Answer
$\mu=\frac{141}{22} \approx 6.409$
$Var(X)=\approx 2.093$
$\sigma \approx 1.45$
Work Step by Step
We are given $f(x)=\frac{1}{11}(1+\frac{3}{\sqrt x})=\frac{1}{11}+\frac{3}{11\sqrt x}; [4;9]$
The expected value
$\mu=\int^{9}_{4}xf(x)dx$
$=\int^{9}_{4}x\frac{1}{11}(1+\frac{3}{\sqrt x})dx$
$=\frac{1}{11}\int^{9}_{4}(x+3\sqrt x)dx$
$=\frac{1}{11}(\frac{x^{2}}{2}+2x^{\frac{3}{2}})|^{9}_{4}$
$=\frac{141}{22}\approx 6.409$
The variance is
$Var(X)=\int^{9}_{4}x^{2}f(x)dx-E(x)^{2}$
$=\frac{1}{11}\int^{9}_{4}(x^{2}+3x^{\frac{3}{2}})dx-(\frac{141}{22})^{2}$
$=\frac{1}{11}(\frac{x^{3}}{3}+\frac{6}{5}x^{\frac{5}{2}})|^{9}_{4}-(\frac{141}{22})^{2}$
$=\frac{7123}{165}-(\frac{141}{22})^{2}\approx2.093$
The standard deviation of X is
$\sigma=\sqrt Var(X)$
$=\sqrt 2.093 \approx 1.45$
