Find the median from the following data:
MarksNo. of studentsBelow 1012Below 2032Below 3057Below 4080Below 5092Below 60116Below 70164Below 80200
Sol:
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
Marks | Frequency fi | C.F |
0-10 | 12 | 12 |
10-20 | 20 | 32 |
20-30 | 25 | 57 |
30-40 | 23 | 80 |
40-50 | 12 | 92 |
50-60 | 24 | 116 |
60-70 | 24 | 116 |
70-80 | 36 | 200 |
N=∑fi=200 |
N =200
=> N/2 = 100
Now,
The cumulative frequency just greater than 100 is 116 and corresponding class is 50-60.
Thus, the median class is 50-60
I = 50, h = 10, f = 24, c = C.F. preceding median class = 92 and N/2 = 100
Median = I+[h×(N/2−c)/f]=50+[10×(100−92)/24]
=50+[10×84/24]=50+[10×84/24]= 50 + 3.33 = 53.33
Hence, Median = 53.33