Chapter Five. The Binomial Probability Distribution and Related Topics

Transcription

1 Chapter Five The Binomial Probability Distribution and Related Topics

2 Section 4 The Geometric and Poisson Probability Distributions

3 Essential Questions How are Geometric and Poisson probability distributions different from the binomial probability distribution? How are they the same?

4 Student Objectives The student will determine the probability of an event using the geometric probability distribution. The student will determine the probability of an event using the Poisson probability distribution. The student will use the Poisson probability distribution to approximate the probability of a binomial experiment. The student will determine the probability of an event using the appropriate commands on their calculator. The student will determine the probability of an event using the appropriate formula. The student will determine the probability of an event using the appropriate chart.

5 Key Terms Geometric probability distribution Negative binomial distribution Poisson probability distribution

6 Geometric Probability Distribution Geomeric Probability Distribution P( n) = p( 1 p) n 1 where n is the number of binomial trials on which the first success occurs (n = 1, 2, 3,...) and p is the probability of success on each trial. Notice: p is the same for each trial. Using some mathematics involving infinite series, it can be shown that the population mean and standard deviation of the geometric distribution are: µ= 1 p and σ = 1 p p = q p

7 Geometric Probability Distribution Geomeric Probability Distribution P( n) = p( 1 p) n 1 Calculator commands or Single probability 2nd DISTR (Option E) ( ) = geometpdf ( p, r) P r Cummulative probability ( 1, 2, 3,..., r) 2nd DISTR (Option F) ( ) = geometcdf ( p, r) P r

8 Geometric Probability Distribution A student must come to the front of the room and correctly answer a randomly chosen question before returning to their seat. The probability of answering the questions has been determined to be (a) What is the mean number of questions that will need to be answered to sit down? (b) What is the standard deviation? (c) What is the probability of having to answer exactly 3 questions before getting to sit down? (Formula and Calculator) (d) What is the probability of being able to sit down by having to answer at most the 5th question? (Calculator)

9 Geometric Probability Distribution A student must come to the front of the room and correctly answer a randomly chosen question before returning to their seat. The probability of answering the questions has been determined to be (a) What is the mean number of questions that will need to be answered to sit down? µ = 1 p µ = µ =

10 Geometric Probability Distribution A student must come to the front of the room and correctly answer a randomly chosen question before returning to their seat. The probability of answering the questions has been determined to be (b) What is the standard deviation? σ = q p σ = σ = σ =

11 Geometric Probability Distribution A student must come to the front of the room and correctly answer a randomly chosen question before returning to their seat. The probability of answering the questions has been determined to be (c) What is the probability of having to answer exactly 3 questions before getting to sit down? (Formula and Calculator) ( ) = p( q) r 1 P r P( 3) ( 0.75) ( 0.25) 2 ( 0.75) ( ) P( r) = geometpdf ( p, r) P( 3) geometpdf ( 0.75, 3)

12 Geometric Probability Distribution A student must come to the front of the room and correctly answer a randomly chosen question before returning to their seat. The probability of answering the questions has been determined to be (d) What is the probability of being able to sit down by having to answer at most the 5th question? (Calculator) P( r k) = geometpdf ( p, k) P( r 5) geometcdf ( 0.75, 5)

13 Sample Question You are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. (a) (b) (c) (d) What is the probability of winning the first time you play the game? The second time? The third time? How many games would you need to play to be 95% certain that you would win?

14 Sample Question You are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. 1 st game (a) What is the probability of winning the first time you play the game? ( ) P ( )

15 Sample Question You are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. 2 nd game (b) The second time? ( ) P

16 Sample Question You are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. 3 rd game (c) The third time? ( ) P

17 Sample Question You are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. (d) How many games would you need to play to be 95% certain that you would win? P( r 8) geometcdf ( 1/ 3, 8)

18 Sample Question You are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. (a) (b) What is the mean number of games required to win the game? What is the standard deviation?

19 Sample Question You are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. (a) What is the mean number of games required to win the game? µ = 1 p µ = µ = 3

20 Sample Question You are playing a carnival game where the winning coin is under one of 3 cups placed upside-down on a table. The cups are moved in a scrambled pattern so you cannot determine where the winning coin is at. Use the formula to answer the first three questions and the calculator to answer last answer. (b) What is the standard deviation? σ = σ = q p σ = ( ) σ = σ =

21 Poisson Probability Distribution Poisson Probability Distribution Let λ (Greek letter lambda) be the mean number of success over time, volume, area, and so forth. Let r, be the number of successes (r = 0, 1, 2, 3,...) in a corresponding interval of time, volume, area, and so forth. Then the probability of r successes in the interval is P( r) = e λ λ r r! where e is approximately equal to Using some mathematics involving infinite series, it can be shown that the population mean and standard deviation of the Poisson distribution are: µ = λ and σ = λ

22 Poisson Probability Distribution Poisson Probability Distribution ( ( ) = e λ λ) k P r = k k! Calculator commands or Single probability 2nd DISTR (Option C) ( ) = poissonpdf ( λ, k) P r = k Cummulative probability ( 0, 1, 2, 3,..., k) 2nd DISTR (Option D) ( ) = poissoncdf ( p, k) P r k

23 Poisson Probability Distribution The typical teenager sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days. (a) What is the mean number of text messages? (b) What is the standard deviation? (c) What is the probability of sending 10 text messages during this time period? (Formula) (d) What is the probability of sending 15 text messages during this time period? (Calculator) (e) What is the probability of sending 12 text messages during this time period? (Chart - Round the value of lambda to the nearest 10 th ) (f) What is the probability of sending at least 5 text messages? (Calculator)

24 Poisson Probability Distribution The typical teenager sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days. (a) What is the mean number of text messages? λ = 3339 messages 1 month λ = messages 2-hours λ = 9.1 messages 2-hours 1 month 30.5 days 1 day 24 hours 2 hours 2 hours

25 Poisson Probability Distribution The typical teenager sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days. (b) What is the standard deviation? σ = λ σ = σ =

26 Poisson Probability Distribution The typical teenager sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days. (c) What is the probability of sending 10 text messages during this time period? (Formula) P( r = 10) ( e )( ) 10 10! ( )( )

27 Poisson Probability Distribution The typical teenager sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days. (d) What is the probability of sending 15 text messages during this time period? (Calculator) Use the stored value (exact value) for lambda when doing the calculations on the calculator or by formula! P( r = 15) poissonpdf ( , 15)

28 Poisson Probability Distribution The typical teenager sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days. (e) What is the probability of sending 12 text messages during this time period? (Chart - Round the value of lambda to the nearest 10 th ) λ = 9.1 ( ) P r =

29 Poisson Probability Distribution The typical teenager sends 3,339 text messages per month. We plan to study the number of text messages during an 2-hour period of a day. For this study we will assume a month consists of 30.5 days. (f) What is the probability of sending at least 5 text messages? (Calculator) P( r 5) 1 P( r 4) 1 poissoncdf ( , 4)

30 Sample Question It has been determined that approximately 14 cars run a red-light in downtown Annville everyday. The police have decided to place a patrol car in the area for a 10-hour period. (a) Round your value of lambda to the nearest tenth and use the chart to determine the probability of 0, 1, and 2 cars running the red light. (b) Use the formula and the exact value of lambda to determine the probability of 5 cars running the red light. (c) Use the calculator to determine the probability of 4 cars running the red light. (d) Use the calculator to determine the probability that more than 7 cars will run the red light.

31 Sample Question It has been determined that approximately 14 cars run a red-light in downtown Annville everyday. The police have decided to place a patrol car in the area for a 10-hour period. λ = (a) Round your value of lambda to the nearest tenth and use the chart to determine the probability of 0, 1, and 2 cars running the red light. 14 cars 1 day 1 day 24 hours λ = cars shift λ = 5.8 cars shift 10 hours 1 shift P( r = 0) P( r = 2) P( r = 1)

32 Sample Question It has been determined that approximately 14 cars run a red-light in downtown Annville everyday. The police have decided to place a patrol car in the area for a 10-hour period. (b) Use the formula and the exact value of lambda to determine the probability of 5 cars running the red light. P( r = 5) ( e )( ) 5 5! ( ) ( )

33 Sample Question It has been determined that approximately 14 cars run a red-light in downtown Annville everyday. The police have decided to place a patrol car in the area for a 10-hour period. (c) Use the calculator to determine the probability of 4 cars running the red light. ( ) P r = 4 ( ) poissonpdf ,

34 Sample Question It has been determined that approximately 14 cars run a red-light in downtown Annville everyday. The police have decided to place a patrol car in the area for a 10-hour period. (d) Use the calculator to determine the probability that 7 or more cars will run the red light. P( r 7) 1 P( r 6) 1 poissoncdf ( , 6)

35 Binomial vs Poisson How to approximate Binomial Probabilities using Poisson Probabilities Suppose you have a binomial distribution with n = number of trials p = probability of succes in each trial, and r = the number of success If n 100 and np < 10, then r has a binomial distribution that is approximated by a Poisson distribution with λ = np. P r ( ) = e λ λ r r! Note: λ = np is the expected value of the binomial distribution.

36 Binomial vs Poisson The population of Cleona borough is approximated at 2,091. Roughly 0.29% of the population owns some type of exotic wildlife. (a) Why is the Poisson probability distribution a good approximation of the binomial distribution? (b) What is the value of lambda? (c) What is the probability that 2 people will own some exotic wildlife? (Formula - Binomial and Poisson) (d) What is the probability of having at least 7 people owning some type of exotic wildlife? (Calculator - Binomial and Poisson)

37 Binomial vs Poisson The population of Cleona borough is approximated at 2,091. Roughly 0.29% of the population owns some type of exotic wildlife. (a) Why is the Poisson probability distribution a good approximation of the binomial distribution? n = 2091 Since n 100, the population is large enough to use the Poisson distribution to approximate the binomial probability. np = 2091( ) np = Since np < 10, the expected value of the binomial distribution is small enoughto use the Poisson distribution to approximate the binomial probability. Yes, the Poisson distribution is a good approximation of the binomial distribution since n 100 and np < 10.

38 Binomial vs Poisson The population of Cleona borough is approximated at 2,091. Roughly 0.29% of the population owns some type of exotic wildlife. (b) What is the value of lambda? ( ) np = np = λ =

39 Binomial vs Poisson The population of Cleona borough is approximated at 2,091. Roughly 0.29% of the population owns some type of exotic wildlife. (c) What is the probability that 2 people will own some exotic wildlife? (Formula - Binomial and Poisson) Binomial vs. Poisson P r = 2 ( ) P( r = 2) e 2091C 2 ( ) 2 ( ) ( 2091) ( ) ( )( 1) ( ) ( ) ( ) ( ) 2 2! ( ( ) ) ( ) ( ) 2 ( ) ( )( )

40 Binomial vs Poisson The population of Cleona borough is approximated at 2,091. Roughly 0.29% of the population owns some type of exotic wildlife. (d) What is the probability of having at least 7 people owning some type of exotic wildlife? (Calculator - Binomial and Poisson) Binomial vs. Poisson P r 7 ( ) P( r 7) 1 P( r 6) 1 P( r 6) 1 binomcdf ( 2091, , 2) 1 poissoncdf ( , 6)

41 Sample Question In a recent study of the high school graduates it has been determined that the probability that a graduate will be in serious trouble with the law within 5 years of the graduate is There are 144 students in the graduating class. (a) Is the Poisson distribution a good approximation of the binomial distribution? Explain. (b) Use the formula for binomial probability to determine the probability that 4 students will get in serious trouble with the law within 5 years of graduation. Repeat with the Poisson formula. (c) Use your calculator to determine the probability that at least 10 students will get in serious trouble with the law. Repeat with the Poisson formulas. Are the answers similar?

42 Sample Question In a recent study of the high school graduates it has been determined that the probability that a graduate will be in serious trouble with the law within 5 years of the graduate is There are 144 students in the graduating class. (a) Is the Poisson distribution a good approximation of the binomial distribution? Explain. n = 144 Since n 100, the population is large enough to use the Poisson distribution to approximate the binomial probability. np = 144( 0.025) np = 3.6 Since np < 10, the expected value of the binomial distribution is small enoughto use the Poisson Yes, the Poisson distribution is a distribution to approximate the binomial probability. good approximation of the binomial distribution since n 100 and np < 10.

43 Sample Question In a recent study of the high school graduates it has been determined that the probability that a graduate will be in serious trouble with the law within 5 years of the graduate is There are 144 students in the graduating class. (b) Use the formula for binomial probability to determine the probability that 4 students will get in serious trouble with the law within 5 years of graduation. Repeat with the Poisson formula. P 4 P( 4) ( ) C ( ( ) ) ( ) 4 ( 0.975) 140 ( 0.025) 4 ( 0.975) 140 ( )( ) ( ) e 3.6 ( 3.6) 4 4! ( ) ( )

44 Sample Question In a recent study of the high school graduates it has been determined that the probability that a graduate will be in serious trouble with the law within 5 years of the graduate is There are 144 students in the graduating class. (c) Use your calculator to determine the probability that at least 10 students will get in serious trouble with the law. Repeat with the Poisson formulas. Are the answers similar? P( r 10) 1 P( r 9) 1 binomcdf ( 144, , 9) ( ) P r 10 1 P( r 9) 1 poissoncdf ( 3.6, 9) Yes, the answers only differ by

45 Negative Binomial Probability Distribution Negative Binomial Distribution Let k 1 be a fixed whole number. The probability that the kth success occurs on trial n is where P( n) = n 1 C k 1 ( p) k ( q) n k n 1 C k 1 = ( n 1)! ( k 1)! ( n k)! n = k, k +1, k + 2,... The expected value and standard deviation of the geometric distribution are: µ = k p and σ = kq p Note: if k = 1, the negative binomial distribution is called the geometric distribution. * Special Note: Fix the formula on page 238. *

46 Negative Binomial Probability Distribution Dylan can leave the final examination when he is finished with the examination. Dylan decided that he needed only to answer 5 questions correct in order to pass the course. There were a total of 100 questions on the final examination. The probability that he can correctly answer any of the questions is (a) (b) (c) What is the mean number of questions that will need to be answered to pass the course? What is the standard deviation? What is the probability that he answers 5 correct in the first 10 questions?

47 Negative Binomial Probability Distribution Dylan can leave the final examination when he is finished with the examination. Dylan decided that he needed only to answer 5 questions correct in order to pass the course. There were a total of 100 questions on the final examination. The probability that he can correctly answer any of the questions is (a) What is the mean number of questions that will need to be answered to pass the course? µ = µ =

48 Negative Binomial Probability Distribution Dylan can leave the final examination when he is finished with the examination. Dylan decided that he needed only to answer 5 questions correct in order to pass the course. There were a total of 100 questions on the final examination. The probability that he can correctly answer any of the questions is (b) What is the standard deviation? σ = ( 5) ( 0.35) 0.65 σ = σ = σ =

49 Negative Binomial Probability Distribution Dylan can leave the final examination when he is finished with the examination. Dylan decided that he needed only to answer 5 questions correct in order to pass the course. There were a total of 100 questions on the final examination. The probability that he can correctly answer any of the questions is (c) What is the probability that he answers 5 correct in the first 10 questions? P( 5 out of 10) 9C 4 ( 0.65) 5 ( 0.35) 5 ( 9) ( 8) ( 7) ( 6) ( 4) ( 3) ( 2) ( 1) ( 0.65 ) 5 ( 0.35) ( 0.65) 5 ( 0.35) 5 ( )( ) ( 126)

50 Negative Binomial Probability and Geometric Distribution The geometric distribution has only one success and it must occur on the last trial. The negative binomial distribution may have multiple successes but the last trial must be a success. Negative Binomial Distribution vs Geometric Distribution k number of successes 1 k 1 successes prior to the last trial 0 n number of trials n p probability of a success p k p kq p µ σ n 1C k 1 p k q n k Probability formula p 1 q n 1 1 p q p

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