Question

Let 𝑌 ∼ 𝑁𝑜𝑟𝑚𝑎𝑙(3, 1), 𝑊 ∼ 𝑁𝑜𝑟𝑚𝑎𝑙(1, 2) and 𝑋 ∼ 𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖 (𝑝 = 0.9) where 𝑋 = 1 is success and 𝑋 = 0 is failure. Let 𝑆 = 𝑋𝑌 + (1 − 𝑋)𝑊. Then 𝐸(𝑆) = _________ (round off to 1 decimal place)

Answer 1

Correct Answer: 2.8

We are given: - \(Y \sim Normal(3,1)\) → so \(E(Y) = 3\) - \(W \sim Normal(1,2)\) → so \(E(W) = 1\) - \(X \sim Bernoulli(0.9)\) → \[ P(X=1)=0.9,\quad P(X=0)=0.1 \] Define: \[ S = XY + (1 - X)W \] This means: - When \(X=1\), \(S = Y\) - When \(X=0\), \(S = W\) So the expectation is: \[ E(S) = E(S \mid X=1)P(X=1) + E(S \mid X=0)P(X=0) \] Now compute: \[ E(S \mid X=1) = E(Y) = 3 \] \[ E(S \mid X=0) = E(W) = 1 \] Therefore: \[ E(S) = 3(0.9) + 1(0.1) \] \[ E(S) = 2.7 + 0.1 = 2.8 \] 

Final Answer (rounded to 1 decimal place): \( \boxed{2.8} \)