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Question
Find the mean deviation about the mean for the data.
| Income per day in ₹ | Number of persons |
| 0-100 | 4 |
| 100-200 | 8 |
| 200-300 | 9 |
| 300-400 | 10 |
| 400-500 | 7 |
| 500-600 | 5 |
| 600-700 | 4 |
| 700-800 | 3 |
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Solution
Let a = 350, h = 100, di = `(x_i - 350)/100`
| Income per day | Mid values xi | di | Frequency fi | fidi | `|x_i - overline x|` | `f_i|x_i - overline x|` |
| 0 - 100 | 50 | −3 | 4 | −12 | 308 | 1232 |
| 100 - 200 | 150 | −2 | 8 | −16 | 308 | 1664 |
| 200 - 300 | 250 | −1 | 9 | −9 | 108 | 972 |
| 300 - 400 | 350 | 0 | 10 | 0 | 8 | 80 |
| 400 - 500 | 450 | 1 | 7 | 7 | 92 | 644 |
| 500 - 600 | 550 | 2 | 5 | 10 | 192 | 960 |
| 600 - 700 | 650 | 3 | 4 | 12 | 292 | 1168 |
| 700 - 800 | 750 | 4 | 3 | 12 | 392 | 1176 |
| Sum | - | - | 50 | 4 | - | 7896 |
`overline x = a + (sumf_id_i)/N xx h`
= `350 + 4/50 xx 100`
= 358
Mean Deviation = `(sumf_i |x_i - overline x|)/N`
= `7856/50`
= 157.92
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