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]}=√(52)=\boxed{2√(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
`√(52)`.