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Statistics and Modelling Course

Statistics and Modelling Course. 2011. Topic 5: Probability Distributions. Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward problems 4 Credits Externally Assessed NuLake Pages 278  322. Binomial Distribution revision question: NCEA 2009 :

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Statistics and Modelling Course

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  1. Statistics and Modelling Course 2011

  2. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward problems 4 Credits Externally Assessed NuLake Pages 278  322

  3. Binomial Distribution revision question: NCEA 2009: Tom and Tane are surfers. Tom successfully rides a wave on average 7 out of every 10 attempts, and Tane 6 out of every 10. Calculate the probability that Tom & Tane each ride 2 of the next 5 waves. What assumptions must be made to use the Binomial Distn here?

  4. Lesson 1: Intro to the Poisson Distribution. Learning outcomes: • Learn about what the Poisson Distribution is and its parameter, l. • Learn the 4 requirements for the use of the Poisson Distribution. • Calculate probabilities using the Poisson Distribution. • Warm-up: 2009 NCEA q. on the Binomial Distn. • The Poisson Distribution: Notes & examples using the formula & tables (most of lesson). HW: Sigma - Old edition: Ex. 6.1 or New: Ex. 16.01

  5. The Poisson Distribution Deals with rate of occurrence in an interval. We use the Poisson Distribution when counting the numberof occurrences of a discrete event over a given continuous interval (time/space). • Number of earthquakes peryear (excluding after-shocks). • Population density - Number of people persquare km. • House burglaries reported perweekin Christchurch.

  6. The Poisson Distribution Deals with rate of occurrence in an interval. We use the Poisson Distribution when counting the numberof occurrences of a discrete event over a given continuous interval (time/space). • Number of earthquakes peryear. • Population density - Number of people persquare km. • House burglaries reported perweek in Christchurch. The Poisson Distribution is also known as the Distribution of Rare Events.

  7. We use the Poisson Distribution when counting the numberof occurrences of a discrete event over a given continuous interval (time/space). • Number of earthquakes peryear. • Population density - Number of people persquare km. • House burglaries reported perweek in Christchurch. The Poisson Distribution is also known as the Distribution of Rare Events. You’ll be using the Poisson Distribution to calculate the probability of an event occurring x times in a given interval. You will always be given the average number of occurrences,l, per interval. lis the Greek symbol “Lambda”

  8. Population density - Number of people per square km. • House burglaries reported per weekin Christchurch. The Poisson Distribution is also known as the Distribution of Rare Events. You’ll be using the Poisson Distribution to calculate the probability of an event occurring x times in a given interval. You will always be given the average number of occurrences,l, per interval. l is the Greek symbol “Lambda” l = Mean rate of occurrence.

  9. You’ll be using Poisson Distribution to calculate the probability of an event occurring x times in a given interval. You will always be given the average number of occurrences,l, per interval. The Poisson Distribution is also known as the Distribution of Rare Events. The 4 requirements for a Poisson Distribution are (RIPS): (1.) Random: Events occur at random (unpredictable). (2.) Independent: Each event occurs completely independently of any others. • l is the Greek symbol “Lambda” • = Mean rate of occurrence.

  10. You’ll be using the Poisson Distribution to calculate the probability of an event occurring x times in a given interval. You will always be given the average number of occurrences,l, per interval. The 4 requirements for a Poisson Distribution are (RIPS): (1.) Random: Events occur at random (unpredictable). (2.) Independent: Each event occurs completely independently of any others. (3.) Proportional: The probability of an event occurring is proportional to the size of the interval. (4.) Simultaneous: Events CANNOT occur simultaneously (at the same time or in exactly the same spot). • l is the Greek symbol “Lambda” • = Mean rate of occurrence.

  11. l is the Greek symbol “Lambda” • = Mean rate of occurrence. The 4 requirements for a Poisson Distribution are (RIPS): (1.) Random: Events occur at random (unpredictable). (2.) Independent: Each event occurs completely independently of any others. (3.) Proportional: The probability of an event occurring is proportional to the size of the interval. (4.) Simultaneous: Events CANNOT occur simultaneously (at the same time or in exactly the same spot). • We define the Poisson variable, X, by the formula: • P(X = x) = e-llx • x!

  12. (1.) Random: Events occur at random (unpredictable). (2.) Independent: Each event occurs completely independently of any others. (3.) Proportional: The probability of an event occurring is proportional to the size of the interval. (4.) Simultaneous: Events cannot occur simultaneously (at the same time or in exactly the same spot). • We define the Poisson variable, X, by the formula: • P(X = x) = e-llx • x! Eg: The average number of accidents at a particular intersection every year is 18. (a) Calculate the probability that there are exactly 2 accidents there this month.

  13. Eg: The average number of accidents at a particular intersection every year is 18. (a) Calculate the probability that there are exactly 2 accidents there this month. There are 12 months in a year, so l = ? = 1.5accidents per month P(X = 2) = = 0.2510(4sf) answer (b) Calculate the probability that there is at least one accident this month.

  14. There are 12 months in a year, so l = = 1.5accidents per month P(X = 3) = = 0.2510(4sf) answer (b) Calculate the probability that there is at least one accident this month. P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So…

  15. There are 12 months in a year, so l = = 1.5accidents per month P(X = 3) = = 0.2510(4sf) answer (b) Calculate the probability that there is at least one accident this month. P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0)

  16. P(X = 3) = = 0.2510(4sf) answer (b) Calculate the probability that there is at least one accident this month. P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0)

  17. = 0.2510 (4sf) answer (b) Calculate the probability that there is at least one accident this month. P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0) = 0.223130…

  18. (b) Calculate the probability that there is at least one accident this month. P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0) = 0.223130… So P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer

  19. P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. So… Take the complement: P(X=0) = 0.223130… So P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer (c) What is the probability that there are more than 2 accidents in a particular month

  20. = 0.223130… So P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer (c) What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 P(X > 2) = 1 – P(X< 2)

  21. So P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer (c) What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ _____ + ______ + ______]

  22. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ _____ + ______ + ______]

  23. If l = 1.5, P(X = 0) = ?

  24. If l = 1.5, P(X = 0) = 0.2231

  25. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ _____ + ______ + ______]

  26. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ 0.2231 + ______ + ______]

  27. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ 0.2231 + ______ + ______]

  28. If l = 1.5, P(X = 0) = 0.2231

  29. If l = 1.5, P(X = 0) = 0.2231 + P(X = 1) = ?

  30. If l = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347

  31. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ 0.2231 + ______ + ______]

  32. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + ______]

  33. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + ______]

  34. If l = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347

  35. If l = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347 + P(X = 2) = ?

  36. If l = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347 + P(X = 2) = 0.2510

  37. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + ______]

  38. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + 0.2510]

  39. If l = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347 + P(X = 2) = 0.2510 P(X<2) =

  40. If l = 1.5, P(X = 0) = 0.2231 + P(X = 1) = 0.3347 + P(X = 2) = 0.2510 P(X<2) = 0.8088

  41. = 1 – 0.223130… = 0.7769 (4sf) answer c What is the probability that there are more than 2 accidents in a particular month l = 1.5 There are an infinite number of cases so instead consider X< 2 b P(X > 2) = 1 – P(X< 2) = 1 – [ P(X =0) + P(X=1) + P(X=2)] Look up Poisson Tables = 1 – [ 0.2231 + 0.3347 + 0.2510] HW: Sigma: Old edition: Ex. 6.1 or New edition: Ex. 16.01 = 1 – 0.8088 = 0.1912

  42. Lesson 2: Poisson probabilities using a Graphics calculator Learning outcome: Practice calculating probabilities of combined events using the Poisson Distribution, using both tables and GC. Work: • HW qs. • Problem on board – how to use your GC. • Do Sigma (old – 2nded): p84 – Ex. 6.2 or in new version: p340 – Ex. 16.02

  43. Poisson probabilities:Tables vs Graphics Calculator E.g. Calculate P (X>3) for l=4 USING TABLES. Do not copy this. P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)]

  44. Poisson probabilities:Tables vs Graphics Calculator E.g. Calculate P (X>3) for l=4 USING TABLES. Do not copy this. P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [0.018315 + 0.073262 + 0.14652 + 0.19536]

  45. Poisson probabilities:Tables vs Graphics Calculator E.g. Calculate P (X>3) for l=4 USING TABLES. Do not copy this. P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [0.018315 + 0.073262 + 0.14652 + 0.19536] = 1 – 0.433457 = 0.5665(4sf)

  46. Poisson probabilities:Tables vs Graphics Calculator E.g. Calculate P (X>3) for l=4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [0.018315 + 0.073262 + 0.14652 + 0.19536] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… • On your Graphics Calculator: (copy)

  47. Poisson probabilities:Tables vs Graphics Calculator E.g. Calculate P (X>3) for l=4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [0.018315 + 0.073262 + 0.14652 + 0.19536] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… • On your Graphics Calculator: (copy) • STAT, DIST (same as before)

  48. Poisson probabilities:Tables vs Graphics Calculator E.g. Calculate P (X>3) for l=4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [0.018315 + 0.073262 + 0.14652 + 0.19536] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… • On your Graphics Calculator: (copy) • STAT, DIST (same as before) • POISN (F6, F1) (same as before)

  49. Poisson probabilities:Tables vs Graphics Calculator E.g. Calculate P (X>3) for l=4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [0.018315 + 0.073262 + 0.14652 + 0.19536] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… • On your Graphics Calculator: (copy) • STAT, DIST (same as before) • POISN (F6, F1) (same as before) • Pcd (c for Cumulative probabilities – more than 1 value)

  50. Poisson probabilities:Tables vs Graphics Calculator E.g. Calculate P (X>3) for l=4. Using tables (do not copy). P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 – [0.018315 + 0.073262 + 0.14652 + 0.19536] = 1 – 0.433457 = 0.5665(4sf) BUT… there is a much faster way… • On your Graphics Calculator: (copy) • STAT, DIST (same as before) • POISN (F6, F1) (same as before) • Pcd (c for Cumulative probabilities – more than 1 value) • Data: Variable (as before)

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