Compute the mean and std of a random variable from its correlation with known random variables? Context
I'm running statistical simulations on IQ distribution among people and I would like to sample the IQ of a child given :


*

*The general distribution of IQ following : 
$$ \mathcal{N}(100, 15.5) \text{ for male}$$
$$ \mathcal{N}(100, 14.5) \text{ for female}$$ 

*The correlation between the average of parents' IQ and their child IQ (0.5).

*Both parents' IQs

*The sex of the child


Question
Is it possible to do that and if it is, how ?
 A: Let $X_1$ and $X_2$ be random variables corresponding to the IQ of the mother and father. The average of their IQs also has a normal distribution with
$$
Y_1 \equiv \frac{X_1 + X_2}{2} \sim \mathcal{N}(100, \frac{1}{4}(15.5 + 14.5 + 2\mathrm{Cov}(X_1, X_2)).
$$
It's probably the case that $\mathrm{Cov}(X_1, X_2) \neq 0$ for parents of the child - you could include this in your simulations if you wished. Anyway, continue by letting $\sigma^2 = \frac{1}{4}(15.5 + 14.5 + 2\mathrm{Cov}(X_1, X_2))$.
Now if $Y_2$ is another random variable for the IQ of the child (say it's a girl) then the joint distribution of $Y_1$ and $Y_2$ is multivariate normal with
$$
\begin{pmatrix}
    Y_1 \\
    Y_2 \\
\end{pmatrix} \sim \mathcal{N}\left(
\begin{pmatrix}
    100 \\
    100 \\
\end{pmatrix},
\begin{pmatrix}
    \sigma^2 & \mathrm{Cov}(Y_1, Y_2) \\
    \mathrm{Cov}(Y_1, Y_2) & 14.5\\
\end{pmatrix} \right).
$$
The off-diagonal covariance terms can be found by the definition of correlation
$$
\frac{\mathrm{Cov}(Y_1, Y_2)}{\sqrt{14.5}\sqrt{\sigma^2}} = 0.5.
$$
Now you can use the formula for the conditional distribution of the multivariate normal to find the distribution of $Y_2 | Y_1 = y_1$ where $y_1$ is the realisation of the parent's average IQ, which you have. This should give you a univariate distribution which can be simulated from easily (say using rnorm if you are using R.)
