8(i)
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 |
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| Marks | Number of students (f) | Class marks (x) | Deviation from assumed mean (A) (d = x - A) | f × d |
|---|---|---|---|---|
| 0-10 | 3 | 5 | -20 | -60 |
| 10-20 | 8 | 15 | -10 | -80 |
| 20-30 | 14 | A = 25 | 0 | 0 |
| 30-40 | 9 | 35 | 10 | 90 |
| 40-50 | 4 | 45 | 20 | 80 |
| 50-60 | 2 | 55 | 30 | 60 |
| Total | Σf = 40 | Σfd = 90 |
By formula,
Mean = A + $\frac{Σfd}{Σf} = 25 + \frac{90}{40}$
= 25 + 2.25
= 27.25
Hence, required mean = 27.25