## Calculus with Applications (10th Edition)

$\mu=\frac{4}{3}$ $Var(X)=\frac{2}{9}$ $\sigma=\frac{\sqrt 2}{3}$
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 $\sigma=\sqrt Var(X)$ $=\sqrt \frac{2}{9}=\frac{\sqrt 2}{3}$