Answer
$\mu=\frac{4}{3}$
$Var(X)=\frac{2}{9}$
$\sigma=\frac{\sqrt 2}{3}$
Work Step by Step
We are given $f(x)=4x^{-5}=\frac{4}{x^{5}}; [1;\infty]$
The expected value
$\mu=\int^{\infty}_{1}xf(x)dx$
$=\int^{\infty}_{1}x(\frac{4}{x^{5}})dx$
$=\int^{\infty}_{1}\frac{4}{x^{4}}dx$
$=\lim\limits_{b \to \infty}\int^{b}_{1}\frac{4}{x^{4}} dx$
$=\lim\limits_{b \to \infty}(\frac{4}{-3x^{3}})|^{b}_{1}$
$=\frac{4}{3}$
The variance is
$Var(X)=\int^{\infty}_{1}x^{2}(\frac{4}{x^{5}})dx-(\frac{4}{3})^{2}$
$=\int^{\infty}_{1}\frac{4}{x^{3}}dx-\frac{16}{9}$
$=\lim\limits_{b \to \infty}\int^{b}_{1}\frac{4}{x^{3}}dx-\frac{16}{9}$
$=\lim\limits_{b \to \infty}(\frac{4}{-2x^{2}})|^{b}_{1}-\frac{16}{9}$
$=2-\frac{16}{9}=\frac{2}{9}$
The standard deviation of X is
$\sigma=\sqrt Var(X)$
$=\sqrt \frac{2}{9}=\frac{\sqrt 2}{3}$
