1. Chapter 7: Random Variables and Probability Distributions

2. Section 7.1: Random Variables

3. Random Variable – a numerical variable whose value depends on the outcome of a chance experiment. A random variable associates a numerical value with each outcome of a chance experiment. Discrete – A random variable is discrete if its set of possible values is a collection of isolated points on the number line. Continuous – The variable is continuous if its set of possible values includes an entire interval on the number line.

4. Example Consider an experiment in which the type of car, new (N) or used (U), chosen by each of three successive customers at a discount car dealership is noted. Define a random variable x by x = number of customers purchasing a new car

5. The eight possible outcomes are listed below: Outcome: UUU NUU UNU UUN NNU NUN UNN NNN x value: 0 1 1 1 2 2 2 3 There are only four possible x values – 0, 1, 2, and 3 – and these are isolated points on the number line. Thus, x is a discrete random variable.

6. Example In an engineering stress test, pressure is applied to a thin 1-ft long bar until the bar snaps. The precise location where the bar will snap is uncertain. Let x be the distance from the left end of the bar to the break. Then x = 0.25 is on possibility, x = 0.9 is another, and in face any number between 0 and 1 is a possible value of x. This set of possible values is an entire interval on the number line, so x is a continuous random variable.

7. Section 7.2: Probability Distributions for Discrete Random Variables

8. Probability distribution of a discrete random variable x – gives the probability associated with each possible x value. Each probability is the limiting relative frequency of occurrence of the corresponding x value when the chance experiment is repeatedly performed. Common ways to display a probability distribution for a discrete random variable are a table, a probability histogram, or a formula.

9. Example Suppose that each of four randomly selected customers purchasing a hot tub at a certain store chooses either an electric (E) or a gas (G) model. Assume that these customers make their choices independently of one another and that 40% of all customers select an electric model. This implies that for any particular one of the four customers, P(E) =.4 and P(G) =.6. One possible outcome is EGGE, where the first and fourth customers select electric models and the other two choose gas models. Because the customers make their choices independently, the multiplication rule for independent events implies…

10. P(EGGE) = P(1 st chooses E and 2 nd chooses G and 3 rd chooses G and 4 th chooses E) = P(E)P(G)P(G)P(E) = (.4)(.6)(.6)(.4) =.0576 Let x = the number of electric hot tubs purchased by the four customer

11. Outcomes and Probabilities OutcomeProbabilityx valueOutcomeProbabilityx value GGGG.12960GEEG.05762 EGGG.08641GEGE.05762 GEGG.08641GGEE.05762 GGEG.08641GEEE.03843 GGGE.08641EGEE.03843 EEGG.05762EEGE.03843 EGEG.05762EEEG.03843 EGGE.05762EEEE.02564

12. The probability distribution of x is easily obtained from this information. Consider the smallest possible x value, 0. The only outcome for which x = 0 is GGGG, so p(0) = P(x = 0) = P(GGGG) =.1296

13. Let’s find the 4 other outcomes: p(1) = P(x = 1) = P(EGGG or GEGG or GGEG or GGGE) = P(EGGG) + P(GEGG) + P(GGEG) + P(GGGE) =.0864 +.0864 +.0864 +.0864 = 4(.0864) =.3456 p(2) = 6(.0576) =.3456 p(3) = 4(.0384) =.1536 p(4) =.0256

14. Properties of Discrete Probability Distributions: 1.For every possible x value, 0 ≤ p(x) ≤ 1.

15. Example A consumer organization that evaluates new automobiles customarily reports the number of major defects on each car examined. Let x denote the number of major defects on a randomly selected car of a certain type. A large number of automobiles were evaluated, and a probability distribution consistent with theses observations is:

16. x 0 1 2 3 4 5 6 7 8 9 10 p(x).041.130.209.223.178.114.061.028.011.004.001