Cumulative distribution function and probabilities I've run into problem with this exercise:

Let $X$ have distribution function:
$F(x)=\begin{cases}
0,  & \text{if $x$}<0 \\
\frac{1}{2}x, & \text{if }0\leq x\leq 2,  \\
1, & \text{if }x>2, 
\end{cases}$

I need to find:

a) $P(x\leq 1)$
b) $P(0.5 \leq x)$

Answers are $\frac{1}{2}$ and $\frac{3}{4}$ respectively. Could anyone explain me how to obtain them? Thanks!

The (cumulative) distribution function $F(x)=F_X(x)$ tells us the probability that $X\le x$. Write this as $\Pr(X\le x)$.   
So for (a), we can read off the answer directly from the given formula. We want $\Pr(X\le 1)$. This is $F(1)$, which is $\frac{1}{2}$.
For (b), we want $\Pr(X\ge 0.5)$, which is equal to $1-\Pr(X\lt 0.5)$. 
But $X$ has a continuous distribution, so $\Pr(X\lt 0.5)=\Pr(X\le 0.5)$. This is $F(0.5)$, which is $0.25$.
It follows that $\Pr(X\ge 0.5)=1-0.25=0.75$.
