Clearly, we can see that given data is not continuous. So, to convert into continuous frequency distribution, we need to subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class interval
Class | fi | Cumulative frequency | xi | |xi−M| | fi|xi−M| |
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 |
Total | 100 | | | | 735 |
Here, N=100N2=1002=50
which occurs in the cumulative frequency 63.
Hence, the median class is 35.5−40.5.
Median M=l+N2−Cf−1fm×h
=35.5+50−3726×5
=35.5+1326×5
⇒M=35.5+2.5=38
Mean deviation about the median is
M.D.(M)=1N8∑i=1fi∣xi−M∣
=1100×735=7.35