_i)/(Ʃ_i f_i)) xx h `
=52.5`+(10/50) xx 15`
= 52.5 + 3
= 55.5
Thus, the mean is 55.5.
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