Answer
$\mu=5$
$Var(X)=\frac{4}{3}$
$\sigma=\frac{2\sqrt 3}{3}$
Work Step by Step
We are given $f(x)=\frac{1}{4}; [3;7]$
The expected value
$\mu=\int^{7}_{3}xf(x)dx$
$=\int^{7}_{3}x(\frac{1}{4})dx$
$=\frac{1}{4}(\frac{x^{2}}{2})|^{7}_{3}$
$=\frac{1}{4}(\frac{49}{2}-\frac{9}{2})=5$
The variance is
$Var(X)=\int^{7}_{3}(x-5)^{2}f(x)dx$
$=\int^{7}_{3}(x^{2}-10x+25)\frac{1}{4}dx$
$=\frac{1}{4}\int^{7}_{3}(x^{2}-10x+25)dx$
$=\frac{1}{4}(\frac{x^{3}}{3}-5x^{2}+25x)|^{7}_{3}$
$=\frac{4}{3}$
The standard deviation of X is
$\sigma=\sqrt Var(X)$
$=\sqrt \frac{4}{3}=\frac{2\sqrt 3}{3}$
