Answer
a) $\mu=4$;
b) $V=0.5$;
c) $\sigma=0.7071$;
d) $m \approx 4.1213$;
e) $F(x)=\frac{1}{9}x^2-\frac{4}{9}x+\frac{4}{9}$
Work Step by Step
We are given $f(x)=\frac{2}{9}(x-2), [2,5]$
a) The expected value is
$\mu=\frac{2}{9}\int^5_2x(x-2)dx$ $=\frac{2}{9}\int^5_2(x^2-2x)dx$
$=\frac{2}{9}(\frac{x^3}{3}-x^2)|^5_2$
$=4$
b) The variance is $V=\frac{2}{9}\int^5_2x^2(x-2)dx-\mu^2$ $=\frac{2}{9}\int^5_2 (x^3-2x^2)dx-(4)^2$
$=\frac{2}{9}(\frac{x^4}{4}-\frac{2x^3}{3})|^5_2 - 16$
$=0.5$
c) The standar deviation is
$\sigma=\sqrt V \approx 0.7071$
d) The median is
$\int^m_2 f(x)dx=\frac{1}{2}$
$\frac{2}{9}(\frac{x^2}{2}-2x)|^m_2=\frac{1}{2}$
$\frac{m^2}{2}-2m+2=\frac{9}{4}$
$m^2-4m-\frac{1}{2}=0$
$m=\frac{4+3\sqrt 2}{2}$ and $m=\frac{4-3\sqrt 2}{2}$
$\rightarrow m =\frac{4+3\sqrt 2}{2} \approx4.1213$
e) The cumulative distribution is
$F(x)=P(X \leq x)= \int^x_2 \frac{2}{9}(t-2)dx$
$= \frac{2}{9}( \frac{t^2}{2}-2t)|^x_2$
$= \frac{2}{9}( \frac{x^2}{2}-2x+2)$
$=\frac{1}{9}x^2-\frac{4}{9}x+\frac{4}{9}$
