Answer
a. $P(0\leq X \leq 2)\approx 0.423$
b. $P(1 \leq X \leq 3) \approx 0.207$
c. $P(X\geq 5)\approx .408$
Work Step by Step
We are given $f(x)=\frac{1}{2}(1+x)^{-3/2}; [0;\infty)$
a. $P(0\leq X \leq 2)$
$=\int^{2}_{0}\frac{1}{2}(1+x)^{-3/2}dx$
$=-(1+x)^{-1/2}|^{2}_{0}$
$\approx 0.423$
b. $P(1 \leq X \leq 3)$
$=\frac{1}{2}\int^{2}_{0}(1+x)^{-3/2}dx$
$=-(1+x)^{-1/2}|^{3}_{1}$
$\approx 0.207$
c. $P(X\geq 5)$
$=\frac{1}{2}\int^{\infty}_{5}(1+x)^{-3/2}dx$
$=-(1+x)^{-1/2}|^{\infty}_{5}$
$\approx .408$
