Answer
a. $P(0\leq X \leq 1)=\frac{1}{21}$
b. $P(1 \leq X \leq 5)=\frac{16}{105}$
c. $P(X \geq 5)=\frac{4}{5}$
Work Step by Step
We are given $f(x)=\frac{20}{(x+20)^{2}}; [0;\infty)$
a. $P(0\leq X \leq 1)$
$ =\int^{1}_{0}\frac{20}{(x+20)^{2}}dx$
$=20(\frac{-1}{x+20})|^{1}_{0}$
$=\frac{1}{21}$
b. $P(1 \leq X \leq 5)$
$ =\int^{5}_{1}\frac{20}{(x+20)^{2}}dx$
$=20(\frac{-1}{x+20})|^{5}_{1}$
$=\frac{16}{105}$
c. $P(X \geq 5)$
$ =\int^{\infty}_{5}\frac{20}{(x+20)^{2}}dx$
$=20(\frac{-1}{x+20})|^{\infty}_{5}$
$=\frac{4}{5}$
