2. Slide 5-2 Copyright © 2008 Pearson Education, Inc. Chapter 5 Probability and Random Variables

3. Slide 5-3 Copyright © 2008 Pearson Education, Inc. Definition 5.1

4. Slide 5-4 Copyright © 2008 Pearson Education, Inc. Figure 5.1 Example 5.3 When two balanced dice are rolled, 36 equally likely outcomes are possible: The sum of the dice can be 11 in two ways. The probability the sum is 11 is f/N = 2/36 = 0.056. Doubles can be rolled in six ways. The probability of doubles is f/N = 6/36 = 0.167.

5. Slide 5-5 Copyright © 2008 Pearson Education, Inc. Figure 5.2 If we toss a balanced coin once, we reason that there is a 50-50 chance the coin will land with heads facing up. The frequentist interpretation is that in a large number of tosses, the coin will land with heads facing up about half the time. We used a computer to perform two simulations of tossing a balanced coin 100 times. Both graphs seem to corroborate the frequentist interpretation.

6. Slide 5-6 Copyright © 2008 Pearson Education, Inc. Key Fact 5.1

7. Slide 5-7 Copyright © 2008 Pearson Education, Inc. Definition 5.2

8. Slide 5-8 Copyright © 2008 Pearson Education, Inc. Definition 5.3 Figure 5.9

9. Slide 5-9 Copyright © 2008 Pearson Education, Inc. Definition 5.4

10. Slide 5-10 Copyright © 2008 Pearson Education, Inc. Figure 5.14 Figure 5.15

11. Slide 5-11 Copyright © 2008 Pearson Education, Inc. Definition 5.5 Probability Notation If E is an event, then P(E) represents the probability that event E occurs. It is read “the probability of E.”

12. Slide 5-12 Copyright © 2008 Pearson Education, Inc. Formula 5.1

13. Slide 5-13 Copyright © 2008 Pearson Education, Inc. Formula 5.2

14. Slide 5-14 Copyright © 2008 Pearson Education, Inc. Formula 5.3

15. Slide 5-15 Copyright © 2008 Pearson Education, Inc. Definition 5.6 Definition 5.7

16. Slide 5-16 Copyright © 2008 Pearson Education, Inc. Definition 5.8

17. Slide 5-17 Copyright © 2008 Pearson Education, Inc. a.Determine the probability distribution of the random variable X. b.Construct a probability histogram for the random variable X. Table 5.6 Example 5.14 Professor Weiss asked his introductory statistics students to state how many siblings they have. Table 5.6 presents a grouped-data table for that information. The table shows, for instance, that 11 of the 40 students, or 27.5%, have two siblings. Let X denote the number of siblings of a randomly selected student.

18. Slide 5-18 Copyright © 2008 Pearson Education, Inc. Table 5.7 Solution Example 5.14 Table 5.7 displays these probabilities and provides the probability distribution and Figure 5.1 displays the probability histogram of the random variable X. Figure 5.21

19. Slide 5-19 Copyright © 2008 Pearson Education, Inc. Key Fact 5.2

20. Slide 5-20 Copyright © 2008 Pearson Education, Inc. Example 5.18 Suppose we repeat the experiment of observing the number of heads, X, obtained in three tosses of a balanced dime a large number of times. Then the proportion of those times in which, say, no heads are obtained (X = 0) should approximately equal the probability of that event [P(X = 0)]. The same statement holds for the other three possible values of the random variable X. Use simulation to verify these facts.

21. Slide 5-21 Copyright © 2008 Pearson Education, Inc. Solution Example 5.18 We used a computer to simulate 1000 observations of the random variable X, the number of heads obtained in three tosses of a balanced dime. Table 5.12 shows the frequencies and proportions for the numbers of heads obtained in the 1000 observations. Table 5.12

22. Slide 5-22 Copyright © 2008 Pearson Education, Inc. Solution Example 5.18 As expected, the proportions in the third column of Table 5.12 are fairly close to the true probabilities in the second column of Table 5.11. Table 5.11 Table 5.12

23. Slide 5-23 Copyright © 2008 Pearson Education, Inc. Solution Example 5.18 This result is more easily seen if we compare the proportion histogram to the probability histogram of the random variable X, as shown in Fig. 5.22. Figure 5.22

24. Slide 5-24 Copyright © 2008 Pearson Education, Inc. Key Fact 5.3

25. Slide 5-25 Copyright © 2008 Pearson Education, Inc. Definition 5.9

26. Slide 5-26 Copyright © 2008 Pearson Education, Inc. Key Fact 5.4

27. Slide 5-27 Copyright © 2008 Pearson Education, Inc. We used a computer to simulate the number of busy tellers at 1:00 P.M. on 100 randomly selected days; that is, we obtained 100 independent observations of the random variable X. The data are displayed in Table 5.16. Table 5.16 The average value of the 100 observations in Table 5.16 is 4.25. This value is quite close to the mean, μ = 4.118, of the random variable X.

28. Slide 5-28 Copyright © 2008 Pearson Education, Inc. Figure 5.23 If we made, say, 1000 observations instead of 100, the average value of those 1000 observations would most likely be even closer to 4.118. Figure 5.23(a) shows a plot of the average number of busy tellers versus the number of observations for the data in Table 5.16.

29. Slide 5-29 Copyright © 2008 Pearson Education, Inc. Definition 5.10

30. Slide 5-30 Copyright © 2008 Pearson Education, Inc. Definition 5.11

31. Slide 5-31 Copyright © 2008 Pearson Education, Inc. Definition 5.12

32. Slide 5-32 Copyright © 2008 Pearson Education, Inc. Definition 5.13

33. Slide 5-33 Copyright © 2008 Pearson Education, Inc. Table 5.18 Example 5.24 According to tables provided by the U.S. National Center for Health Statistics in Vital Statistics of the United States, a person aged 20 has about an 80% chance of being alive at age 65. Suppose that three people aged 20 are selected at random. As Table 5.18 indicates, eight outcomes are possible. Solution All eight possible outcomes and their probabilities are shown in Table 5.19. Table 5.19

34. Slide 5-34 Copyright © 2008 Pearson Education, Inc. Figure 5.24 Solution A tree diagram is useful for organizing and summarizing the possible outcomes of this experiment and their probabilities.

35. Slide 5-35 Copyright © 2008 Pearson Education, Inc. Key Fact 5.5

36. Slide 5-36 Copyright © 2008 Pearson Education, Inc. Formula 5.4

37. Slide 5-37 Copyright © 2008 Pearson Education, Inc. Procedure 5.1

38. Slide 5-38 Copyright © 2008 Pearson Education, Inc. Figure 5.26 Generally, a binomial distribution is right skewed if p < 0.5, is symmetric if p = 0.5, and is left skewed if p > 0.5. Figure 5.26 illustrates these facts for three different binomial distributions with n = 6.

39. Slide 5-39 Copyright © 2008 Pearson Education, Inc. Formula 5.5

40. Slide 5-40 Copyright © 2008 Pearson Education, Inc. Key Fact 5.6