1. ©The McGraw-Hill Companies, Inc. 2008McGraw-Hill/Irwin Probability Distributions Chapter 6

2. 2 Define the terms probability distribution and random variable. Distinguish between discrete and continuous probability distributions. Calculate the mean, variance, and standard deviation of a discrete probability distribution. Describe the characteristics of and compute probabilities using the binomial probability distribution. GOALS

3. 3 What is a Probability Distribution? Experiment: Toss a coin three times. Observe the number of heads. The possible results are: zero heads, one head, two heads, and three heads. What is the probability distribution for the number of heads?

4. 4 Probability Distribution of Number of Heads Observed in 3 Tosses of a Coin

5. 5 Characteristics of a Probability Distribution

6. 6 Random Variables Random variable - a quantity resulting from an experiment that, by chance, can assume different values.

7. 7 Types of Random Variables Discrete Random Variable can assume only certain clearly separated values. It is usually the result of counting something Continuous Random Variable can assume an infinite number of values within a given range. It is usually the result of some type of measurement

8. 8 Discrete Random Variables - Examples The number of students in a class. The number of children in a family. The number of cars entering a carwash in a hour. Number of home mortgages approved by Coastal Federal Bank last week.

9. 9 Continuous Random Variables - Examples The distance students travel to class. The time it takes an executive to drive to work. The length of an afternoon nap. The length of time of a particular phone call.

10. 10 Features of a Discrete Distribution The main features of a discrete probability distribution are: The sum of the probabilities of the various outcomes is 1.00. The probability of a particular outcome is between 0 and 1.00. The outcomes are mutually exclusive.

11. 11 The Mean of a Probability Distribution MEAN The mean is a typical value used to represent the central location of a probability distribution. The mean of a probability distribution is also referred to as its expected value.

12. 12 The Variance, and Standard Deviation of a Probability Distribution Variance and Standard Deviation Measures the amount of spread in a distribution The computational steps are: 1. Subtract the mean from each value, and square this difference. 2. Multiply each squared difference by its probability. 3. Sum the resulting products to arrive at the variance. The standard deviation is found by taking the positive square root of the variance.

13. 13 Mean, Variance, and Standard Deviation of a Probability Distribution - Example John Ragsdale sells new cars for Pelican Ford. John usually sells the largest number of cars on Saturday. He has developed the following probability distribution for the number of cars he expects to sell on a particular Saturday.

14. 14 Mean of a Probability Distribution - Example

15. 15 Computational Formula for Variance An equivalent formula for the variance of a discrete distribution (and one that is simpler to use) is the following: After we have calculated the mean, , we calculate the above sum and subtract the square of the mean to get the variance.

16. 16 Variance and Standard Deviation of a Probability Distribution - Example In the previous example, we already found that the mean was 2.10. We calculate the following sum: and subtract the square of the mean to get the variance of the probability distribution. The standard deviation is then

17. 17 A Special Distribution There are certain types of probability distributions that are encountered often in many real-world situations. In this chapter, we concentrate on a single particular probability distribution that occurs in many real-world situations, the binomial distribution.

18. 18 Binomial Probability Distribution Characteristics of a Binomial Experiment The experiment consists of a fixed number of trials, The trials are identical – they are performed the same way. The trials are independent of each other, Each trial has two possible outcomes: Success or Failure, The probability of Success is the same for each trial. The random variable is the number of Successes.

19. 19 Example 1: Presidential polls Suppose that today is October 15, 2008, and I want to predict the outcome of the 2008 Presidential election. I select a random sample of n = 1067 voters from the population of all registered and likely voters in the United States, and I ask each voter the following question: “Do you intend to vote for Senator John McCain in the Presidential election?”

20. 20 Example 1: (continued) 1) The experiment consists of 1067 trials, 2) The trials are identical, since each one consists of randomly selecting a registered and likely voter and asking the same question. 3) The trials are independent of each other, due to random sampling, 4) Each trial results in one of two possible outcomes: Success = {voter says, “Yes”} or Failure = {voter says, “No”}, 5) P(Success) is the same for each trial, due to random sampling. Therefore this is a binomial experiment.

21. 21 Example 1: (continued) If I let X = number of voters in the sample who answer “Yes”, then X is a random variable that has a binomial distribution with n = 1067 and  = Senator McCain’s level of support in the voting population, expressed as a fraction. Using this probability distribution, I can calculate probabilities of various outcomes of the poll.

22. 22 Example 1: (continued) For example, a recent poll say that the Senator’s level of support in the voting population is  = 0.441, with Senator Obama’s level of support being 0.496. In our experiment, then, X has a binomial distribution with n = 1067 and  = 0.441. How likely is it that a majority of voters in the sample will be supporters of Senator McCain? We will answer this question later, after we acquire some more tools for calculating binomial probabilities.

23. 23 Binomial Probability Formula The probability that a binomial random variable X will be found to take on the value x when we perform the binomial experiment can be found using the following equation:

24. 24 Binomial Probability – Example 2 There are five flights daily from Pittsburgh via US Airways into the Bradford, Pennsylvania, Regional Airport. Suppose the probability that any flight arrives late is.20. What is the probability that none of the flights is late today?

25. 25 Binomial Dist. – Mean and Variance The mean, or expectation, of a binomial Random variable X gives the expected number of Successes out of the n trials of the experiment. The variance and standard deviation give us an idea about how spread out the probability distribution is.

26. 26 Mean and Variance, Example 1: Let’s go back to our poll example. We have n = 1067 and  = 0.441. Then the expected number of “Yes” answers in our poll is  = n  = (1067)(0.441) = 470.547. The variance of the distribution is  2 = n  (1-  ) = 263.0358, and the standard deviation is  = 16.2184.

27. 27 For the example regarding the number of late flights, recall that  =.20 and n = 5. What is the average number of late flights? What is the variance of the number of late flights? Binomial Dist. – Mean and Variance: Example 2

28. 28 Binomial Distribution - Table Five percent of the worm gears produced by an automatic, high- speed Carter-Bell milling machine are defective. What is the probability that out of six gears selected at random none will be defective? Exactly one? Exactly two? Exactly three? Exactly four? Exactly five? Exactly six out of six?

29. 29 Binomial Distribution - MegaStat Five percent of the worm gears produced by an automatic, high-speed Carter-Bell milling machine are defective. What is the probability that out of six gears selected at random none will be defective? Exactly one? Exactly two? Exactly three? Exactly four? Exactly five? Exactly six out of six?

30. 30 Binomial – Shapes for Varying  (n constant)

31. 31 Binomial – Shapes for Varying n (  constant)

32. 32 Cumulative Binomial Probability Distributions A study in June 2003 by the Illinois Department of Transportation concluded that 76.2 percent of front seat occupants used seat belts. A sample of 12 vehicles is selected. What is the probability the front seat occupants in at least 7 of the 12 vehicles are wearing seat belts?

33. 33 Example 1: (continued) We will use MegaStat to calculate the probability that a majority of those polled support Sen. McCain. We want the probability that X is at least 534. By the complement rule, this is 1 minus the probability that X is no more than 533. From the table given by MegaStat, we find that P(X  533) = 0.99995, so that P(X > 534) = 0.00005. It is very unlikely that a majority of the sample will say, “Yes.” (Of course, the latest poll also includes voters who are still undecided.)

34. 34 End of Chapter 6