This question is an end-of-chapter problem from Probability & Statistics for Engineers & Scientists, 9th Edition:
Where the cumulative distribution function is defined as $F(x) = P(X \leq x)$.
Here is how I proceeded with a solution:
$P(X \leq 0) = F(0) = f(0) = 0.41$
$P(X \leq 1) = F(1) = f(0) + f(1) = 0.78$
$P(X \leq 2) = F(2) = f(0) + f(1) + f(2) = F(1) + f(2) = 0.94$
$P(X \leq 3) = F(3) = F(2) + f(3) = 0.99$
$P(X \leq 4) = F(4) = F(3) + f(4) = 1.00$
Now, the solutions to these problems are typically given in the form of a piecewise function. And here's the book's solution:
I'm having a hard time understanding how they arrived at those intervals. The chapter did give an example problem like this, but it did not explain the logic behind how they arrived at the piecewise breakdown.
If someone could please guide me through those steps, I would really appreciate the help. Thank you!
Best Answer
You have a discrete RV $X$ that can only take five values : $0 \ 1 \ 2 \ 3 \ 4$ with respective probabilities $$P(X=0) = 0.41$$ $$P(X=1) = 0.37$$ $$P(X=2) = 0.16$$ $$P(X=3) = 0.05$$ $$P(X=4) = 0.01$$ Suppose that those values are written of the faces of a very strange dice. You throw that dice and ask yourself:
$$P(X \le x) = F(x) = 0 \ for \ x< 0$$
I could go on until $4$ but I think you got it.