Answer
$\mu=\frac{17}{6}$
$Var(X)=\frac{103}{180}$
$\sigma=\frac{\sqrt 515}{30}$
Work Step by Step
We are given $f(x)=1-\frac{1}{\sqrt x}; [1;4]$
The expected value
$\mu=\int^{4}_{1}xf(x)dx$
$=\int^{4}_{1}x(1-\frac{1}{\sqrt x})dx$
$=\int^{4}_{1}(x-\sqrt x)dx$
$=(\frac{x^{2}}{2}-\frac{x^{\frac{3}{2}}}{\frac{3}{2}})|^{4}_{1}$
$=\frac{17}{6}$
The variance is
$Var(X)=\int^{4}_{1}(x-\frac{17}{6})^{2}f(x)dx$
$=\int^{4}_{1}(x^{2}-\frac{17}{3}x+\frac{289}{36})(1-\frac{1}{\sqrt x})dx$
$=\int^{4}_{1}(x^{2}-x\sqrt x-\frac{17}{3}x+\frac{17}{3}\sqrt x+\frac{289}{36}-\frac{289}{36\sqrt x})dx$
$=\frac{x^{3}}{3}-\frac{2}{5}x^{\frac{5}{2}}-\frac{17}{6}x^{2}+\frac{34}{9}x^{\frac{3}{2}}+\frac{289}{36}x-\frac{289}{18}x^{\frac{1}{2}}|^{4}_{1}$
$=\frac{103}{180}$
The standard deviation of X is
$\sigma=\sqrt Var(X)$
$=\sqrt \frac{103}{180}=\frac{\sqrt 515}{30}$
