Question

Let X be a random variable with mean X = 25 and X = 6 and let Y be a random variable with mean Y = 30 and Y = 4. It is known that X and Y are independent random variables. Suppose the random variables X and Y are added together to create new random variable W (i.e., W = X + Y). What is the standard deviation of W?

Answer 1

I'm guessing you intended to say $X$ has mean $\mu_X=E[X]=25$ and standard deviation $\sigma_x=\sqrt{\mathrm{Var}[X]}=6$, and $Y$ has means $\mu_Y=E[Y]=30$ and standard deviation $\sigma_Y=\sqrt{\mathrm{Var}[Y]}=4$.

If $W=X+Y$, then $W$ has mean

$E[W]=E[X+Y]=E[X]+E[Y]=55$

and variance

$\mathrm{Var}[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2$

Given that $\mathrm{Var}[X]=36$ and $\mathrm{Var}[Y]=16$, we have

$\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]=36+25^2=661$

$\mathrm{Var}[Y]=E[Y^2]-E[Y]^2\implies E[Y^2]=16+30^2=916$

Then

$E[W^2]=E[(X+Y)^2]=E[X^2]+2E[XY]+E[Y^2]$

$X$ and $Y$ are independent, so $E[XY]=E[X]E[Y]$, and

$E[W^2]=E[X^2]+2E[X]E[Y]+E[Y^2]=661+2\cdot25\cdot30+916=3077$

so that the variance, and hence standard deviation, are

$\mathrm{Var}[W]=3077-55^2=52$

$\implies\sqrt{\mathrm{Var}[W]}=\sqrt{52}=\boxed{2\sqrt{13}}$

# # #

Alternatively, if you've already learned about the variance of linear combinations of random variables, that is

$\mathrm{Var}[aX+bY]=a^2\mathrm{Var}[X]+b^2\mathrm{Var}[Y]$

then the variance of $W$ is simply the sum of the variances of $X$ and $Y$, $\mathrm{Var}[W]=36+16=52$, and so the standard deviation is again $\sqrt{52}$.