Find $c$ such that $P(Z^2 > c) = 0.95$

Question

I was wondering if any of you could help me with this statistics problem.

Answer 1

If $Z\sim N(0,\,1)$, you can proceed in one of two ways:

1. Since $Z^2\sim\chi_1^2$, a $\chi_\nu^2$ table will tell you that for $\nu=1$ the value with CDF $1-0.95=0.05$ is $0.00393$.
2. Let $\Phi$ denote the CDF of $Z\sim N(0,\,1)$. For $c\ge0$,$$P(Z^2\le c)=P(-\sqrt{c}\le Z\le\sqrt{c})=\Phi(\sqrt{c})-\Phi(-\sqrt{c})=2\Phi(\sqrt{c})-1,$$i.e. $P(Z^2>c)=1-P(Z^2\le c)=2(1-\Phi(\sqrt{c}))$. Let $\Phi^{-1}$ denote the inverse of $\Phi$, so$$\Phi(\sqrt{c})=1-\frac{0.95}{2}=0.525\implies c=(\Phi^{-1}(0.525))^2=0.00393.$$Again, a table - this time, for $N(0,\,1)$ - will tell you what you need, i.e. $\Phi^{-1}(0.525)=0.0627$.