JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
Let μ be the mean and σ be the standard deviation of the distribution
where Σ fᵢ=62. If [x] denotes the greatest integer ≤ x, then [μ²+σ²] is equal to :
| xᵢ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| fᵢ | k + 2 | 2k | k² - 1 | k² - 1 | k² + 1 | k - 3 |
- (a) 9
- (b) 8
- (c) 6
- (d) 7
Answer and solution
Answer (b) 8
Solution
Maths sir · 3 steps
Do not find μ and σ separately: μ² + σ² is just the mean of the squares.
- Find k. Adding the frequencies gives 3k² + 4k - 2 = 62, so 3k² + 4k - 64 = 0, which factorises as (k - 4)(3k + 16) = 0. A frequency count must be a whole number and k - 3 cannot be negative, so k = 4.
xᵢ fᵢ fᵢ xᵢ fᵢ xᵢ² 0 6 0 0 1 8 8 8 2 15 30 60 3 15 45 135 4 17 68 272 5 1 5 25 Total 62 156 500 - Use the variance formula. Variance is the mean of the squares minus the square of the mean, σ² = (Σ fᵢ xᵢ²)/N - μ². Move μ² to the other side and it adds straight back.
μ² + σ² = (Σ fᵢ xᵢ²)/(Σ fᵢ) = 500/62 = 250/31 ≈ 8.06
- So. [μ² + σ²] = [8.06] = 8, which is option (b).
The idea to keep. Whenever a question asks for μ² + σ², it is asking for (Σ fᵢ xᵢ²)/N. You never need μ on its own, so skip that work.
Answer from the NTA final answer key. Solution by PrepWiser.