Answer to Question #222339 in Statistics and Probability for maximila

Question

Suppose X and Y are 2 independent normal random variable such that x~N(65,28) and Y~N(85,36).Find p(134<X+Y<166) and p(Y>x)

Answer 1

The distribution of a sum of two normally distributed independent variables $X\sim N(\mu_1, \sigma_1^2)$ and $Y\sim N(\mu_2, \sigma_2^2)$ is another normal distribution $(X+Y)\sim N(\mu_1+\mu_2, \sigma_1^2+\sigma_2^2).$

Given $X\sim N(65, 28), Y\sim N(85, 36).$ Then $(X+Y)\sim N(65+85, 28+36).$

$$P(134<X+Y<166)$$

$$=P(X+Y<166)-P(X+Y\leq134)$$

$$=P(Z<\dfrac{166-150}{\sqrt{64}})-P(Z\leq\dfrac{134-150}{\sqrt{64}})$$

$$=P(Z<2)-P(Z\leq-2)$$

$$\approx0.97725-0.02275=0.9545$$

The distribution of a difference of two normally distributed independent variables $Y\sim N(\mu_2, \sigma_2^2)$ and $X\sim N(\mu_1, \sigma_1^2)$ is another normal distribution $(Y-X)\sim N(\mu_2-\mu_1, \sigma_1^2+\sigma_2^2).$

Given $X\sim N(65, 28), Y\sim N(85, 36).$ Then $(Y-X)\sim N(85-65, 28+36).$

$$P(Y>X)=P(Y-X>0)=1-P(Y-X\leq0)$$

$$=1-P(Z\leq\dfrac{0-20}{\sqrt{64}})=1-P(Z\leq -2.5)$$

$$\approx0.9938$$