# Finding Mean using Cumulative Distributive Function

I understand that to the expected value for the random variable X but with the cumulative distribution function (shown below), I am unable to differentiate it to find the probability density function to solve for the expected value.

$$F_X(x)=\begin{cases}0 & for \ \ x < 0.8\ \\0.2 & for \ \ 0.8 ≤ x < 1.2\ \\0.45 & for \ \ 1.2 ≤ x < 1.8\ \\0.75 & for \ \ 1.8≤ x < 2.5\ \\0.9 & for \ \ 2.5≤ x < 4\ \\ 1 & for \ \ x≥4 \end{cases}$$

Is there another way for me to find the expected value of X without having to find the probability density function?

I would greatly appreciate any help that can point me in the right direction for this question.Thank you!

First note that there is an error in your question, since the first condition is $$x<0.8$$ and the second condition is $$0.7 \le x < 1.2$$. They should be disjoint.
$$P(X=1.2)= P(X \le 1.2)-P(X < 1.2)=0.45-0.2=0.25$$
• Are not you assuming that $X$ is a discrete r.v.? If it is continuous, $P(X=1.2)=0$. – farruhota Oct 6 at 9:51