Statistics - Cumulative Distribution Function $$\begin{equation}F(x)=\begin{cases}0 &\quad x<-10 \\ 0.25 &   \quad -10\leqslant x <30 \\
0.75 &\quad 30 \leqslant x <50 \\ 1 &\quad 50 \leqslant x \end{cases}\end{equation}$$
$a)\,\, P(X \leqslant 50)$
$b)\,\, P(X \leqslant 40)$
$c)\,\, P(40 \leqslant X \leqslant 60)$
$d)\,\, P(X < 0)$
$e)\,\, P(0\leqslant X < 10)$
$f)\,\, P(-10<X < 10)$
I'm having trouble wrapping my head about how the answers in the back of the book are obtained:
$a)\,\, 1$
$b)\,\, .75$
$c)\,\, .25$
$d)\,\, .25$
$e)\,\, 0$
$f)\,\, 0$
What I do not understand is how to correctly plug in the different values of $X$ and determine the probability of each...in particular when $X$ is between a range of two integers, yet the jumps in probabilities are at different values.
Would anyone care to explain the methodology behind finding these solutions? Thank you
 A: The $F(x)$ given to you is the cumulative distribution function. Before $-10$, it is $0$. At $-10$, it suddenly jumps to $0.25$, and stays there for a while. So there must be a weight of $0.25$ at $x=-10$. That is, $\Pr(X=-10)=0.25$.
Similarly, there is a sudden jump from $0.25$ to $0.75$ at $30$. Thus $\Pr(X=30)=0.50$.
Finally, $\Pr(X=50)=0.25$.
So we have the complete probability distribution function of $X$. The random variable $X$ takes on the values $-10$, $30$, and $50$, with probabilities $0.25$, $0.50$, and $0.25$. Now you can answer any question.
We could also answer the questions "algebraically." The only nuisance is that, because of the jumps, we need to carefully distinguish between $\lt$ and $\le$, also between $\gt$ and $\ge$.
Here is a sample. For (e), we want $\Pr(0\le x\lt 10)$. This is $\Pr(X\lt 10)-\Pr(X\lt 0)$.
Look at the cumulative distribution function $F(x)$ of $X$. Just short of $10$, it has value $0.25$. Just short of $0$, it has value $0.25$. Subtract: the result is $0$.
But it is intuitively clearer to observe that $X$ "takes on" no values between $0$ and $10$.
Remark: Occasionally, one even bumps into random variables that have a character which is a hybrid of the discrete and the continuous. For example, after the weight of $0.25$ at $x=-10$, the cumulative distribution function could after that climb smoothly to $1$. 
A: It's a step function that jumps at $-10$, $30$, and $50$. So at these steps you gain the increments indicated by $F$: $0.25, 0.75-0.25 = 0.5,$ and $1 - 0.75 = 0.25$ (the difference between before and after). Note also that if you hit any of the thresholds, you will get the full increase, which matters to distinguish $<$ from $\leq$ in the questions (c., eg, (e) below as you have a right end of the interval of $< 10$, not $\leq 10$). For instance,  
(a) $\mathbf{P}(X \leq 50)=1$ as you jumped $50$ to reach the final $1$
(c) $\mathbf{P}(40 \leq X < 60)= 0.25$ as you jumped at 50 by 0.25
(e) $\mathbf{P}(0 \leq X < 10) = 0$ as you last jumped at $-10$, and will again at $10$.
A: Note that, by definition of a c.d.f. $P(X\le x)=F(x)$. Hence,
$P(X\le 50)=F(50)=1$
$P(X \le 40)=F(40)=0.75$
$P(40 \le X \le 60)=P(X\le 60)-P(X < 40)=F(60)-F(40^{-})=1-0.75=0.25$
For the strict inequality, since as $X$ gets close to $0$ from below, it satisfies the $-10<X<30$ condition:
$P(X<0)=0.25$
$P(0\le X < 10)=P(X < 10) - P(X < 0)=0.25-0.25=0$
$P(-10 < X < 10)=P(X<10)-P(X<-10)=0.25-0.25=0$
