Adding and Subtracting Normal Distributions

Question

I think I know how to do this, but I'm not sure. I'm just hoping to check myself here before I do a bunch of work incorrectly.

Suppose you have three independent normal distributions:

Distribution $A$: Mean = $30.48$, SD = $8.15$
Distribution $B$: Mean = $16.58$, SD = $7.99$
Distribution $C$: Mean = $10.51$, SD = $4.43$

What is the probability that $B$ and $C$ will combine for a greater result than $A$?

I believe the first step is to combine distributions $B$ and $C$ into one distribution through addition. So...

Distribution $D$: Mean = $27.09$, SD = $9.135918$

Next I think I have to combine $D$ and $A$ through subtraction to get the final distribution.

Distribution $E$: Mean = $-3.39$, SD = $12.24286$

Given these values, the $z$-score using the standard normal distribution is $-.28$, so the probability of distributions $B$ and $C$ combining for a value greater than distribution $A$ is $.3897$ I think.

Any help is appreciated.

Answer 1

More generally, if $X_1, \ldots, X_n$ are independent normal random variables with means $\mu_i$ and variances $\sigma^2_i$, and $c_1, \ldots, c_n$ are constants, $Y = c_1 X_1 + \ldots + c_n X_n$ is normal with mean $\mu_Y = c_1 \mu_1 + \ldots + c_n \mu_n$ and variance $c_1^2 \sigma_1^2 + \ldots + c_n^2 \sigma_n^2$. Here you're doing the case $n=3$ with $c_1 = -1, c_2 = 1, c_3 = 1$.

Answer 2

You did this right. The short way to look at it is that $B+C-A$ is normally distributed with mean being $\mu = \mu_B+\mu_C-\mu_A$ and $\sigma^2 = \sigma^2_B + \sigma^2_C + \sigma^2_A$. The key point you need to know is that a variate made of the sum of two independent normal variates is itself normally distributed, even if the means of those two variates are not the same.

If you were given the analagous problem with, say, Poisson deviates instead, the answer would be different, and the technique for getting the answer would be more involved.