Question

Let $X_1, X_2,$ and $X_3$ be independent random variables such that $X_1 \sim N(47, 10)$, $X_2 \sim N(55, 15)$, and $X_3 \sim N(60, 14)$. Then $P(X_1 + X_2 \ge 2X_3)$ equals ............ (round off to two decimal places).

Hint:

When comparing linear combinations of normal variables, compute the difference and standardize it using mean and variance to apply the standard normal distribution.

Answer 1

Correct Answer: 0.01

Step 1: Define a new variable.
Let $Y = X_1 + X_2 - 2X_3$. Since $X_1, X_2, X_3$ are independent normal variables, $Y$ is also normal.

Step 2: Compute mean and variance.
\[ E(Y) = 47 + 55 - 2(60) = -18. \] \[ \mathrm{Var}(Y) = 10 + 15 + 4(14) = 10 + 15 + 56 = 81. \] So, \[ Y \sim N(-18, 81) $\Rightarrow$ \sigma_Y = 9. \]

Step 3: Required probability.
We need $P(Y \ge 0) = P\left(\frac{Y - (-18)}{9} \ge \frac{0 - (-18)}{9}\right) = P(Z \ge 2) = 1 - \Phi(2). \] \[ \Phi(2) = 0.9772 $\Rightarrow$ P(Y \ge 0) = 1 - 0.9772 = 0.0228. \] Hence, approximately \[ \boxed{P(X_1 + X_2 \ge 2X_3) = 0.02.} \] If variance rounding corrected, $P \approx 0.03$.