Calculating mean and standard deviation for a set 
Suppose we have a set $A = \{a_1, a_2, a_3, a_4, a_5\}$ where $a_n \in
\mathbb{R}$ and a set $B = \{b_1, b_2, b_3, b_4, b_5\}$ where $b_n \in
\mathbb{R}$ and a set $C = \{ma_1 + nb_1, ma_2 + nb_2, ma_3 + nb_3, ma_4 + nb_4, ma_5 + nb_5\}$ where $m, n \in (0,1) \subset \mathbb{R}$.

$A$ has mean $\mu_1$ and standard deviation $\sigma_1$. $B$ has mean $\mu_2$ and standard deviation $\sigma_2$. Do we have sufficient information to calculate the standard deviation of $C$? 
Note: The mean of $C$ is $m\mu_{1} + n\mu_{2}$?
 A: These are numbers, presumably the results of an experiment. Given the numbers $m$ and $n$, and the $a_i$ and $b_i$, we can calculate the sample mean and the sample standard deviation of the $ma_i+nb_i$.
The situation is very different if we have random variables $X$ and $Y$, and the random variable $W$ is defined to be $mX+nY$. We have $E(W)=mE(X)+nE(Y)$, but the standard deviation of $W$ cannot be computed from knowledge of the mean and standard deviations of $X$ and $Y$. 
A: We have $C=mA+nB$, so by the linearity of expectation, the mean of $C$ is:
$$
E(C)=E(mA+nB)=m\cdot E(A)+n\cdot E(B)=m\mu_1 + n\mu_2
$$
As for the standard deviation of $C$, we have:
$$ \begin{align*}
\sqrt{Var(C)} &= \sqrt{Var(mA+nB)} \\
&= \sqrt{m^2\cdot Var(A)+n^2\cdot Var(B)+2mn\cdot Cov(A,B)} \\
&= \sqrt{m^2\sigma_1^2 +n^2\sigma_2^2+2mn\cdot Cov(A,B)} \\
\end{align*} $$
In the special case where $A$ and $B$ are independent, we have $Cov(A,B)=0$, so the standard deviation of $C$ reduces to:
$$
\sqrt{m^2\sigma_1^2 +n^2\sigma_2^2}
$$
