A Solution to Riddle 14

This is my solution to the riddle “Should You Shoot Free Throws Underhand?”  I apologize for the ugly mathematical mark-up.  
We see C=\{x, y \in \mathbb{R}: x^2+y^2\leq 1\}, and V=(X,Y) where X and Y are normal with \mu=0 and \sigma unknown.
The question asks for <img alt="P(-1<Y, but without knowledge of \sigma, we cannot compute the probability. So, we need to find \sigma.
We are given that P(V\in C)=0.75, which translates to P(X^2+Y^2\leq 1)=0.75. If we divide both sides of the inequality inside the probability by \sigma^2, then we have the sum of two standard normal random variables squared, which is a chi-square random variable with 2 degrees of freedom (you can find a reference for that here).
Thus, with \chi^2_0=\frac{X^2}{\sigma^2}+\frac{Y^2}{\sigma^2}, we have P(\chi^2_0\leq 1/\sigma^2)=0.75. Using some tables or technology, we can find the 75th percentile of a chi-square distribution with 2 df to be 2.772589, which when set equal to 1/\sigma^2, will reveal that \sigma = .6005612.
Now, <img alt="P(|Y|<1) = P(-1<Y<1) = P(-1.6651< Z .

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