Answer
$\mu=\frac{13}{2}$
$Var(X)=\frac{25}{12}$
$\sigma=\frac{5\sqrt 3}{6}$
$m=\frac{13}{2}$
The cumulative distribution is
$F(x)=\frac{1}{5}x-\frac{4}{5}$
Work Step by Step
We are given $f(x)=\frac{1}{5}; [4;9]$
The expected value
$\mu=\int^{9}_{4}xf(x)dx$
$=\int^{9}_{4}x(\frac{1}{5})dx$
$=\frac{1}{5}(\frac{x^{2}}{2})|^{9}_{4}$
$=\frac{1}{5}(\frac{81}{2}-8)=\frac{13}{2}$
The variance is
$Var(X)=\int^{9}_{4}(x-\frac{13}{2})^{2}f(x)dx$
$=\int^{9}_{4}(x^{2}-13x+\frac{169}{4})\frac{1}{5}dx$
$=\frac{1}{5}\int^{9}_{4}(x^{2}-13x+\frac{169}{4})dx$
$=\frac{1}{5}(\frac{x^{3}}{3}-\frac{13}{2}x^{2}+\frac{169}{4}x)|^{9}_{4}$
$=\frac{25}{12}$
The standard deviation of X is
$\sigma=\sqrt Var(X)$
$=\sqrt \frac{25}{12}=\frac{5\sqrt 3}{6}$
The median is
$\int^{m}_{4}f(x)dx=\frac{1}{2}$
$\frac{1}{5}m-\frac{4}{5}=\frac{1}{2}$
$m=\frac{13}{2}$
The cumulative distribution is
$F(x)=P(X\leq x)=\int ^{x} _{4}\frac{1}{5}dt=\frac{1}{5}t| ^{x} _{4}=\frac{1}{5}x-\frac{4}{5}$
