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The probability distribution function (or Cumulative Distributions Function) of a discrete random variable $X$ is given by $$\begin{equation} F_X(x) = \begin{cases} 0, & \text{for $x<-2.5$}.\\ 0.3, & \text{for $-2.5\le x< 0$}.\\ 0.6, & \text{for $0\le x< 1$}.\\ 1, & \text{for } x\ge 1\\ \end{cases} \end{equation}$$ a) Find the probability mass function for this random variable, i.e $P(X = x_j ) = p (x_j)$
b) $𝑃(𝑋<1 ) =$ ?
c) Find the Expected value of $X$. ($𝐸(𝑋)=$?)
d) Find the variance and standard deviation of X. ( $\operatorname {𝑣𝑎𝑟}(𝑋)=$? and $\sigma_𝑋=$?)

here what I tried:
a) $$\begin{equation} p(x) = \begin{cases} 0, & \text{for $x<-2.5$}.\\ 0.3, & \text{for $-2.5\le x< 0$}.\\ 0.3, & \text{for $0\le x< 1$}.\\ 0.4, & \text{for } x\ge 1\\ \end{cases} \end{equation}$$

b) $P(X<1 )= P(X<1 )=0.3+0.3=0.6$
c) I know this : $E(X) = p*f(x_i) +... $ but i dont think it is suitable neither this: $\int_{a}^{b}xf(x)dx$
so to find expected value what method i can use ? and for question a) and b) are correct right ? Thank you

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    $\begingroup$ Part (a) isn't right. The pmf for a discrete random variable should be defined by point masses, not over intervals. Once you fix that, it should help you with (c) and (d). Part (b) is ok. $\endgroup$ – Mick A Dec 22 '15 at 20:37
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$X$ is a discrete random variable. So it assumes discrete integral values only.

Hence your $p(x)$ can be correctly written as: $$\begin{equation} p(x) = \begin{cases} 0.3, & \text{for $x=-2$}.\\ 0.3, & \text{for $x=-1$}.\\ 0.3, & \text{for $x= 0$}.\\ 0.4, & \text{for $x= 1$}.\\ \end{cases} \end{equation}$$

As a result, $E(X)=-0.6-0.3+0.4=-0.5$

And so on, you will get the other results as well.

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