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प्रश्न
Calculate the mean deviation about the mean for the following frequency distribution:
| Class interval | 0 – 4 | 4 – 8 | 8 – 12 | 12 – 16 | 16 – 20 |
| Frequency | 4 | 6 | 8 | 5 | 2 |
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उत्तर
| Class-interval | `f_i` | `x_i` | `f_ix_i` | `d_i = |x_i - barx|` | `f_i d_i` |
| 0 – 4 | 4 | 2 | 8 | 7.2 | 28.8 |
| 4 – 8 | 6 | 6 | 36 | 3.2 | 19.2 |
| 8 – 12 | 8 | 10 | 80 | 0.8 | 6.4 |
| 12 – 16 | 5 | 14 | 70 | 4.8 | 24.0 |
| 16 – 20 | 2 | 18 | 36 | 8.8 | 17.6 |
| N = 25 | `sumf_ix_i` = 230 | `sumf_i d_i` = 96.0 |
Mean = `(sumf_ix_i)/N = 230/25` = 9.2
and Mean deviation = `(sumf_i d_i)/N = 96/25` = 3.84
Hence, the required M.D. = 3.84
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