I have this statement:
Let $f$ the probability density function of the continuous random $X$ defined by:
$ f(x)=\left\{ \begin{array}{lr} \frac{x}{8} & 0≤x ≤4\\ 0 & \text{otherwise} \end{array} \right. $
My current development is:
$ F(x)=\left\{ \begin{array}{lr} 1 & x >4\\ 0 & x \leq 0\\ ?& 0≤x ≤4 \end{array} \right. $
I only need the $F(x)$ between $0 \leq x \leq 4$, but i don't know how to solve it.
PD: I can't use calculus to solve it, so, how could I do it then?
Thanks in advance.
You don't need calculus if you know how to calculate the surface area of a triangle.
Per definitionem, the pdf tells the probability that a random variable falls into an interval by its surface area below the pdf and above the interval.
See our pdf below:
The cdf at an $0\leq x \leq 4$ qives the following probability
$$P(X\leq x)=\text{the area of the red triangle} =\frac{x^2}{16}.$$
(And of course, for $x$'s below $0$ the cdf is zero and above $4$ the cdf is $1$ which is the surface area of the full triangle.)