JEE Main 2023 · 6 April · evening · Numerical answer · +4 / −0
If the mean and variance of the frequency distribution
are 9 and 15.08 respectively, then the value of α²+β²-αβ is ___________.
| xᵢ | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|---|---|---|
| fᵢ | 4 | 4 | α | 15 | 8 | β | 4 | 5 |
Answer and solution
Answer 25
Solution
Maths sir · 4 steps
Use the mean first to get α = β, then take deviations from 9 so the variance becomes one equation in α.
- Use the mean. Total frequency is N = 40 + α + β and Σ fᵢ xᵢ = 360 + 6α + 12β. Setting Σ fᵢ xᵢ = 9N gives 360 + 6α + 12β = 360 + 9α + 9β, so 3β = 3α and α = β.
- Deviations from mean. With dᵢ = xᵢ - 9, the squares dᵢ² are 49, 25, 9, 1, 1, 9, 25, 49. Then Σ fᵢ dᵢ² = 196 + 100 + 9α + 15 + 8 + 9β + 100 + 245 = 664 + 18α, using β = α.
σ² = (Σ fᵢ dᵢ²)/N = (664 + 18α)/(40 + 2α) = 15.08
- Solve for alpha. Cross-multiplying gives 664 + 18α = 603.2 + 30.16α, so 12.16α = 60.8 and α = 5. Check: N = 50 and 754/50 = 15.08.
- So. α = β = 5, so α² + β² - αβ = 25 + 25 - 25 = 25.
The idea to keep. When the mean is given, take deviations from that mean. Then the variance is just (Σ f d²)/N, with no x̄² term to subtract.
Answer from the NTA final answer key. Solution by PrepWiser.