Find the median from the following data: Class1−56−1011−1516−2021−2526−3031−3536−4041−45Frequency710163224161152
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 |
0.5-5.5 | 7 | 7 |
5.5-10.5 | 10 | 17 |
10.5-15.5 | 16 | 33 |
15.5-20.5 | 32 | 65 |
20.5-25.5 | 24 | 89 |
25.5-30.5 | 16 | 105 |
30.5-35.5 | 11 | 116 |
35.5-40.5 | 5 | 121 |
40.5-45.5 | 2 | 123 |
N=∑fi=123 |
N =123
=> N/2 = 61.5
Now,
The cumulative frequency just greater than 61.5 is 65 and corresponding class is 15.5-20.5.
Thus, the median class is 15.5-20.5
I = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33 and N/2 = 61.5
Median = I+[h×(N/2−c)/f]=15.5+[5×(61.5−33)/32]
= 15.5+4.45 = 19.95
Hence, median = 19.95