Let X be a random variable. Let $E(X)$ and $Var (X)$ denote the mean and the variance of X respectively. Then match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(IV), (B)-(III), (C)-(I), (D)-(II) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) |
(A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II) Matching with Explanation(A) If Var(X) = α, then Var(2X+3) → (III) 4α Explanation: $Var(aX + b) = a^2 Var(X) = 2^2 \cdot \alpha = 4\alpha$ (B) If E(X) = α, then E(2X) → (IV) 2α Explanation: $E(aX) = a E(X) = 2 \cdot \alpha = 2\alpha$ (C)If Var(X) = α, then Var(3X−α) + Var( $\sqrt{2} X$ + β ) → (I) 11α Explanation: $Var(3X-\alpha) = 3^2 \cdot \alpha = 9\alpha$, $Var(\sqrt{2} X + \beta) = (\sqrt{2})^2 \cdot \alpha = 2\alpha$, sum = $9\alpha + 2\alpha = 11\alpha$ (D) If $E(X) = \frac{5\alpha}{12}$, then E(12X+α) → (II) 6α Explanation: $E(12X + \alpha) = 12 \cdot \frac{5\alpha}{12} + \alpha = 5\alpha + \alpha = 6\alpha$ |