JEE Main 2024 · 30 January · morning · Single correct · +4 / −1
Let M denote the median of the following frequency distribution
Then 20M is equal to :
| Class | 0 - 4 | 4 - 8 | 8 - 12 | 12 - 16 | 16 - 20 |
|---|---|---|---|---|---|
| Frequency | 3 | 9 | 10 | 8 | 6 |
- (a) 104
- (b) 52
- (c) 208
- (d) 416
Answer and solution
Answer (c) 208
Solution
Maths sir · 3 steps
Build the cumulative frequency column, find where the 18th value sits, then use the median formula.
- Total and half. The total frequency is N = 3 + 9 + 10 + 8 + 6 = 36, so N/2 = 18. The median class is the first class whose cumulative frequency reaches 18.
Class Frequency Cumulative frequency 0 to 4 3 3 4 to 8 9 12 8 to 12 10 22, first to pass 18 12 to 16 8 30 16 to 20 6 36 - Read the values. For the median class 8 to 12, the lower limit is l = 8 and the class frequency is f = 10. The cumulative frequency before it is cf = 12, and the class width is h = 4.
M = l + (N/2 - cf)/f × h = 8 + (18 - 12)/10 × 4 = 8 + 2.4 = 10.4
- So. 20M = 20 × 10.4 = 208, which is option (c).
The idea to keep. For grouped data, find N/2, then the class where the cumulative frequency first crosses it. In the formula, cf belongs to the class before the median class.
Answer from the NTA final answer key. Solution by PrepWiser.