JMB Chapter 5 Part 2 EGR Spring 2011 Slide 1 Multinomial Experiments  What if there are more than 2 possible outcomes? (e.g., acceptable, scrap, rework)  That is, suppose we have: n independent trials k outcomes that are  mutually exclusive (e.g., ♠, ♣, ♥, ♦)  exhaustive (i.e., ∑ all k p i = 1)  Then f(x 1, x 2, …, x k ; p 1, p 2, …, p k, n) =

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 2 Problem 22, pg. 152  Convert ratio 8:4:4 to probabilities  f( __, __, __; ___, ___, ___, __) =_______ f( 5,2,1; 0.5, 0.25, 0.25, 8) = (8 choose 5,2,1)(0.5) 5 (0.25) 2 (0.25) 1 = 8!/(5!2!1!)* )(0.5) 5 (0.25) 2 (0.25) 1 = 21/256 or x 1 =_______p 1 =0.50 x 2 =_______p 2 =0.25n = _____ x 3 =_______p 3 =0.25

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 3 Binomial vs.Hypergeometric Distribution  Replacement and Independence Binomial (assumes sampling “with replacement”) and hypergeometric (sampling “without replacement”) Binomial assumes independence, while hypergeometric does not.  Hypergeometric: The probability associated with getting x successes in the sample (given k successes in the lot.)

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 4  Example from Complete Business Statistics, 4th ed (McGraw-Hill) Automobiles arrive in a dealership in lots of 10. Five out of each 10 are inspected. For one lot, it is known that 2 out of 10 do not meet prescribed safety standards.What is probability that at least 1 out of the 5 tested from that lot will be found not meeting safety standards?  This example follows a hypergeometric distribution: A random sample of size n is selected without replacement from N items. k of the N items may be classified as “successes” and N-k are “failures.”  The probability associated with getting x successes in the sample (given k successes in the lot.) Hypergeometric Example

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 5 Solution: Hypergeometric Example  In our example, k = number of “successes” = 2 n = number in sample = 5 N = the lot size = 10 x = number found = 1 or 2 P(X > 1) = = 0.778

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 6 Expectations: Hypergeometric Distribution  The mean and variance of the hypergeometric distribution are given by  What are the expected number of cars that fail inspection in our example? What is the standard deviation? μ = nk/N = 5*2/10 = 1 σ 2 = (5/9)(5*2/10)(1-2/10) = σ = 0.667

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 7 Your turn … A worn machine tool produced defective parts for a period of time before the problem was discovered. Normal sampling of each lot of 20 parts involves testing 6 parts and rejecting the lot if 2 or more are defective. If a lot from the worn tool contains 3 defective parts: 1.What is the expected number of defective parts in a sample of six from the lot? N = 20 n = 6 k = 3 μ = nk/N = 6*3/20 =18/20=0.9 2.What is the expected variance? σ 2 = (14/19)(6*3/20)(1-3/20) = What is the probability that the lot will be rejected? P(X>2) = 1 – [P(0)+P(1)]

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 8 Binomial Approximation  Note, if N >> n, then we can approximate the hypergeometric with the binomial distribution.  Example: Automobiles arrive in a dealership in lots of out of each 100 are inspected. 2 /10 (p=0.2) are indeed below safety standards. What is probability that at least 1 out of 5 will be found not meeting safety standards?  Recall: P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0) Hypergeometric distributionBinomial distribution 1 - h(0;100,5,20) = b(0;5,0.2) = (Compare to example 5.14, pg. 129)

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 9 Negative Binomial Distribution b*  A binomial experiment in which trials are repeated until a fixed number of successes occur.  Example: Historical data indicates that 30% of all bits transmitted through a digital transmission channel are received in error. An engineer is running an experiment to try to classify these errors, and will start by gathering data on the first 10 errors encountered. What is the probability that the 10 th error will occur on the 25 th trial?

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 10  This example follows a negative binomial distribution: Repeated independent trials. Probability of success = p and probability of failure = q = 1-p. Random variable, X, is the number of the trial on which the k th success occurs.  The probability associated with the k th success occurring on trial x is given by, Where, k = “success number” = 10 x = trial number on which k occurs = 25 p = probability of success (error) = 0.3 q = 1 – p = 0.7 Negative Binomial Example

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 11 Negative Binomial Distribution  In our example, k = “success number” = 10 x = trial number on which k occurs = 25 p = probability of success (error) = 0.3 q = 1 – p = 0.7

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 12 Geometric Distribution  Example: In our example, what is the probability that the 1 st bit received in error will occur on the 5 th trial?  This is an example of the geometric distribution, which is a special case of the negative binomial in which k = 1. The probability associated with the 1 st success occurring on trial x is given by = (0.3)(0.7) 4 = 0.072

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 13 Your turn … A worn machine tool produces 1% defective parts. If we assume that parts produced are independent: 1.What is the probability that the 2nd defective part will be the 6th one produced? 2.What is the probability that the 1st defective part will be seen before 3 are produced? 3.How many parts can we expect to produce before we see the 1st defective part? (Hint: see Theorem 5.4, pg. 161)

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 14 Poisson Process  The number of occurrences in a given interval or region with the following properties: “memoryless” ie number in one interval is independent of the number in a different interval P(occurrence) during a very short interval or small region is proportional to the size of the interval and doesn’t depend on number occurring outside the region or interval. P(X>1) in a very short interval is negligible

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 15 Poisson Process Situations  Number of bits transmitted per minute.  Number of calls to customer service in an hour.  Number of bacteria present in a given sample.  Number of hurricanes per year in a given region.

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 16 Service Call Example - Poisson Process  Example An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. To determine personnel and equipment needs to maintain a desired level of service, the plant manager needs to be able to determine the probabilities associated with numbers of service calls.

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 17 Poisson Distribution Probabilities  The probability associated with the number of occurrences in a given period of time is given by, Where, λ = average number of outcomes per unit time or region t = time interval or region

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 18 Our Example: λ = 2.7 and t = 1 minute  What is the probability that fewer than 2 calls will be received in any given minute?  The probability that fewer than 2 calls will be received in any given minute is P(X < 2) = P(X = 0) + P(X = 1)  The mean and variance are both λt, so μ = λt =________________  Note: Table A.2, pp , gives Σ t p(x;μ)

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 19 Service Call Example - Part 2  If more than 6 calls are received in a 3-minute period, an extra service technician will be needed to maintain the desired level of service. What is the probability of that happening? μ = λt = (2.7) (3)= is not in the table; use basic equation  Suppose λt = 8; see table with μ = 8 and r = 6 P(X > 6) = 1 – P(X < 6) = =

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 20 Poisson Distribution

JMB Chapter 5 Part 2 EGR Spring 2011 Slide 21 Poisson Distribution The effect of λ on the Poisson distribution