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प्रश्न
Calculate the Mean deviation about median of the following data.
| Class interval: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 |
| Frequency: | 6 | 7 | 15 | 16 | 4 | 2 |
उत्तर
| x | f | Cumulative Frequency |
| 0 - 10 | 6 | 6 |
| 10 - 20 | 7 | 13 |
| 20 - 30 | 15 | 28 |
| 30 - 40 | 16 | 44 |
| 40 - 50 | 4 | 48 |
| 50 - 60 | 2 | 50 |
| N = 50 |
`"N"/2 = 50/2` = 25
∴ Median lies in the interval (20 - 30) and its corresponding values are L = 20, f = 15, pcf = 13, C = 10
∴ Median = `"L" + (("N"/2 - "pcf")/"f") xx "C"`
= `20 + (25 - 13)/15 xx 10`
= `20 + 120/15`
= 20 + 8
= 28
| x | f | Mid x | |D| = |X − 28| | f|D| |
| 0 - 10 | 6 | 5 | 23 | 138 |
| 10 - 20 | 7 | 15 | 13 | 91 |
| 20 - 30 | 15 | 25 | 3 | 45 |
| 30 - 40 | 16 | 35 | 7 | 112 |
| 40 - 50 | 4 | 45 | 17 | 68 |
| 50 - 60 | 2 | 55 | 27 | 54 |
| ∑f|D| = 508 |
Mean deviation about median = `(sum "f"|"D"|)/"N"`
= `508/50`
= 10.16
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