Let $X_1, X_2, X_3, X_4$ be independent standard normal random variables and $$Y = X_1^2 + X_2^2 + X_3^2 + X_4^2$$ Find the probability that $Y \leq 3$.

For this problem I know that the distribution is $N(0,1)$ for each $X$. I tried combining them to equal $Y$ by finding the distribution of $Y = N(0^2 + 0^2 + 0^2 + 0^2, 1^2 + 1^2 + 1^2 + 1^2) = N(0,4)$. I then tried calculating $P(Z> (0-\mu)/\sigma)$ where $\mu$ is the mean and $\sigma$ is the variance, but obtained $0$ which I know is incorrect. I was then unsure of where to go with this or if my initial strategy was even correct.


The distribution of $Y$ cannot be $N(0,4)$ for the simple reason that $Y$ is never negative, while $N(0,4)$ is negative half the time.

The sum of squares of independent random variables is described by the $\chi^2$ distribution.


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