Two random variables $X$ and $Y$ are i.i.d. normal$(\mu, \sigma^2)$. If $P(X > 3) = 0.8413$ find $P((X+Y)/2 > 3)$.
The result must be exact number, so normal distribution parameters are considered unknown. I've tried to solve this problem but I got stuck.
$$P(X > 3) = \frac{1}{2}\left(1 + \mathrm{erf}\left(\frac{\mu - 3}{\sigma \sqrt{2}}\right)\right),$$ so $\mu = 3 + \sigma \sqrt{2}\, \mathrm{erf}^{-1}(2 c - 1)$, where $c = 0.8413$.
$(X + Y)/2$ has normal distribution with $\mathsf{E} (X+Y)/2 = \mu$ and $\mathrm{Var}((X + Y)/2) = \sigma^2 / 2$. Therefore, $P((X + Y)/2 > 3) = \frac{1}{2} (1 + \mathrm{erf}(\frac{\mu - 3}{\sigma}))$. Substituting $\mu$ from above, $$ P((X + Y)/2 > 3) = \frac{1}{2}\left(1 + \mathrm{erf}\left(\frac{3 + \sigma \sqrt{2}\, \mathrm{erf}^{-1}(2 c - 1) - 3}{\sigma}\right)\right) = \frac{1}{2}\left(1 + \mathrm{erf}\left(\sqrt{2}\, \mathrm{erf}^{-1}(2 c - 1)\right)\right) \approx 0.9213. $$