Discrete Random Variables Help?

Question

Delivery truck is required to make five stops on its morning route. Let X denote # of on-time deliveries for a given morning, let f denote probability mass function of X, and let F denote the cumulative distribution function of X. Historical data suggests that:

f_X (5) = 0.88, f_X (4) = 0.06, f_X (3) = 0.03, f_X (2) = 0.02, f_X (1) = 0.008, f_X (0) = 0.002

1. Construct cumulative distribution function of X.
2. Determine probability that truck makes at least 4 on-time deliveries.

Any help or advice on how to start solving this problem would be great. My professor is not great at explaining and most of the class is lost in this subject.

Answer 1

The cumulative distribution function is just derived by adding the individual probabilities; F(x) = P(X <= x).
So F(0) = f_X(0) = 0.002
F(1) = f_X(0) + f_X(1) = 0.01
F(2) = F(1) + f_X(2) = 0.03
F(3) = F(2) + f_X(3) = 0.06
F(4) = F(3) + f_X(4) = 0.12
F(5) = F(4) + f_X(5) = 1.
If you want to make this a continuous function, recall the definition I gave above: F(x) = P(X <= x). You can see, for instance, that for 1 < x < 2 we will have F(x) = P(X <= x) = P(X <= 1) = F(1). So we get a step type function that has jumps at the possible values of X; it will be 0 for x < 0 and 1 for x >= 5.

P(X >= 4) = P(X = 4 or X = 5)
= f_X(4) + f_X(5)
= 0.06 + 0.88
= 0.94.

Answer 2

The idea is the for probability mass functions you have specific values (that is when "x is equal to something") however, for cumulative you have "x is greater than or equal etc)

The prob
F(1>X=>0) == zero on time ===0.002 == f(0)
F(2>X) == at most 1 on time == 0.01 == f(0) +f(1)
F(3>X) == at most 2 on time == 0.03 == f(0) +f(1) +f(2)
F(4>X) == at most 3 on time == 0.06 == f(0) +f(1)+f(2) + f(3)
F(5>X) == at most 4 on time == 0.12 == f(0) +f(1)+f(2) + f(3)+ f(4)
F(X>5) == at most 5 on time == 1 == f(0) +f(1)+f(2) + f(3)+f(4)+f(5)

At least 4 on time means
.. 4 on time deliveries + 5 on time deliveries
= 0.06 + 0.88
= 0.94