Mathematics
Calculate the mean of the distribution given below using the short cut method.
| Marks | No. of students |
|---|---|
| 11 - 20 | 2 |
| 21 - 30 | 6 |
| 31 - 40 | 10 |
| 41 - 50 | 12 |
| 51 - 60 | 9 |
| 61 - 70 | 7 |
| 71 - 80 | 4 |
Measures of Central Tendency
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Answer
Construct the table as under, taking assumed mean as 45.5
| Marks (Classes) | Class mark (yi) | Deviation (di = yi - a) | No. of students (Frequency (fi)) | fidi |
|---|---|---|---|---|
| 11 - 20 | 15.5 | -30 | 2 | -60 |
| 21 - 30 | 25.5 | -20 | 6 | -120 |
| 31 - 40 | 35.5 | -10 | 10 | -100 |
| 41 - 50 | 45.5 | 0 | 12 | 0 |
| 51 - 60 | 55.5 | 10 | 9 | 90 |
| 61 - 70 | 65.5 | 20 | 7 | 140 |
| 71 - 80 | 75.5 | 30 | 4 | 120 |
| Total | 50 | 70 |
idi}{∑f_i} \\[1em] = 45.5 + \dfrac{70}{50} \\[1em] = 45.5 + 1.4 \\[1em] = 46.9
Hence, mean of the following distribution is 46.9 marks
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