# How can I find $\rho$, given $P(4 < Y < 16|X=5)=0.9544$?

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.

• How would you integrate a standard normal density between $a$ and $b$? Commented Nov 5, 2019 at 4:52
• I see that the most common way is to use the Z table, but I was wondering if there was a more rigorous way to compute this. This problem is presented in my math stats class before we reach Z tables. Commented Nov 5, 2019 at 4:53
• You can't integrate it in "closed form" ... Doing this problem will rely on facts about the normal; in this case knowing how much of the probability is within 2 sd's of the mean. I bet you covered that much... Commented Nov 5, 2019 at 4:57
• We did. Thank you for confirming my suspicions about the integration. I appreciate the guided thought! Commented Nov 5, 2019 at 5:00
• @Glen_b-ReinstateMonica why 2 sd's of the mean? It doesn't explicitly mentions that Commented Feb 13, 2020 at 13:59

Its integral cannot be expressed by elementary functions by definition. You'll use z-table. Let the RV represented by the marginal density of $$Y$$ given $$X=5$$ be denoted as $$W$$. We ask for \begin{align}P(4 Then, $$P(Z\leq a)=0.9772\rightarrow a=2\rightarrow\rho=0.8$$