1. ## Cumulative distribution function

Dear all, I have a question
Given this distribution function:
0 , x<0
5/12x - 1/24x^2 , 0<x<4
1 , x>4

I want to find P(X<2)
The answer gives it as 0.667 but why?
Thank you

2. ## Re: Cumulative distribution function

X= 2 is between 0 and 4 so we evaluate $\frac{5}{12}(2)- \frac{1}{24}(4)= \frac{5}{6}- \frac{1}{6}= \frac{4}{6}= \frac{2}{3}$ which, reduced to three decimal places, is approximately 0.667.

3. ## Re: Cumulative distribution function

Originally Posted by bbear123
Dear all, I have a question
Given this distribution function:
0 , x<0
5/12x - 1/24x^2 , 0<x<4
1 , x>4
I want to find P(X<2)
The answer gives it as 0.667 but why?
$CDF(x)=\begin{cases}0 &: x\le 0 \\ \frac{5}{12}x-\frac{1}{24}x^2 &: 0<x\le 4\\ 1 &:4<x\end{cases}$ This is the cumulative distribution function.
The idea is that the function increases(accumulates) from zero to one from left to right.
We define $\mathscr{P}(X)\le x)=CDF(x)$. More generally $\mathscr{P}(a\le X\le b)$ is $CDF(b)-CDF(a)$.

Thus to answer your question: $\mathscr{P}(X\le 2)=CDF(2)$