Mathematics
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 |
|---|---|
| 0 – 10 | 3 |
| 10 – 20 | 8 |
| 20 – 30 | 14 |
| 30 – 40 | 9 |
| 40 – 50 | 4 |
| 50 – 60 | 2 |
Measures of Central Tendency
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Answer
| Marks | Number of students (f) | class marks (x) | Deviation d = x - A | fd |
|---|---|---|---|---|
| 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 | ∑f = 40 | ∑fd = 90 |
By formula,
Hence, required mean = 27.25
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