Question -
| Age (on nearest birthday) | 17-19.5 | 20-25.5 | 26-35.5 | 36-40.5 | 41-50.5 | 51-55.5 | 56-60.5 | 61-70.5 |
| No. of persons | 5 | 16 | 12 | 26 | 14 | 12 | 6 | 5 |
Calculate the meandeviation from the median age.
Answer -
To find the mean deviation from the median, firstly let uscalculate the median.
N = 96
So, N/2 = 96/2 = 48
The cumulative frequency just greater than 48 is 59, and thecorresponding value of x is 38.25
So, Median = 38.25
| Class Interval | xi | fi | Cumulative Frequency | |di| = |xi – M| | fi |di| |
| 17-19.5 | 18.25 | 5 | 5 | 20 | 100 |
| 20-25.5 | 22.75 | 16 | 21 | 15.5 | 248 |
| 36-35.5 | 30.75 | 12 | 33 | 7.5 | 90 |
| 36-40.5 | 38.25 | 26 | 59 | 0 | 0 |
| 41-50.5 | 45.75 | 14 | 73 | 7.5 | 105 |
| 51-55.5 | 53.25 | 12 | 85 | 15 | 180 |
| 56-60.5 | 58.25 | 6 | 91 | 20 | 120 |
| 61-70.5 | 65.75 | 5 | 96 | 27.5 | 137.5 |
| | Total = 96 | | | Total = 980.5 |
N = 96
MD=
= 1/96 × 980.5
= 10.21
∴ Themean deviation is 10.21