Solution
= Solution
{c}
Since $Y=Z^2/2$ is nonnegative, its <cumulative distribution function> is zero for $y<0$. For $y\geq0$,
$$
\begin{aligned}
F_Y(y)&=\mathbb P(Z^2\leq2y)=\mathbb P(-\sqrt{2y}\leq Z\leq\sqrt{2y})\\
&=\Phi(\sqrt{2y})-\Phi(-\sqrt{2y})=2\Phi(\sqrt{2y})-1.
\end{aligned}
$$
The last equality uses the symmetry of the <standard normal distribution>. Thus
$$
\boxed{F_Y(y)=\begin{cases}2\Phi(\sqrt{2y})-1,&y\geq0,\\0,&y<0.\end{cases}}
$$