Suppose there are two boxes, $A$ and $B$. Box $A$ contains three tickets: $1$, $3$, and $5$. Box $B$ has two tickets: $2$ and $4$. One number is drawn from box $A$ and one number is drawn from box $B$. Let $X$ be the number from $A$, and $Y$ be the number from $B$.
What is expected value and standard error of $X+Y$.
Here is my thought: the box of possibility is $\{3,5,5,7,7,9\}$
For expected value, we have (average of box)$\times$(number of draws)$=6\times 1=6$. For standard error, we have SE=$\sqrt{number \ of\ draws}\times$ SD=$1\times\sqrt{\frac{22}{6}}$
But I think this method is pretty slow and I am not sure the validity of the result. Is there a better way to solve this?
$X$ and $Y$ are independent. Therefore $Var(X+Y)=Var(X)+Var(Y)$
And $Var(X)=\sum_{i=1}^3 p_i (x_i-\mu_x)^2 =\frac13 (1-2)^2+\frac13 (2-2)^2+\frac13 (3-2)^2$
Similar for $Var(Y)$
The standard errors are
$\sigma_{\overline X}=\frac{\sigma_X}{\sqrt n_x}=\sigma_X$
$\sigma_{\overline Y}=\frac{\sigma_Y}{\sqrt n_y}=\sigma_Y$
$n_x$ and $n_y$ are the number of draws. In both cases it is 1.
$Var(X)=\sigma_X^2, \ Var(Y)=\sigma_Y^2$
Thus the standard error of $X+Y$ is $\sigma_{\overline X +\overline Y}=\sqrt{Var(X)+Var(Y)}$