Determining variance from sum of two random correlated variables I understand that the variance of the sum of two independent normally distributed random variables is the sum of the variances, but how does this change when the two random variables are correlated?  

Let's work this out from the definitions. Let's say we have 2 random variables $x$ and $y$ with means $\mu_x$ and $\mu_y$. Then variances of $x$ and $y$ would be:
$${\sigma_x}^2 = \frac{\sum_i(\mu_x-x_i)(\mu_x-x_i)}{N}$$
$${\sigma_y}^2 = \frac{\sum_i(\mu_y-y_i)(\mu_y-y_i)}{N}$$
Covariance of $x$ and $y$ is:
$${\sigma_{xy}} = \frac{\sum_i(\mu_x-x_i)(\mu_y-y_i)}{N}$$
Now, let us consider the weighted sum $p$ of $x$ and $y$:
$$\mu_p = w_x\mu_x + w_y\mu_y$$
$${\sigma_p}^2 = \frac{\sum_i(\mu_p-p_i)^2}{N} = \frac{\sum_i(w_x\mu_x + w_y\mu_y - w_xx_i - w_yy_i)^2}{N} = \frac{\sum_i(w_x(\mu_x - x_i) + w_y(\mu_y - y_i))^2}{N} = \frac{\sum_i(w^2_x(\mu_x - x_i)^2 + w^2_y(\mu_y - y_i)^2 + 2w_xw_y(\mu_x - x_i)(\mu_y - y_i))}{N} \\ = w^2_x\frac{\sum_i(\mu_x-x_i)^2}{N} + w^2_y\frac{\sum_i(\mu_y-y_i)^2}{N} + 2w_xw_y\frac{\sum_i(\mu_x-x_i)(\mu_y-y_i)}{N} \\ = w^2_x\sigma^2_x + w^2_y\sigma^2_y + 2w_xw_y\sigma_{xy}$$
