A quantity x has cumulative distribution function p(x) = x − x2/4 for 0 ≤ x ≤ 2 and p(x) = 0 for x < 0 and p(x) = 1 for x > 2. find the mean and median of x.

[tex]F_X(x)=\begin{cases}0&\text{for }x<0\\\\x-\dfrac{x^2}4&\text{for }0\le x\le2\\\\1&\text{for }x>2\end{cases}[/tex]

Recall that the PDF is given by the derivative of the CDF:

[tex]f_X(x)=\dfrac{\mathrm dF_X(x)}{\mathrm dx}=\begin{cases}1-\dfrac x2\\\\0&\text{otherwise}\end{cases}[/tex]

The mean is given by

[tex]\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^2\left(x-\dfrac{x^2}2\right)\,\mathrm dx=\frac23[/tex]

The median is the number [tex]M[/tex] such that [tex]F_X(x)=\mathbb P(X\le M)=\dfrac12[/tex]. We have

[tex]F_X(M)=M-\dfrac{M^2}4=\dfrac12\implies M=2\pm\sqrt2[/tex]

but both roots can't be medians. As a matter of fact, the median must satisfy [tex]0\le M\le2[/tex], so we take the solution with the negative root. So [tex]M=2-\sqrt2[/tex] is the median.