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Question
Find out the coefficient of mean deviation about median in the following series.
| Age in years | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 |
| No. of persons | 8 | 12 | 16 | 20 | 37 | 25 | 19 | 13 |
Solution
| X | f | cf |
| 0 - 10 | 8 | 8 |
| 10 - 20 | 12 | 20 |
| 20 - 30 | 16 | 36 |
| 30 - 40 | 20 | 56 |
| 40 - 50 | 37 | 93 |
| 50 - 60 | 25 | 118 |
| 60 - 70 | 19 | 137 |
| 70 - 80 | 13 | 150 |
| N = 150 |
`"N"/2 = 150/2` = 75
The class interval corresponding to cumulative frequency 75 is (40 - 50).
So, the corresponding values from the median class are L = 40, pcf = 56, f = 37, C = 10, `"N"/2` = 75.
Median = `"L" + (("N"/2 - "pcf"))/"f" xx "C"`
= `40 + ((75 - 56)/37) xx 10`
= `40 + 190/37`
= 40 + 5.1351
= 45.1351
= 45.14
Now we calculate the mean deviation about the median of 45.11
| X | f | M | |D| = |M − 45.14| | f|D| |
| 0 - 10 | 8 | 5 | 40.14 | 321.12 |
| 10 - 20 | 12 | 15 | 30.14 | 361.68 |
| 20 - 30 | 16 | 25 | 20.14 | 322.24 |
| 30 - 40 | 20 | 35 | 10.14 | 202.8 |
| 40 - 50 | 37 | 45 | 0.14 | 5.18 |
| 50 - 60 | 25 | 55 | 9.86 | 246.50 |
| 60 - 70 | 19 | 65 | 19.86 | 377.34 |
| 70 - 80 | 13 | 75 | 29.86 | 388.18 |
| N = 150 | ∑f|D| = 2225.04 |
Mean deviation about median = `(sum"f"|"D"|)/"N" = 2225.04/150` = 14.83
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