The Shaw Show: A Solution to Riddle 14

Tuesday, March 15, 2016

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 $P(-1<Y<1)$ , 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 $\sigma = .6005612$ .