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Question
The following table gives the marks scored by a set of students in an examination. Calculate the mean of the distribution by using the short cut method.
| Marks | Number of Students (f) |
| 0 – 10 | 3 |
| 10 – 20 | 8 |
| 20 – 30 | 14 |
| 30 – 40 | 9 |
| 40 – 50 | 4 |
| 50 – 60 | 2 |
Solution
Let assumed mean A = 25
| C.I. marks |
xi | fi | di = xi – A | fidi |
| 0 – 10 | 5 | 3 | – 20 | – 60 |
| 10 – 20 | 15 | 8 | – 10 | – 80 |
| 20 – 30 | 25 = A | 14 | 0 | 0 |
| 30 – 40 | 35 | 9 | 10 | 90 |
| 40 – 50 | 45 | 4 | 20 | 80 |
| 50 – 60 | 55 | 2 | 30 | 60 |
| n = 40 | `sumf_i d_i` = 90 |
Mean = `A + (sumf_i d_i)/n`
= `25 + 90/40`
= 25 + 2.25
= 27.25 marks
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