Answer
$\mu=\frac{14}{3}$
$Var(X)=\frac{8}{9}$
$\sigma=\frac{2\sqrt 2}{3}$
Work Step by Step
We are given $f(x)=\frac{x}{8}-\frac{1}{4}; [2;6]$
The expected value
$\mu=\int^{6}_{2}xf(x)dx$
$=\int^{6}_{2}x(\frac{1}{8}x-\frac{1}{4})dx$
$=\frac{1}{8}.\frac{x^{3}}{3}-\frac{x^{2}}{8}|^{6}_{2}$
$=\frac{14}{3}$
The variance is
$Var(X)=\int^{6}_{2}(x-\frac{14}{3})^{2}f(x)dx$
$=\int^{6}_{2}(x^{2}-\frac{28}{3}x+\frac{196}{9})(\frac{1}{8}x-\frac{1}{4})dx$
$=\int^{6}_{2}(\frac{1}{8}x^{3}-\frac{1}{4}x^{2}-\frac{7}{6}x^{2}+\frac{7}{3}x+\frac{49}{18}x-\frac{49}{9})dx$
$=\int^{6}_{2}(\frac{1}{8}x^{3}-\frac{17}{12}x^{2}+\frac{91}{18}x-\frac{49}{9})dx$
$=\frac{x^{4}}{32}-\frac{17x^{3}}{36}+\frac{91}{36}x^{2}-\frac{49}{9}x|^{6}_{2}$
$=\frac{8}{9}$
The standard deviation of X is
$\sigma=\sqrt Var(X)$
$=\sqrt \frac{8}{9}=\frac{2\sqrt 2}{3}$
