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
The above distribution is discontinuous, converting into continuous distribution, we get :
Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2
=
= 0.5
Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.
Let assumed mean (A) be 45.5
| Marks (Classes before adjustment) | Marks (Classes after adjustment) | Class mean (x) | d = x - A | No. of students (frequency) | fd |
|---|---|---|---|---|---|
| 11 - 20 | 10.5 - 20.5 | 15.5 | -30 | 2 | -60 |
| 21 - 30 | 20.5 - 30.5 | 25.5 | -20 | 6 | -120 |
| 31 - 40 | 30.5 - 40.5 | 35.5 | -10 | 10 | -100 |
| 41 - 50 | 40.5 - 50.5 | 45.5 | 0 | 12 | 0 |
| 51 - 60 | 50.5 - 60.5 | 55.5 | 10 | 9 | 90 |
| 61 - 70 | 60.5 - 70.5 | 65.5 | 20 | 7 | 140 |
| 71 - 80 | 70.5 - 80.5 | 75.5 | 30 | 4 | 120 |
| Total | 50 | 70 |
n = Σf = 50
Mean = A +
=
= 45.5 + 1.4
= 46.9
Hence, mean = 46.9
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