(a)
\[
P(X\leqslant 1)=F(1)=\frac{1}{4}=0.25
\]
(b)
\[
P(.5\leqslant X\leqslant 1.5)=F(1.5)-F(.5)=\frac{1.5^2}{4}-\frac{0.5^2}{4}=\frac{1}{4}\biggl[(\frac{3}{2})^2-(\frac{1}{2})^2\biggr]=\frac{1}{2}=0.5
\]
\[
P(.5\leqslant X\leqslant 1)=F(1)-F(.5)=\frac{1^2}{4}-\frac{0.5^2}{4}=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}=0.1875
\]
(c)
\[
P(X > .5)=1-P(X\leqslant .5)=1-F(0.5)=1-\frac{0.5^2}{4}=1-\frac{1}{16}=\frac{15}{16}=0.9375
\]
(d)
Let $p=0.5$, the median checkout duration $\tilde{\mu}=\eta(p)$ satisfies
\[
p=F(\eta{p})=F(\tilde{\mu})=\int_{-\infty}^{\tilde{\mu}}f(x)dx
\]
Now, the expression of cdf $F(x)$ has been given. For $p=0.5$, $F(\tilde{\mu})=\frac{\tilde{\mu}^2}{4}$. Thus,
\[
\frac{\tilde{\mu}^2}{4}=0.5\Rightarrow\tilde{\mu}^2=2\Rightarrow\tilde{\mu}=\sqrt{2}\approx 1.41421356
\]
(e)
For $x < 0$, $f(x)=F'(x)=0$, since function $F(x)$ is constant while $x < 0$.
For $x=0$,
\[
F'_{+}(0)=\lim_{x\rightarrow 0^+}\frac{F(x)-F(0)}{x-0}=\lim_{x\rightarrow 0^+}\frac{\frac{x^2}{4}-\frac{0^2}{4}}{x}=\lim_{x\rightarrow 0^+}\frac{x}{4}=0,
\]
\[
F'_{-}(0)=\lim_{x\rightarrow 0^-}\frac{F(x)-F(0)}{x-0}=\lim_{x\rightarrow 0^-}\frac{0-\frac{0^2}{4}}{x}=0.
\]
Hence, the derivative of the function $F(x)$ at the point $x=0$ exists and it is equal to $0$. So $f(0)=F'(0)=0$.
For $x\in(0,2)$,
\[
f(x)=F'(x)=\biggl(\frac{x^2}{4}\biggr)'=\frac{x}{2}.
\]
For $x=2$,
\[
\begin{split}
F'_{-}(2)&=\lim_{x\rightarrow 2^-}\frac{F(x)-F(2)}{x-2}=\lim_{x\rightarrow 2^-}\frac{\frac{x^2}{4}-1}{x-2}=\lim_{x\rightarrow 2^-}\frac{\frac{1}{4}(x^2-4)}{x-2}\\
&=\lim_{x\rightarrow 2^-}\frac{x+2}{4}=1,
\end{split}
\]
\[
F'_{+}(2)=\lim_{x\rightarrow 2^+}\frac{F(x)-F(2)}{x-2}=\lim_{x\rightarrow 2^+}\frac{1-1}{x-2}=0.
\]
Hence, the derivative of the function $F(x)$ at the point $x=2$ doesn't exist. So $f(2)$ is not defined.
For $x > 2$, $f(x)=F'(x)=0$, since $F(x)$ is constant for $x\in(2,+\infty)$.
