The median of the following data is 525. Find he missing frequency, if it is given that there are 100 observations in the data:
Class Interval | Frequency | Class interval | Frequency |
0-100 | 2 | 500-600 | 20 |
100-200 | 5 | 600-700 | f2 |
200-300 | f1 | 700-800 | 9 |
300-400 | 12 | 800-900 | 7 |
400-500 | 17 | 900-1000 | 4 |
Class interval | Frequency | Cumulative frequency |
0-100 | 2 | 2 |
100-200 | 5 | 7 |
200-300 | f1 | 7 + f1 |
300-400 | 12 | 19 + f1 |
400-500 | 17 | 36 + f1 (F) |
500-600 | 20 (f) | 56 + f1 |
600-700 | f2 | 56 + f1 + f2 |
700-800 | 9 | 65 + f1 + f2 |
800-900 | 7 | 72 + f1 + f2 |
900-1000 | 4 | 76 + f1 + f2 |
N = 100 |
Given, Median = 525
Then median class = 500-600
l = 500, f = 20, F = 36 + f1, h = 100
Median = l + 
525 = 500 + 
* 100
525 – 500 = 
* 100
25 = (14 – f1) 5
5f1 = 45
f1 = 9
Given, sum of frequencies = 100
= 2 + 5 + f1 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100
= 2 + 5 + 9 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100
=85 + f2 = 100
f2 = 15
Therefore, f1 = 9 and f2 = 15