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
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List-I |
List-II |
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(A) If $Var (X) = α$, then $Var (2X+ 3)$ is |
(I) $11α$ |
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(B) If $E(X) = α$, then $E(2X)$ is |
(ΙΙ) $6α$ |
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(C) If $Var (X) = α$, then $Var(3X-α) + Var (\sqrt{2}X+β)$ is |
(III) $4α$ |
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(D) If $E(X) =\frac{5α}{12}$, then $E (12X+ α)$ is |
(IV) $2α$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
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: $E(12X + \alpha) = 12 \cdot \frac{5\alpha}{12} + \alpha = 5\alpha + \alpha = 6\alpha$