Find probability based on cumulative functions

Question

I have a CDF:

$F(x) = \begin{cases} 0, & x\le-2 \\ \frac 18, & -2< x\le -1 \\ \frac 38, &-1< x\le 0 \\ \frac 58,& 0 < x \le 1 \\ \frac 78,& 1 < x \le 2 \\ 1, &x> 2\end{cases}$

Find $P(-1\le X \le1)$, $P(X \le -1$ or $X = 2)$.

I'm not sure whether $P(-1\le X \le1) = F(1)-F(-1)=\frac 12$ or $P(-1\le X \le1) =F(2)-F(-1)=\frac 78 - \frac 18=\frac 68$

Please help me, I have been searching all sources and found no positive results.

Answer 1

Remark:

• The proposed CDF is not a valid cdf, a cdf should be right-continous, which is not the case here.

If we subtract $F(-1)$, we would not have included $P(X=-1)$.

Assuming that the typos are just with the inequality that is rather than $(-2<x\le -1)$, it should be $(-2 \le x < -1)$.

The answer should be \begin{align}Pr(-1 \le X \le 1) &= Pr(X \le 1) - Pr(X < -1) \\ &= F(1) - F(-2)\\ &=\frac78 - \frac18\\ &= \frac68\end{align}