# Find out the probability $P(U>V)?$

Let $X_1, X_2, X_3, X_4, X_5$ be independent random variable with $X_1$ follow $N(200, 8)$, $X_2$ follow $N(104, 8)$, $X_3$ follow $N(108, 15)$, $X_4$ follow $N(120, 15)$, $X_5$ follow $N(210, 15)$. Let $U=\frac {X_1 + X_2}{2}$ and $V=\frac{X_3 + X_4 + X_5}{3}$. Then $P(U>V)?$

I find $U$ follow $N(152, 2)$ and $V$ follow $N(146, 5)$ Is this correct and after that what is the probability?

• hint: $U$ and $V$ are independent – Martín Vacas Vignolo Jan 9 '18 at 13:59
• Why do you think $U$ has variance $2$ rather than $4$? – Henry Jan 9 '18 at 14:06
• Can you find the distribution of $V-U$? – Henry Jan 9 '18 at 14:06
• Variance of $U$ is 4 but S.d is 2 – 1256 Jan 9 '18 at 14:07

If $$U \sim N(\mu_1,\sigma_1^2) \text{ and } V\sim N(\mu_2,\sigma_2^2)$$ and they are independent, then $$U-V\sim N(\mu_1-\mu_2, \sigma_1^2 + \sigma_2^2),$$ so \begin{align} \Pr(U>V) & = \Pr(U-V>0) \\[10pt] & = \Pr\left( \frac{(U-V) - (\mu_1-\mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}} > \frac{0 - (\mu_1-\mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}} \right) \\[10pt] & = 1- \Phi\left( \frac{0 - (\mu_1-\mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}} \right) \end{align} where $\Phi$ is the standard normal c.d.f.