Question -
Answer -
The given data isconverted into continuous frequency distribution by subtracting 0.5 from thelower limit and adding the 0.5 to the upper limit of each class intervals andappend other columns after calculations.
| Age | Number fi | Cumulative frequency (c.f.) | Mid – points xi | |xi – Med| | fi|xi – Med| |
| 15.5 – 20.5 | 5 | 5 | 18 | 20 | 100 |
| 20.5 – 25.5 | 6 | 11 | 23 | 15 | 90 |
| 25.5 – 30.5 | 12 | 23 | 28 | 10 | 120 |
| 30.5 – 35.5 | 14 | 37 | 33 | 5 | 70 |
| 35.5 – 40.5 | 26 | 63 | 38 | 0 | 0 |
| 40.5 – 45.5 | 12 | 75 | 43 | 5 | 60 |
| 45.5 – 50.5 | 16 | 91 | 48 | 10 | 160 |
| 50.5 – 55.5 | 9 | 100 | 53 | 15 | 135 |
| 100 | | | | 735 |
The class intervalcontaining Nth/2 or 50th item is 35.5 –40.5
So, 35.5 – 40.5 is themedian class.
Then,
Median = l + (((N/2) –c)/f) × h
Where, l = 35.5, c =37, f = 26, h = 5 and N = 100
Median = 35.5 + (((50– 37))/26) × 5
= 35.5 + 2.5
= 38
