A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes on value x. Why is this a discrete probability distribution function (two reasons)?

We are given,

X = the number of times a patient rings the nurse during a 12-hour shift.

$x=0,1,2,3,4,5$

P(x) = the probability that X takes on value x .

A discrete probability distribution counts events with countable outcomes. Discrete distributions, on the other hand, depict data with a countable number of outcomes, i.e., the alternative outcomes may be listed. A discrete distribution has two characteristics:

1. Each probability is between zero and one.
2. The sum of the probability is one.

In the given problem, we are given the probability distribution of random variables of the number of times the average post-op patient will ring the nurse during a 12-hour shift. Let's see these characteristics:

We can clearly see that all probabilities of a given random variable are from zero to one.

The sum of probabilities is:

$\frac{4}{50}+\frac{8}{50}+\frac{16}{50}+\frac{14}{50}+\frac{6}{50}+\frac{2}{50}+\frac{50}{50}=1$

Hence, verified.