Is finding/using the MGF an appropriate strategy for this problem?

Question

Problem statement: Suppose we have a random sample $\{X_i:1\le i\le10\}$ from a normal distribution with mean $\mu=20$ and variance $\sigma^2=100$. Find the probability, $$P\left(\sum_{i=1}^6X_i>X_7+2X_8+2X_9+X_{10}+16\right)$$ My initial thoughts: For any $X_i$, we know the MGF is $M_{X_i}(t)=20t+50t^2$. I defined a new random variable $Y$ denoted by the linear combination in which every term in the inequality is on one side, i.e. $$Y=X_1+\cdots+X_6-X_7\cdots-X_{10}-16$$ so that now the problem is finding $P(Y>0)$. I find that the MGF of $Y$ is $$M_Y(t)=e^{-16t}\left(20t+50t^2\right)^6\left(-20t+50t^2\right)^2\left(-40t+200t^2\right)^2$$ Is this making the problem more difficult than it needs to be? I don't think this looks like the MGF of a distribution I might be familiar with.

Answer 1

Well, one can use the mgf to show that if $X_i$ are independent normal, then $\sum a_iX_i$ is normal. But that is a basic result, so I would just use it. Let $$W=\sum_1^6 X_i-X_7-2X_8-2X_9-X_{10}-16.$$ Then $W$ has normal distribution. Compute the mean and variance of $W$. You should get mean $-16$ and variance $1600$, and now it's almost over.