1 / 11

Lesson 6 - 3

Lesson 6 - 3. Poisson Probability Distribution. Objectives. Understand when a probability experiment follows a Poisson process Compute probabilities of a Poisson random variable Find the mean and standard deviation of a Poisson random variable. Vocabulary.

janna
Download Presentation

Lesson 6 - 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lesson 6 - 3 Poisson Probability Distribution

  2. Objectives • Understand when a probability experiment follows a Poisson process • Compute probabilities of a Poisson random variable • Find the mean and standard deviation of a Poisson random variable

  3. Vocabulary • Poisson process – used to computer probabilities of experiments in which the random variable X counts the number of occurrences (successes) of a particular event with in a specified interval (usually time or space) If we examine the binomial distribution as the number of trials gets larger and larger while the probability of success p gets smaller and smaller, we observe the Poisson Distribution (Simeon Denis Poisson). This distribution deals with the probabilities of rare events that occur infrequently in space, time, distance, area, volume, and so forth.

  4. Criteria for a Poisson Probability Experiment An experiment is said to be a Poisson experiment provided: • The probability of two or more successes in any sufficiently small subinterval* is 0 • The probability of success is the same for any two intervals of equal length • The number of successes in any interval is independent of the number of successes in any other interval provided the intervals are not overlapping * - for example, the fixed interval might be any time between 0 and 5 minutes. A subinterval could be any time between 1 and 2 minutes

  5. Poisson PDF If X is the number of successes in an interval of fixed length t, then probability formula for X is (λt)x P(x) = --------- e-λt x = 0, 1, 2, 3, … x! where λ (the Greek letter lamda) represents the average number of occurrences of the event in some interval of length 1 and e = 2.71828.... (Euler's constant) If X is the number of successes in an interval of fixed length and X follows a Poisson process with mean μ, the probability distribution function (PDF) for X is μx P(x) = --------- e-μ x = 0, 1, 2, 3, … x!

  6. Poisson PDF (cont) Mean (or Expected Value) and Standard Deviation : A random variable X that follows a Poisson process with parameter λ has a mean (or expected value) and standard deviation given by the formulas below: Mean: μx = λt Standard Deviation: σx = √λt = √μx where t is the length of the interval Note: At least probabilities must be computed using the Complement rule for Poisson probabilities Calculator: 2nd VARS poissonpdf(μ,x) poissoncdf(μ,x)

  7. Examples of a Poisson a) The number of accidents that occur per month (or week or day, etc) at a given intersection or in a manufacturing plant. b) The number of arrivals per minute (or hour, etc.) at a medical clinic or other servicing facility such as a garage, hospital, bank airport, etc. Number of cars arriving at a toll booth per day ( or hour or minute) c) The number of defects detected by quality control per day(per foot, per yard of material, per batch, etc.)

  8. Example 1 The number of accidents in an office building during a four week period is averages 2. What is the probability that there will be one or fewer accidents in the next four-week period? μ = 2 μx P(x) = --------- e-μ x! P(1 or fewer) = P(0) + P(1) 20 P(0) = --------- e-2 0! = e-2 = 0.13534 21 P(1) = --------- e-2 1! = 2e-2 = 0.27067 P(1 or fewer) = P(0) + P(1) = 0.13534 + 0.27067 = 0.40602

  9. Example 2 μx P(x) = --------- e-μ x! Daily demand for a certain replacement part for a VCR averages 0.9. a) What is the probability that no replacements parts will be needed on a particular day? b) That no more than three will be needed? μ = 0.9 (0.9)0 P(0) = --------- e-(0.9) 0! = e-0.9 = 0.4066 P(no more than 3) = P(x ≤ 3) = P(0) + P(1) + P(2) + P(3) = 0.9866 (0.9)2 P(2) = --------- e-(0.9) 2! (0.9)1 P(1) = --------- e-(0.9) 1! = 0.1647 = 0.3659 (0.9)3 P(3) = --------- e-(0.9) 3! = 0.0494

  10. Example 3 μx P(x) = --------- e-μ x! The number of calls to a police department between 8pm and 8:30pm on Friday averages 3.5. a) What is the probability of no calls during this period? b) Is it likely that the police will get 7 calls? c) What is the mean and standard deviation of the number of calls? μ = 3.5 3.50 P(0) = --------- e-3.5 0! = 0.0302 3.57 P(7) = --------- e-3.5 7! = 0.0386 < 5% so not likely! σ = 3.5 = 1.8708 μ = 3.5

  11. Summary and Homework • Summary • Poisson process has 3 criteria • Used for occurrences over time or space • Calculator has pdf and cdf functions • Homework • pg 348 - 350: 3, 6, 9, 10, 14, 17

More Related