Find $P(796.2 \leq \sum_{i=1}^{16}(X_i-50)^2 \leq 2630)$ with a sample of $n=16$ and $X \sim N(50,100)$

Question

If $X_1,X_2, ..., X_{16}$ is a random sample of size $n=16$ from the Normal Distribution $N(50,100)$, determine:

$$P(796.2 \leq \sum_{i=1}^{16}(X_i-50)^2 \leq 2630)$$

Okay well I know that $\overline X \sim N(50, \frac{100}{16})$ and I think that the summation in this is the sample variance so $\frac{\sigma^2}{16}$?

Is that correct? I know how to work with the Normal Distribution and Z tables but I'm confused with this setup because we do not have an X value any more?

Answer 1

Start by transforming $X_i-50$ to $N(0,1)$. Then what distribution is the sum of the squares of standard normal distributions?