# 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!

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What is the definition of $F(x)$? Perhaps you could edit your question to include this information. –  Dilip Sarwate Apr 25 '13 at 22:19
Please do not confuse $x$ and $X$. –  Did Apr 25 '13 at 22:21

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 $\Pr(X\ge 0.5)=1-0.25=0.75$.

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Now I get it. Thanks for help! BTW -- shouldn't there be $Pr(X\ge 0.5)=1-0.25=0.75$ instead of $Pr(X\ge 0.5)=1-0.5=0.75$ –  atomoutside Apr 25 '13 at 22:36
You are welcome. Thanks for catching the typo. –  André Nicolas Apr 25 '13 at 22:42