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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!

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Guide:

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.

Note that from the CDF, we are able to compute the pmf, for example,

$$P(X=1.2)= P(X \le 1.2)-P(X < 1.2)=0.45-0.2=0.25$$

After you compute the pmf, you can compute the expected value.

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  • $\begingroup$ Are not you assuming that $X$ is a discrete r.v.? If it is continuous, $P(X=1.2)=0$. $\endgroup$ – farruhota Oct 6 at 9:51
  • $\begingroup$ yes, the function is discontinous at those points. $\endgroup$ – Siong Thye Goh Oct 6 at 10:00

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