Normal distribution sum probability

Question

Let $X_{1},X_{2},.......,X_{7}$ are independent with $X_{i} \in N(0,1)$.How can I find $P(X_{1}^2+X_{2}^2+......+X_{7}^2 <9.04)$

What I have tried to do: $X_{1}^2+X_{2}^2+......+X_{7}^2 \in N(0,7)$, so we need to find $Φ(\frac{9.04-0}{\sqrt{7}})$

Answer 1

It's not quite true that $X_1^2 + X_2^2 + \dots + X_7^2 \sim N(0, 7)$; this would be true if you weren't squaring each of the variables, but you are. (Note in particular that that expression can't be negative, but any true normally-distributed variable must have that ability.)

Instead, that expression has another distribution entirely. Do you happen to know what that is already?

Answer 2

The sum of squares of normal random variables follow a chi-square distribution. You need to use the density of that distribution to find your answer.