Standard deviation of the sum of two normally distributed random variables $X\sim N(52,6)$, $Y\sim (40,8)$. What's the standard deviation of $Z=X+Y$?
I'm considering to transform the linear relationship to matrix form 
$$Z=\begin{pmatrix}
1& 1\\
\end{pmatrix}\begin{pmatrix}
X\\
Y
\end{pmatrix}$$
and apply $$\Sigma_{z}=A \Sigma^{-1} A'$$
However, I don't know the next step. How to calculate $\Sigma$?
Thank you! 
 A: It's interesting that you are trying to use the matrix form when this is more of a elementary problem.
Remember
$$
\mbox{Var}(X + Y) = \mbox{Var}(X) + \mbox{Var}(Y) + 2  \mbox{Cov}(X,Y).
$$
You don't say anything about the relation between $X$ and $Y$, so either you assume a covariance of zero or the problem is not solvable due to missing informations.
The matrix form of this is
$$
\mbox{Cov}(AX) = A \cdot \mbox{Cov}(X)\cdot A'.
$$
This covariance matrix seems to be your $\Sigma$. I don't know why you inverted it in your formula. Perhaps I misunderstand your equation? You know that $\Sigma_{1,1} = 6$ and $\Sigma_{2,2} = 8$. These are just the variances of $X$ and $Y$ respectively. Again, since you do not have a covariance and since you cannot calculate it from the information given, the question is either unsolvable or you have have to assume independence, which implies $\Sigma_{1,2}=\Sigma_{2,1}=0$. This will then give the same result as the equation for $\mbox{Var}(X+Y)$. Remember to take the root to get the standard deviations.
A: As distribution of $\begin{pmatrix}
X\\
Y
\end{pmatrix}$
is not specified, I would probably assume that $\mbox{cov}(X,Y)=0$ then 
$$\Sigma =\begin{pmatrix} 6 & 0\\ 0 & 8 \end{pmatrix}.$$ 
Where diagonal elements of $\Sigma$ are variances of $X$ and $Y$. If the $\mbox{cov}(X,Y)$ is available you should insert that number instead of $0$'s.
A: If $X$ and $Y$ are independent random variables that are normally distributed, then their sum is also normally distributed. i.e., if
$X\sim N(\mu_X, \sigma_X^2)$
$Y \sim N(\mu_Y, \sigma_Y^2)$
$Z=X+Y$
and $X$ and $Y$ are independent, then
$Z \sim N(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2).$ 
You can prove it using both characteric functions and convolutions. I will refer you to read this page for the proofs.
In the event that the variables $X$ and $Y$ are jointly normally distributed correlated random variables, then $Z=X + Y$ is still normally distributed 
and the mean is the sum of the means. However, the variances are not additive 
due to the correlation. Indeed,
$\sigma_{X+Y} = \sqrt{\sigma_X^2+\sigma_Y^2+2\rho\sigma_X \sigma_Y}$
This one is also shown in the previous link. 
You can also find this using the fact that $Q^{'}=[X \; Y]$ is bivariate normal with variables $X$ and $Y$ and distributed as $N_{2}(\mu,\Sigma).$ Then $Z=AQ$ with $A=[1\;1]$ is distributed as $N_{q}(A\mu,A \Sigma A^{'})$, where $q$ is the number of linear combinations (or the number of rows in $A$). Each row in $A$ corresponds to a linear combination of the variables $X$ and $Y$. Here we have only 1 row. So, $q=1$. Put $\sigma_X^2$, $\sigma_Y^2$ and $\sigma_{XY}$ in $\Sigma_{(2\times2)}$ and find the result for the correlated case.
