Answer
$\mu=\frac{16}{5}$
$Var(X)=-13.048$
Work Step by Step
We are given $f(x)=\frac{x^{3}}{12}$ if $0 \leq x \leq 2$
$f(x)=\frac{16}{3x^{3}}$ if $x\gt 2$
$\mu=\int^{a}_{b}xf(x)dx$
$=\int^{2}_{0}(\frac{x^{4}}{12})dx+\int^{\infty}_{2}(\frac{16}{3x^{2}})dx$
$=\frac{1}{12}(\frac{x^{5}}{5})|^{2}_{0}+\frac{16}{3}(\frac{1}{-1}x^{-1})|^{\infty}_{2}$
$=\frac{8}{15}+\frac{8}{3}$
$=\frac{16}{5}$
$Var(X)=\int^{2}_{0}x^{2}(\frac{1}{12}x^{3})dx+\int^{\infty}_{2}(\frac{16}{3x})dx - (\frac{16}{5})^{2}$
$=\frac{1}{12}\int^{2}_{0}(x^{5})dx + \frac{16}{3}\int^{\infty}_{2}(\frac{1}{x})dx - \frac{256}{25}$
$=\frac{1}{12}\frac{x^{6}}{6}|^{2}_{0} + \frac{16}{3}\ln|x||^{\infty}_{2}-\frac{256}{25}$
$=\frac{8}{9}-3.697-\frac{256}{25}$
$=-13.048$
$\sigma = \sqrt Var(X)=\sqrt -13.048 \rightarrow$ Because Var(X) is negative, we can't find the standard deviation
