$\newcommand{\Cov}{\operatorname{Cov}}$$\newcommand{\Var}{\operatorname{Var}}$$\newcommand{\E}{\mathbb{E}}$$\newcommand{\P}{\mathbb{P}}$We have that $X$ and $Y$ are random variables with a multivariate normal density with $\E[X]=2$, $\E[Y]=-3$, $\Var[X]=4$, $\Var[Y]=25$ and $\Cov[X,Y]=-3$. And they ask me for $\P[X≤3|Y=1]$.
So what I did first was to get the conditional expectation value for X and the conditional variance.
This is what I got: $$\E[X|Y=1]=2+(-3/10)(2/5)(1-(-3))=1.52$$
$$\Var[X|Y=1]=4(1-(-3/10))=3.64$$
Then I got that $Z=(3-1.52)/3.64=0.40659$, and that $\P[X≤3|Y=1]=0.6591$.
But the correct answer is $.76$. What was my mistake?
Here are the formulas that I used to get the variance and the expectation value.
\begin{align} \E[X_2\mid X_1=x_1] &= \mu_2 + \rho\frac{\sigma_2}{\sigma_1}(x_1-\mu_1)\\ \Var[X_2\mid X_1=x_1] &= \sigma_2^2(1-\rho^2). \end{align}