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Question
Using step-deviation method, calculate the mean marks of the following distribution
| Class Interval | 50 - 55 | 55 - 60 | 60 - 65 | 65 - 70 | 70 - 75 | 75 - 80 | 80 - 85 | 85 - 90 |
| Frequency | 5 | 20 | 10 | 10 | 9 | 6 | 12 | 8 |
Sum
Solution
| C.I. | `f` | `x` | `mu = (x - "A")/i` | `f . u` | |
| 50 - 55 | 5 | 52·5 | -3 | -15 | |
| 55 - 60 | 20 | 57·5 | -2 | -40 | |
| 60 - 65 | 10 | 62·5 | -1 | -10 | |
| 65 - 70 | 10 | 67·5 | A = 67·5 | 0 | 0 |
| 70 - 75 | 9 | 72·5 | 1 | 9 | |
| 75 - 80 | 6 | 77·5 | 2 | 12 | |
| 80 - 85 | 12 | 82·5 | 3 | 36 | |
| 85 - 90 | 8 | 87·5 | 4 | 32 | |
| `sumf = 80` | `sumf . u = 24` |
∴ Mean `bar"X" = "A" + (sumf · u)/(sumf) xx i` ..[i = length of C.I.]
= 67·5 + `(24)/(80) xx 5`
= 67·5 + 1·5
= 69.
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Mean of Continuous Distribution
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