JEE Main 2021 · 25 July · morning · Numerical answer · +4 / −0
Consider the following frequency distribution :
If the sum of all frequencies is 584 and median is 45, then | α - β | is equal to _______________.
| Class : | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|
| Frequency : | α | 110 | 54 | 30 | β |
Answer and solution
Answer 164
Solution
Maths sir · 4 steps
Two unknowns need two equations: one from the total frequency, one from the median formula.
- Total frequency. α + 110 + 54 + 30 + β = 584, so α + β = 390.
- Median class. The median is 45, so the median class is 40-50. Here l = 40, h = 10, f = 30 and N/2 = 292. The cumulative frequency before this class is C = α + 110 + 54 = α + 164.
45 = 40 + (292 - (α + 164))/30 × 10
- Solve for alpha. This gives 5 = (128 - α)/3, so 128 - α = 15 and α = 113. Then β = 390 - 113 = 277.
- So. |α - β| = |113 - 277| = 164.
The common slip: 56. Leaving out the 54 of the 30-40 class from the cumulative frequency gives α = 167, β = 223 and 56, not 164.
The idea to keep. For grouped median, the median value tells you the median class. The cumulative frequency C is the sum of all classes before it, not including the median class itself.
Answer from the NTA final answer key. Solution by PrepWiser.