Probability Distribution Function (PDF) for a Discrete Random Variable · Try It 1

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)?

$$\begin{array}{cc}\text{x}& \text{P(x)}\\ \text{0}& \text{P(x = 0) = }\frac{4}{50}\\ \text{1}& \text{P(x = 1) = }\frac{8}{50}\\ \text{2}& \text{P(x = 2) = }\frac{16}{50}\\ \text{3}& \text{P(x = 3) = }\frac{14}{50}\\ \text{4}& \text{P(x = 4) = }\frac{6}{50}\\ \text{5}& \text{P(x = 5) = }\frac{2}{50}\end{array}$$

Its source prints no answer for this problem, so there's nothing to check against.

OpenStax, Introductory Statistics 2e · §4.1 Probability Distribution Function (PDF) for a Discrete Random Variable · Try It 1