I was reviewing for an exam and got the following question :
Consider the following mixed distribution :
$$
F(x) =
\begin{cases}
0 &\quad\text{if x}\le 1\\
0.13 &\quad\text{if }1\leq \text{x}\lt 2\\
0.33 &\quad\text{if }2\leq \text{x}\lt 3\\
0.5 &\quad\text{if }x = 3\\
0.5 + \frac{1}{4}(x-3) &\quad\text{if }3\lt \text{x}\lt 5\\
1 &\quad\text{if x}\gt 5\\
\end{cases}
$$
- $P((−\infty,3))$?
- $P([2,3))$?
The result for the first problem is 0.5 but I cant understand why, here is my reasoning :
$$
P((-\infty,3)) = F(3) - F(-\infty) - P(X=3) = F(3) - F(-\infty) -(F(3) -F(3^-)) =
$$
$$
F(3^-) - F(-\infty) = 0.33 - 0 = 0.33
$$
For the second one the result is 0.17 yet i get :
$$
P([2,3)) = F(3) - F(2) + P(X=2) - P(X=3) =
$$
$$
F(3) - F(2) +(F(2)-F(2^-)) - (F(3) -F(3^-)) =
$$
$$
F(3^-) - F(2^-) = 0.33 - 0.13 = 0.2
$$
Am I doing things wrong?