Answer
a. $P(0 \leq X \leq 1)=\frac{4}{9}$
b. $P(X\gt1)=\frac{5}{9}$
c. $P(0 \leq X \leq 2)=\frac{139}{144}$
Work Step by Step
We are given $\frac{20x^{4}}{9}$ if $0 \leq x \leq1$
and $f(x)=\frac{20x}{9x^{5}}$ if $x \gt 1$
$\int^{1}_{0}\frac{20}{9}x^{4}dx + \int^{\infty}_{1} \frac{20}{9}\frac{1}{x^{5}}dx$
$=\frac{20}{9}(\frac{x^{5}}{5}|^{1}_{0}+\frac{x^{-4}}{-4}|^{\infty}_{1})$
$=\frac{20}{9}\frac{1}{5}+\frac{20}{9}\frac{1}{4}$
$=\frac{4}{9}+\frac{5}{9}=1$
a. $P(0 \leq X \leq 1)=\frac{4}{9}$
b. $P(X\gt1)=\frac{5}{9}$
c. $P(0 \leq X \leq 2)=\int^{1}_{0}\frac{20}{9}x^{4}dx + \int^{2}_{1} \frac{20}{9}\frac{1}{x^{5}}dx$
$=\frac{20}{9}(\frac{x^{5}}{5}|^{1}_{0}+\frac{x^{-4}}{-4}|^{2}_{1})$
$=\frac{20}{9}\frac{1}{5}+\frac{20}{9}\frac{15}{64}$
$=\frac{139}{144}$
