Answer
a. $m=\frac{2-\sqrt 2}{2}$
b. $=\sqrt 2 +\frac{3-2\sqrt 2}{2} -\frac{13}{9}$
Work Step by Step
We are given $f(x)=2(1-x); [0;1]$
The expected value $\mu= \frac{1}{3}$
a. The median is
$\int^{m}_{0}f(x)dx=\frac{1}{2}$
$\int^{m}_{0}2(1-x)dx=\frac{1}{2}$
$2x-x^{2}|^{m}_{0}=\frac{1}{2}$
$2m-m^{2}=\frac{1}{2}$
$m=\frac{2-\sqrt 2}{2}$
b. The probability between mean and median is
$P(\frac{2-\sqrt 2}{2} \leq X \leq \frac{1}{3})=\int ^{\frac{1}{3}}_{\frac{2-\sqrt 2}{2}} 2(1-x)dx=2x-x^{2}|^{\frac{1}{3}}_{\frac{2-\sqrt 2}{2}}=\frac{5}{9}-2+\sqrt 2+\frac{3-2\sqrt 2}{2}=\sqrt 2 +\frac{3-2\sqrt 2}{2} -\frac{13}{9}$
