Answer
$\mu=\frac{10}{9}$
$Var(X)=\frac{110}{567}$
$\sigma = 0.44$
Work Step by Step
We are given $f(x)=\frac{20x^{4}}{9}$ if $0 \leq x \leq 1$
$f(x)=\frac{20}{9x^{5}}$ if $x\gt1$
$\mu=\int^{a}_{b}xf(x)dx$
$=\int^{1}_{0}(\frac{20}{9}x^{5})dx+\int^{\infty}_{1}\frac{20}{9x^{4}}$
$=\frac{20}{9}(\frac{x^{6}}{6})|^{1}_{0}+\frac{20}{9}(\frac{1}{-3}x^{-3})|^{\infty}_{1}$
$=\frac{10}{27}+\frac{20}{27}$
$=\frac{10}{9}$
$Var(X)=\int^{1}_{0}x^{2}(\frac{20}{9}x^{4})dx+\int^{\infty}_{1}(\frac{20}{9x^{3}})dx - (\frac{10}{9})^{2}$
$=\frac{20}{9}\int^{1}_{0}(x^{6})dx + \frac{20}{9}\int^{\infty}_{1}(\frac{1}{x^{3}})dx - \frac{100}{81}$
$=\frac{20}{9}(\frac{x^{7}}{7}|^{1}_{0} + \frac{1}{-2}x^{-2}|^{\infty}_{1}-\frac{5}{9})$
$=\frac{20}{9}(\frac{1}{7}+\frac{1}{2}-\frac{5}{9})$
$=\frac{110}{567}$
$\sigma = \sqrt Var(X)=\sqrt \frac{110}{567} = \frac{\sqrt 770}{63}\approx 0.44$
