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Question
If µ and σ2 are the mean and variance of the discrete random variable X and E(X + 3) = 10 and E(X + 3)2 = 116, find µ and σ2
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Solution
Given E(X + 3) = 10
E(aX + b) = aE(X) + b
E(X + 3) = 10
⇒ E(X) + 3 = 10
⇒ µ = E(X) = 7
E(X + 3)2 = 116
E(X2 + 6X + 9) = 116
E(X2) + 6E(X) + 9 = 116
E(X2) + 6(7) + 9 = 116
E(X2) = 65
σ² = Var (X)
= E(X2) – [E(X)]2
= 65 – 49
= 16
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