2. Copyright © 2009 Pearson Education, Inc. Chapter 17 Probability Models

3. Copyright © 2009 Pearson Education, Inc. Slide 1- 3 NOTE on slides / What we can and cannot do The following notice accompanies these slides, which have been downloaded from the publisher’s Web site: “This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from this site should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.” We can use these slides because we are using the text for this course. Please help us stay legal. Do not distribute these slides any further. The original slides are done in orange / brown and black. My additions are in red and blue. Topics in green are optional.

4. Copyright © 2009 Pearson Education, Inc. Slide 1- 4 Topics in this chapter Probability Models Bernoulli Trials The Binomial Model Assumptions and Conditions The Geometric Model The Poisson Model Approximating distributions with the Normal Success / Failure condition Note: “Optional” topics begin with Slide 50.

5. Copyright © 2009 Pearson Education, Inc. Division of Mathematics, HCC Course Objectives for Chapter 17 After studying this chapter, the student will be able to: Tell if a situation involves Bernoulli trials. Know the appropriate conditions for using a Binomial or Normal model. Find the mean and standard deviation of a Binomial model. Calculate binomial probabilities, perhaps with a Normal model. Find the mean and standard deviation of a Geometric model. Calculate and interpret Geometric probabilities. Calculate and interpret Poisson probabilities.

6. Copyright © 2009 Pearson Education, Inc. Slide 1- 6 Probability Models so far The Normal Model

7. Copyright © 2009 Pearson Education, Inc. Slide 1- 7 Probability Models so far Our Insurance Model Now we are going to look at some more probability models. OutcomePayoutProbability Death$10,0001 in 1000 Disability$5,0002 in 1000 Uninjured0997 in 1000

8. Copyright © 2009 Pearson Education, Inc. Slide 1- 8 Example – Joe Flacco pictures A cereal manufacturer puts pictures of athletes in their cereal boxes. Distribution is 20% Joe Flacco 30% LaBron James 50% Pablo Sandoval You really want a Joe Flacco picture

9. Copyright © 2009 Pearson Education, Inc. Slide 1- 9 Recall our cereal box example Consider the event “Getting a Joe Flacco picture”. We said that 20% of the boxes have Joe Flacco pictures. We can view an attempt to get a Joe Flacco picture, either through buying the cereal or simulation, as a trial with two possible outcomes (success / failure) All trials have the same probability of success (0.2). The trials are independent (there is a way around this one). This is an example of something called a Bernoulli Trial.

10. Copyright © 2009 Pearson Education, Inc. Slide 1- 10 Bernoulli Trials The basis for the probability models we will examine in this chapter is the Bernoulli trial. We have Bernoulli trials if: there are two possible outcomes (success and failure). the probability of success, p, is constant. the trials are independent. (*)

11. Copyright © 2009 Pearson Education, Inc. Slide 1- 11 Example – Cereal Boxes Recall the example of the cereal boxes. We’ll focus on Joe Flacco and the 20% of the boxes that contain his picture. Question: If we buy 5 boxes, what is the probability that we get 2 Joe Flacco pictures?

12. Copyright © 2009 Pearson Education, Inc. Slide 1- 12 Maybe we should check…independence? This can be tricky. There are only a finite number of boxes and we sample without replacement! Strictly speaking, the samples are not independent. But they are “close enough”. Let’s see why.

13. Copyright © 2009 Pearson Education, Inc. Independence for Cereal Boxes Suppose they made 100,000 boxes of cereal. One is bought. There are 99,999 left. If the first one had a Joe Flacco picture, what is the probability of the second box having a Joe Flacco picture? 19,999/99,999 = 0.199992 If the first one did not have a Joe Flacco picture, what is the probability of the second box having a Joe Flacco picture? 20,000/99,999 = 0.200002 They are not independent But does it really matter? Slide 1- 13

14. Copyright © 2009 Pearson Education, Inc. Slide 1- 14 Independence One of the important requirements for Bernoulli trials is that the trials be independent. When we don’t have an infinite population, the trials are not independent. But, there is a rule that allows us to pretend we have independent trials: The 10% condition: Bernoulli trials must be independent. If that assumption is violated, it is still okay to proceed as long as the sample is smaller than 10% of the population.

15. Copyright © 2009 Pearson Education, Inc. Slide 1- 15 Remember this Example – Independence Good Grades PopularityGood at sports Total Boy1175060227 Girl1309130251 Total24714190478 Question 4: Are “being a girl” and “wanting good grades” independent? If so, then P(Grades | Girl) would be the same as P(Grades). But P(Grades | Girl) = 130 or 0.518 251 and P(Grades) = 247 or 0.517 478 Therefore, not independent – but very close!

16. Copyright © 2009 Pearson Education, Inc. Remember this Example – Independence 10% condition – we have sampled less than 10% of all possible students. In Howard County in 2008 there were: 3490 students enrolled in Grade 3 3441 students enrolled in Grade 4 3730 students enrolled in Grade 5 Total:10661 – our percentage: 4.48% We could treat these data as independent if they had been gathered in Howard County. Slide 1- 16

17. Copyright © 2009 Pearson Education, Inc. Slide 1- 17 Back to Cereal Boxes – The Binomial Model If we buy 5 boxes, what is the probability of two Joe Flacco pictures? Again, p = 0.2 and q = 0.8. Try.2 *.2 *.8 *.8 *.8. = 0.02048. No good – this gives the probability that the first two had Joe pictures! We want the probability that any two of the five had Joe pictures. Joe pictures can appear in (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4),(3,5) or (4,5) The probability is 10*.02048 = 0.2048. Is there an easier way?

18. Copyright © 2009 Pearson Education, Inc. Slide 1- 18 The Binomial Model A Binomial model tells us the probability for a random variable that counts the number of successes in a fixed number of Bernoulli trials. Two parameters define the Binomial model: n, the number of trials; and, p, the probability of success. We denote this Binom(n, p). First discussed by James Bernoulli (1654 – 1705) but not published until 1713.

19. Copyright © 2009 Pearson Education, Inc. Slide 1- 19 The Binomial Model (cont.) In n trials, there are ways to have k successes. Read n C k as “n choose k,” and is called a combination. Note: n! = n x (n – 1) x … x 2 x 1, and n! is read as “n factorial.”

20. Copyright © 2009 Pearson Education, Inc. Slide 1- 20 *HOLD IT!! What is this n C k thing? It is the combination of five things taken two at a time. Example: Joe Flacco – 2 occurrences out of 5 trials Same example: You have five (congenial) employees. You have an assignment for which you need two. How many combinations can you pick – same problem. 5 C 2 = __5!__ = 5 * 4 * 3 * 2 * 1 2! * 3! 2 * 1 * 3 * 2 * 1, or 10

21. Copyright © 2009 Pearson Education, Inc. Slide 1- 21 *What is this 5 C 2 thing? On the TI Type 5 – do not hit “enter”. Go to MATH, tab over to PRB, then select #3. Then type 2. The line will look like “5 nCr 2”. Now hit ENTER.

22. Copyright © 2009 Pearson Education, Inc. Binomial Model (graphically) Slide 1- 22 Source: Wikipedia http://en.wikipedia.org/wiki/Binomial_distribution

23. Copyright © 2009 Pearson Education, Inc. Slide 1- 23 *Cereal Box example For 2 successes in 5 tries, 5 C 2 = 5! _ = 5 * 4 * 3 * 2 * 1 2! * 3! 2 * 1 * 3 * 2 * 1, or 10 So P[2 successes in 5 trials] = 10 * (.2) 2 * (.8) 3 =.2048

24. Copyright © 2009 Pearson Education, Inc. Slide 1- 24 The Binomial Model (cont.) Binomial probability model for Bernoulli trials: Binom(n,p) n = number of trials p = probability of success q = 1 – p = probability of failure X = number of successes in n trials

25. Copyright © 2009 Pearson Education, Inc. Slide 1- 25 Back to the Cereal Box example For the TI, binompdf(n,p,k) = binompdf(5,.2,2) To find this, hit 2 nd, then VARS. This brings up DISTR. Then select choice A.

26. Copyright © 2009 Pearson Education, Inc. On the TI Slide 1- 26

27. Copyright © 2009 Pearson Education, Inc. Next question – “at most” What is the probability of at most two Joe Flacco boxes? Hard way: Figure it for 0, then 1, then 2 Binompdf(5,0.2,0)+binomcdf(5,0.2,1)+binomcdf(5,0.2,2) Easier way: binomcdf(5,0.2,2). binomcdf measures “at most”, or equivalently “no more than”. Slide 1- 27

28. Copyright © 2009 Pearson Education, Inc. On the TI – “at most” 2.55 Operating SystemOld Operating System Slide 1- 28

29. Copyright © 2009 Pearson Education, Inc. Next question – “at least” What is the probability of at least two Joe Flacco boxes? Binompdf measures “exactly” Binomcdf measures “at most” Notice that “at least 2” is the complement of “at most 1”. Then do binomcdf(5,0.2,1), but subtract from 1. Slide 1- 29

30. Copyright © 2009 Pearson Education, Inc. On the TI – “at least” 2.55 Operating System 1 – Old Operating System Slide 1- 30

31. Copyright © 2009 Pearson Education, Inc. Slide 1- 31 StatCrunch For StatCrunch, select “Stat”, then “Calculations”, then “Binomial”. Select “Probability”. Number of trials (n) is 5, probability of success (p) is 0.2. Input constant (ProbX) is 2. We want exactly two, so select “=” in the pull-down. Select “Go”.

32. Copyright © 2009 Pearson Education, Inc. Slide 1- 32

33. Copyright © 2009 Pearson Education, Inc. Slide 1- 33

34. Copyright © 2009 Pearson Education, Inc. What we get The answer for n = 2. A picture (and therefore an idea of the answer) for 0, 1, 3, 4, 5 Slide 1- 34

35. Copyright © 2009 Pearson Education, Inc. *Normal approximation to the Binomial Not required for the test or Final. If the Binomial involves a very large number of trials, the normal distribution will serve as a good approximation. This topic is not as important now as it was 30 years ago, when computers could not handle Binomial calculations with a large number of trials. See the “Optional” slides for details.

36. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll A poll was taken of 1040 Americans 18 and over on February 20-21, 2012. Question: “As you may know, two years ago, Congress passed a bill that restructures the nation’s health care system. All in all, do you think it was a good thing or a bad thing that Congress passed this bill?” Source: Gallup web site http://www.gallup.com/poll/152969/Americans- Divided-Repeal-2010-Healthcare-Law.aspx http://www.gallup.com/poll/152969/Americans- Divided-Repeal-2010-Healthcare-Law.aspx Slide 1- 36

37. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll

38. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll 45% support the passage of a health care reform bill. 55% are either neutral or think it was a bad thing. (Actually, 44% are against, the rest have no opinion. However, we need two and only two outcomes. Therefore, we will pool the “against” and the “neural.”) We’ll use the phrased “liked” as synonymous with “thought it was a good thing!” (It saves writing!) Slide 1- 38

39. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll Two outcomes: success (45%) and failure (55%). Again, we achieved this by combining “neutral” and “against.” Probability of success is constant (p = 0.45). We’ve sampled less than 10% of the US population. Slide 1- 39

40. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll Question: What is the probability that (exactly) 468 (45%) people in a survey of 1040 “liked” the passage of the health care reform bill? Binomial: binompdf(1040,0.45,468) = 0.0249. The probability of any one number coming up is small. We are more interested in “regions.” Slide 1- 40

41. Copyright © 2009 Pearson Education, Inc. Binomial Slide 1- 41

42. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll What is the probability that, using the results of our survey, out of 1040 in the survey, at most 500 liked the new health care plan. The “binomcdf” function measures “at most”. Binomcdf(1040,0.45,500) = 0.978464 How about “at most half”, i.e. 520. Binomcdf(1040,0.45,520) = 0.999452. Slide 1- 42

43. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll What is the probability that, using the results of our survey, out of 1040 in the survey, at least 500 liked the new health care plan. Two items we have to realize: The “binomcdf” measures “at most”. We therefore have to subtract what we get from 1 “At least 500” is the complement of “at most 499.” Binomial: 1 - binomcdf(1040,0.45,499) = 0.249. Slide 1- 43

44. Copyright © 2009 Pearson Education, Inc. **The Binomial Model with EXCEL What we want TIEXCEL ExactlyBinompdf(n,p,k)=BINOMDIST(k,n,p,0) At mostBinomcdf(n,p,k)=BINOMDIST(k,n,p,1) At least1-binomcdf(n,p,k-1)=1 - BINOMDIST(k-1,n,p,1) Smallest # of successes for n and p for a given probability of success (none in TI) BINOM.INV(n,p,level) = k Slide 1- 44

45. Copyright © 2009 Pearson Education, Inc. Stay tuned for Chapters 19 to 22!

46. Copyright © 2009 Pearson Education, Inc. Slide 1- 46 What Can Go Wrong? Be sure you have Bernoulli trials. You need two outcomes per trial, a constant probability of success, and independence. Remember that the 10% Condition provides a reasonable substitute for independence. Don’t confuse Geometric and Binomial models. Don’t use the Normal approximation with small n. You need at least 10 successes and 10 failures to use the Normal approximation.

47. Copyright © 2009 Pearson Education, Inc. Slide 1- 47 An Example – Binomial People with O-negative blood are called “universal donors” since they can give to anyone else. Only about 6% of the people have O-negative blood. (Denote O-) Question 1: If 20 people are tested, what is the expected number of universal donors (with standard deviation)? Question 2: What is the probability of either 2 or 3 universal donors?

48. Copyright © 2009 Pearson Education, Inc. An Example – Binomial Before we get to the questions, do we even have Bernoulli trials? Recall the three conditions: there are two possible outcomes (success and failure).In this case, success is finding an O-, failure is in not finding an O-. the probability of success, p, is constant. This is given as 0.06, which is constant since people have lined up as random. the trials are independent. We fall back on the 10% condition – less than 10% of the population came, so we can assume independence for all practical purposes. Slide 1- 48

49. Copyright © 2009 Pearson Education, Inc. Slide 1- 49 Our example – Question 1 Question 2: If 20 people are tested, what is the expected number of universal donors (with standard deviation)? This deals with the number of successes in 20 trials where p[success] = 0.06, which suggests the Binomial(20,0.06) distribution. E(X) = np = 20*.06 = 1.2 SD(X) = SQRT(npq) = SQRT(20 *.06 *.94) = 1.06. Conclusion: In groups of 20 randomly selected blood donors, we expect to find an average of 1.2 univeral donors with a standard deviation of 1.06.

50. Copyright © 2009 Pearson Education, Inc. Slide 1- 50 Our example – Question 2 Question 2: What is the probability of either 2 or 3 universal donors? (recall p = 0.06) We’re asking two questions. Out of 20, what is the probability of 2 successes? 3 successes? Then we add them. We can compute ( 20 C 2 (.06) 2 (.94) 18 )+( 20 C 3 (.06) 3 (.94) 17 ) Or use the TI or StatCrunch: TI: Binomialpdf(20,.06,2) = 0.2246 Binomialpdf(20,.06,3) = 0.0860 Sum is 0.3106..

51. Copyright © 2009 Pearson Education, Inc. StatCrunch – our example k = 2K = 3 Slide 1- 51

52. Copyright © 2009 Pearson Education, Inc. Slide 1- 52 Topics in this chapter Probability Models Bernoulli Trials The 10% Condition The Geometric Model The Binomial Model Approximating distributions with the Normal Success / Failure condition Assumptions and Conditions

53. Copyright © 2009 Pearson Education, Inc. Division of Mathematics, HCC Course Objectives for Chapter 17 After studying this chapter, the student will be able to: Tell if a situation involves Bernoulli trials. Know the appropriate conditions for using a Binomial or Normal model. Find the mean and standard deviation of a Binomial model. Calculate binomial probabilities, perhaps with a Normal model. Find the mean and standard deviation of a Geometric model. Calculate and interpret Geometric probabilities. Calculate and interpret Poisson probabilities.

54. Copyright © 2009 Pearson Education, Inc. *Optional Topics There are two other distributions that are not covered in this course. I have included them in the slide presentation in the event that you want to learn them on your own. They may be required in upper-level courses at four-year colleges. Geometric Distribution: Expected number of trials until the occurrence of an event.. Poisson Distribution: Approximates Binomial(n,p) when n is large and p is small (lottery, insurance). Finally, the Binomial can be approximated by the Normal distribution.

55. Copyright © 2009 Pearson Education, Inc. Slide 1- 55 *The Normal Model to the Rescue! When dealing with a large number of trials in a Binomial situation, making direct calculations of the probabilities becomes tedious (or outright impossible). Fortunately, the Normal model comes to the rescue…

56. Copyright © 2009 Pearson Education, Inc. Slide 1- 56 *The Normal Model to the Rescue (cont.) As long as the Success/Failure Condition holds, we can use the Normal model to approximate Binomial probabilities. Success/failure condition: A Binomial model is approximately Normal if we expect at least 10 successes and 10 failures: np ≥ 10 and nq ≥ 10. In the cereal box example, we did not even come close! We had only 5 trials (n = 5)!

57. Copyright © 2009 Pearson Education, Inc. Slide 1- 57 Continuous Random Variables When we use the Normal model to approximate the Binomial model, we are using a continuous random variable to approximate a discrete random variable. So, when we use the Normal model, we no longer calculate the probability that the random variable equals a particular value, but only that it lies between two values.

58. Copyright © 2009 Pearson Education, Inc. *Normal Approximation to the Binomial We use the same mean and standard deviation as for the binomial, i.e. We can work our binomial examples with the Normal approximation and get “approximately” the same answers. The larger the number of trials, the closer the approximation. Again, we should have np > 10 and nq > 10, i.e. at least 10 success and at least 10 failures.

59. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll A poll was taken of 1005 Americans 18 and over on March 22, 2010. Question: “As you may know, yesterday, the U.S. House of Representatives passed a bill that restructures the nation’s health care system. All in all, do you think it was a good thing or a bad thing that Congress passed this bill?” Source: Gallup web site http://www.gallup.com/poll/126929/Slim-Margin- Americans-Support-Healthcare-Bill-Passage.aspx http://www.gallup.com/poll/126929/Slim-Margin- Americans-Support-Healthcare-Bill-Passage.aspx Slide 1- 59

60. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll Two outcomes: success (49%) and failure (51%). Probability of success is constant (0.49). Success / Failure: np = 1005 * 0.49 = 492.45; nq = 1005 *0.51 = 512.55. Both are greater than 10! We’ve sampled less than 10% of the US population. Slide 1- 60

61. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll Question: What is the probability that (exactly) 500 people in a survey of 1005 “liked” the passage of the health care reform bill? Binomial: binompdf(1005,0.49,500) = 0.022464. Normal: µ = np 492.45; σ = SQRT(npq) = 15.8477 Normalpdf(500,492.45,15.8477) = 0.022472 Slide 1- 61

62. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll This was really close because it happened that we got 49%, which is close to the middle. Had it been, say 25%, the approximation would not have been as good (it still would have been OK, though.) We could also employ a “continuity correction”. Normalcdf(499.5,500.5,492.45, 15.8477) = 0.02247. Slide 1- 62

63. Copyright © 2009 Pearson Education, Inc. Normal with StatCrunch – two steps The normal calculation will not do 500. Do P(x < 500.5) =0.6931521 Do P(x < 499.5) = 0.67064846 Subtract to get 0.02250364 Slide 1- 63

64. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll What is the probability that, using the results of our survey, out of 1005 in the survey, at least 505 liked the new health care plan. Two items we have to realize: The “binomcdf” measures “at most”. We therefore have to subtract what we get from 1 “At least 505” is the complement of “at most 504.” Slide 1- 64

65. Copyright © 2009 Pearson Education, Inc. Another example: Gallup Poll What is the probability that, using the results of our survey, out of 1005, at least 505 liked a new health care plan. Binomial: 1 - binomcdf(1005,0.49,504) = 0.2235. Normal: Recall that µ = 492.45; σ =15.8477 504 is a little less then 1 σ away from µ, so we expect somewhat more than 0.16 by the 69-95-99.7 rule. normalcdf(505,99999,492.45,15.8477) = 0.2142 Slide 1- 65

66. Copyright © 2009 Pearson Education, Inc. Using StatCrunch BinomialNormal Slide 1- 66

67. Copyright © 2009 Pearson Education, Inc. Slide 1- 67 *Example – Cereal Boxes Recall the example of the cereal boxes. We’ll focus on Joe Flacco and the 20% of the boxes that contain his picture. Question: What is the expected number of boxes that we need to buy before a picture of Joe Flacco appears? (We did a simulation before.) This is different than the question that was answered by the Binomial distribution: What is the probability of getting x Joe Flacco pictures.

68. Copyright © 2009 Pearson Education, Inc. Slide 1- 68 *First Question What is the expected number of boxes to examinebefore a picture appears. We can look at There must be an easier way! See the pattern? Joe Flacco Probability 1(.2) 2.8*.2 3.8*.8*.2 4.8*.8*.8*.2 5.8*.8*.8*.8*.2 6(0.8) 5 *(0.2) 7(0.8) 6 *(0.2)

69. Copyright © 2009 Pearson Education, Inc. Slide 1- 69 *The Geometric Model A single Bernoulli trial is usually not all that interesting. A Geometric probability model tells us the probability for a random variable that counts the number of Bernoulli trials until the first success. In the Geometric probability model, our random variable X is the event of success on the x’th try (and not before). Note that we have to fail x - 1 times. First discussed by French mathematician Blaise Pascal (1623-1662). Geometric models are completely specified by one parameter, p, the probability of success, and are denoted Geom(p).

70. Copyright © 2009 Pearson Education, Inc. Slide 1- 70 *The Geometric Model (cont.) Geometric probability model for Bernoulli trials: Geom(p) p = probability of success q = 1 – p = probability of failure X = number of trials until the first success occurs P(X = x) = q x-1 p

71. Copyright © 2009 Pearson Education, Inc. *Geometric Model (contd.) Slide 1- 71 Source: Wikipedia http://en.wikipedia.org/wiki/File:Geometricpdf1.png

72. Copyright © 2009 Pearson Education, Inc. Slide 1- 72 *Cereal Boxes – Geometric distribution Number of boxes until first Joe Flacco p = 0.2, so q = 0.8. E(X) = 1/.2 = 5 Sd(x) = SQRT(.8/(.2) 2 ) = 4.47 We an expect to buy 5 boxes of cereal before a Joe Flacco picture appears. The standard deviation is 4.47. Now, what is the probability of getting Joe Flacco for the first time on the third try?

73. Copyright © 2009 Pearson Education, Inc. *The example solved, or Two out of three isn’t bad! By hand, P(X = 3) =.8*.8*.2 = 0.128 With the TI [DISTR]down to D Geometpdf(.2,3) = 0.128 Slide 1- 73

74. Copyright © 2009 Pearson Education, Inc. Slide 1- 74 *One more model – the Poisson In a recent year, there were 30,000 diagnoses of leukemia in the USA, which had a population of 280,000,000 at the time. Thus, if p is the probability of being diagnosed with leukemia, then p = 0.000011. Whyburn, MA had 8 leukemia cases out of a population of 35,000. If we use the binomial – try it with the TI or MINITAB! Can’t use the Normal approximation since np < 10

75. Copyright © 2009 Pearson Education, Inc. Slide 1- 75 *Poisson (optional) If we use the binomial, with the TI, do the following: (30000/280000000)[STO]P (store). This will give P the above value. Then binomcdf(35000,P,7) = 0.9623871303 Then use the Complement Rule to get 1 – [Ans] = 0.0376128697 We can’t use the Normal approximation because Np = 35000P = 3.75 < 10 Store Np as L (you’ll see why.)

76. Copyright © 2009 Pearson Education, Inc. Slide 1- 76 *The Poisson Model (optional) The Poisson probability model was originally derived in 1838 by French mathematician Simeon-Denis Poisson to approximate the Binomial model when the probability of success, p, is very small and the number of trials, n, is very large. The parameter for the Poisson model is λ. To approximate a Binomial model with a Poisson model, just make their means match: λ = np.

77. Copyright © 2009 Pearson Education, Inc. Slide 1- 77 *The Poisson Model (cont.)(optional) Poisson probability model for successes: Poisson(λ) λ = mean number of successes X = number of successes e is an important mathematical constant (approximately 2.71828)

78. Copyright © 2009 Pearson Education, Inc. *Poisson Distribution (optional) Slide 1- 78 Source: Wikipedia http://en.wikipedia.org/wiki/Poisson_distribution

79. Copyright © 2009 Pearson Education, Inc. Slide 1- 79 *The Poisson Model λ = mean number of successes = 35000 *.0000107142571 = 3.75 Text has 3.85 because of rounding. X = 8 e = 2.71828 So P(X = 8) = (3.75 8 )*exp(-3.75)/8! To get P(X < 7), poissoncdf(3.75,7) = 0.9623786576. The binomial is 0.9623871303. The Poisson can approximate the binomial. Because of advantages in computer technology, this is less important now, though.

80. Copyright © 2009 Pearson Education, Inc. Slide 1- 80 *The Poisson Model On the TI For exactly 8 cases, poissonpdf(0.375,8). Go to DISTR as before, then tab to poissonpdf. For 8 or more, poissoncdf(.375,7) will get you 0, 1, 2, …, 7. Then subtract from 1 (Complement Rule!)

81. Copyright © 2009 Pearson Education, Inc. Slide 1- 81 *The Poisson Model (cont.) Although it was originally an approximation to the Binomial, the Poisson model is also used directly to model the probability of the occurrence of events for a variety of phenomena. It’s a good model to consider whenever your data consist of counts of occurrences. It requires only that the events be independent and that the mean number of occurrences stays constant.

82. Copyright © 2009 Pearson Education, Inc. Slide 1- 82 *The Poisson is good at modJoeng Radioactive Decay Queuing (arrival of people in a line) such as a bank or checkout line Users logging on to the Internet Earthquakes Deaths from various causes

83. Copyright © 2009 Pearson Education, Inc. Slide 1- 83 What have we learned? Geometric model When we’re interested in the number of Bernoulli trials until the next success. Binomial model When we’re interested in the number of successes in a certain number of Bernoulli trials. Normal model To approximate a Binomial model when we expect at least 10 successes and 10 failures. Poisson model To approximate a Binomial model when the probability of success, p, is very small and the number of trials, n, is very large.