Find expected value from given PDF (CDF) 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) $(<1 ) =$ ?
c) Find the Expected value of $X$. ($()=$?) 
d) Find the variance and standard deviation of X. ( $\operatorname {}()=$? and $\sigma_=$?) 
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 

$X$ 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.
