Answer
$\frac{1}{63}x^{3}-\frac{27}{98}$
Work Step by Step
We are given $f(x)=\frac{3}{98}x^{2}$
for $3\leq x \leq 5$
The cumulative distribution function is given by
$F(x)=P(X\leq x)=\int^{x}_{3}(\frac{3}{98}t^{2})dt$
$=\frac{1}{98}t^{3}|^{x}_{3}$
$=\frac{1}{63}x^{3}-\frac{27}{98}$
