Let $X$ and $Y$ have a bivariate normal distribution with $\mu_X=5, \mu_Y=10, \sigma^2_X=1, \sigma^2_Y=25, \rho >0$.

If $P(4 < Y < 16|X=5)=0.9544$, I would like to find $\rho$.

I know that conditional marginals of a bivariate normal distribution are normal distributions. Given this knowledge, I can obtain the distribution $Y|X=5 \sim N(10,25(1-\rho^2)).$ However, integrating this pdf between $4$ and $16$ seems impossible. I have the following:

$f_{Y|X=5}(y)=(5\sqrt{2\pi(1-\rho^2)})^{-1}exp\{-(x-10)^2/(50(1-\rho^2))\}$, where $y \in R$.

$.9544=\int_4^{16}(5\sqrt{2\pi(1-\rho^2)})^{-1}exp\{-(x-10)^2/(50(1-\rho^2))\}dx,$

which does not seem possible to integrate. Is there a more efficient to solving this problem? Thank you.