Answer
a. $m=\sqrt[3] 2 \approx 1.26$
b. $\approx 0.2$
Work Step by Step
We are given $f(x)=3x^{-4}; [1;\infty)$
The expected value
$\mu= \frac{3}{2}$
a. The median is $\int^{m}_{1}f(x)dx=\frac{1}{2}$
$\int^{m}_{1}3x^{-4}dx=\frac{1}{2}$
$-x^{-3}|^{m}_{1}=\frac{1}{2}$
$1-m^{-3}=\frac{1}{2}$
$m=\sqrt[3] 2 \approx 1.26$
b. The probability between mean and median is
$P(\sqrt[3] 2 \leq X \leq \frac{3}{2})=\int ^{\frac{3}{2}}_{\sqrt[3] 2} 3x^{-4}dx=-x^{-3}|^{\frac{3}{2}}_{\sqrt[3] 2} \approx 0.2$
