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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 $Y\leq 3$.

I thought that you would be using some kind of expectation, but I'm unsure how to proceed with that given that there are four random variables. Any direction would be greatly appreciated.

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I think this question was asked the other day, so this might be a duplicate. In any case, you must observe that $$ Y \sim \chi^2_4 $$ see here.

UPDATE

As requested in the comments, here is some more detail: using the formulae found in the link above, you see the density is $$ f_Y(y) = \frac{1}{4} y e^{-y/2}, \quad y \geq 0 $$ and the distribution is $$ F_Y(y) = \frac{\gamma(4/2,y/2)}{\Gamma(4/2)} = \int_0^{y/2} t e^{-t}dt = \left( -te^{-t} - e^{-t} \right)^{y/2}_0 = 1-\left(\frac{y+2}{2}\right)e^{-y/2}. $$ So $$ P[Y \leq y] = 1-\left(\frac{3+2}{2}\right)e^{-3/2} \approx 0.44217. $$

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  • $\begingroup$ I am still confused $\endgroup$ – user143875 Apr 17 '14 at 20:17
  • $\begingroup$ Okay, I've added more detail $\endgroup$ – user139388 Apr 17 '14 at 20:34
  • $\begingroup$ A lot of books have $\chi^2$ tables in the back. $\endgroup$ – Batman Apr 17 '14 at 20:43

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