Answer
a. $m=\sqrt[4] 2 \approx1.189$
b. $\approx 0.184$
Work Step by Step
We are given $f(x)=4x^{-5}; [1;\infty)$
The expected value
$\mu= \frac{4}{3}$
a. The median is
$\int^{m}_{1}f(x)dx=\frac{1}{2}$ $\int^{m}_{1}4x^{-5}dx=\frac{1}{2}$
$-x^{-4}|^{m}_{1}=\frac{1}{2}$
$-m^{-4}+1=\frac{1}{2}$
$m^{-4}=\frac{1}{2}$
$m^{4}=2$
$m=\sqrt[4] 2 \approx1.189$
b. The probability between mean and median is
$P(\sqrt[4] 2 \leq X \leq \frac{4}{3})=\int ^{\frac{4}{3}}_{\sqrt[4] 2} 4x^{-5}dx=-x^{-4}|^{\frac{4}{3}}_{\sqrt[4] 2} \approx 0.184$
