Question

The following is the p.d.f. (Probability Density Function) of a continuous random variable $X$:
$F\left(x\right)=\frac{x}{32},0<x<8;=0$, otherwise
Also find its value at $x=0.5$ and $9$.

Answer 1

Values of $F\left(x\right)$ at different values of $x$.
At $x=0.5$

$F\left(x\right)=F\left(0.5\right)=\frac{(0.5{)}^2}{64}$.............(integration of F(x))

$=\frac{0.25}{64}$

$=\frac{1}{256}$
For any value of $x$ greater than or equal to $8,F\left(x\right)=1$
$\therefore F\left(9\right)=1$