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