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**IMSE 213: Probability and Statistics for Engineers**

IMSE 213 Prob&Statistics 4/1/2017 IMSE 213: Probability and Statistics for Engineers Chapter 5: Some Discrete Probability Distributions Instructor: Steven E. Guffey, PhD, CIH © © 2002 Steven E. Guffey, PhD, CIH

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**Introduction Will cover most common discrete probability distributions**

IMSE 213 Prob&Statistics 4/1/2017 Introduction Will cover most common discrete probability distributions Simplest distribution: Equal probability distribution I.e., all events equally likely If there are k values with equal probability, then: Examples: Toin toss Random selection from a group © 2002 Steven E. Guffey, PhD, CIH

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**Theorem 5.1: Mean and Variance of Discrete Uniform Distribution**

IMSE 213 Prob&Statistics 4/1/2017 Theorem 5.1: Mean and Variance of Discrete Uniform Distribution f(i) = n/N = 1/k © 2002 Steven E. Guffey, PhD, CIH

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**5.3 Binomial Distribution**

IMSE 213 Prob&Statistics 4/1/2017 5.3 Binomial Distribution Bernoulli Process: Experiment consists of n repeated trials Each trial outcome can be classified as “success” or “failure” (i.e., only 2 possible outcomes) Probability of success, p, remains constant from trial to trial The repeated trials are independent Number of successes, X, is called a binomial random variable b(x; n, p) Its distribution is called the binomial distribution © 2002 Steven E. Guffey, PhD, CIH

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**Binomial Distribution**

IMSE 213 Prob&Statistics Binomial Distribution 4/1/2017 Each success has probability, p Each failure has probability, q q = 1 - p x = no. successes n - x = no. failures Since independent, combination probabilities are multiplied: Probability of a specific order = px qn-x Total number of sample points that have n trials and x successes and n-x failures is © 2002 Steven E. Guffey, PhD, CIH

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Graph of Binomial f(x) 6 5 8 7 10 9 12 11 2 1 4 3 13

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IMSE 213 Prob&Statistics Example 4/1/2017 The probability that a component will survive a given shock test is ¾. Find the probability that exactly 2 of the next 4 components will survive. Solution: Assuming the tests are independent And p = ¾ for each of the 4 tests, then: 2 4 3/4 © 2002 Steven E. Guffey, PhD, CIH

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**Sum of Probabilites Must Equal Unity**

IMSE 213 Prob&Statistics 4/1/2017 Sum of Probabilites Must Equal Unity As with all other distributions, the sum of all probabilities must equal unity: © 2002 Steven E. Guffey, PhD, CIH

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**Table A.1: Binomial Cumulative Distribution**

IMSE 213 Prob&Statistics Table A.1: Binomial Cumulative Distribution 4/1/2017 Proportion, p Trials Success 0.1 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.9000 0.8000 0.7500 0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 1.0000 2 0.8100 0.6400 0.5625 0.4900 0.3600 0.2500 0.1600 0.0900 0.0400 0.0100 0.9900 0.9600 0.9375 0.9100 0.8400 0.5100 0.1900 3 0.7290 0.5120 0.4219 0.3430 0.2160 0.1250 0.0640 0.0270 0.0080 0.0010 0.9720 0.8960 0.8438 0.7840 0.6480 0.3520 0.1040 0.0280 0.9990 0.9920 0.9844 0.9730 0.9360 0.8750 0.6570 0.4880 0.2710 4 0.6561 0.4096 0.3164 0.2401 0.1296 0.0625 0.0256 0.0081 0.0016 0.0001 0.9477 0.8192 0.7383 0.6517 0.4752 0.3125 0.1792 0.0837 0.0272 0.0037 0.9963 0.9728 0.9492 0.9163 0.8208 0.6875 0.5248 0.3483 0.1808 0.0523 0.9999 0.9984 0.9961 0.9919 0.9744 0.8704 0.7599 0.5904 0.3439 © 2002 Steven E. Guffey, PhD, CIH

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**Example Disease P(X > 10) = 1- P(X < 10) What is included? x 15**

IMSE 213 Prob&Statistics Example Disease 4/1/2017 The probability that a patient recovers from a disease is p= 0.4. For n=15 with the disease, what is the probability: A) at least 10 survive? P(X > 10) = 1- P(X < 10) What is included? 9 x 15 0.4 = Trials Success 0.1 0.2 0.25 0.3 0.4 0.5 15 5 0.9978 0.9389 0.8516 0.7216 0.4032 0.1509 6 0.9997 0.9819 0.9434 0.8689 0.6098 0.3036 7 1.0000 0.9958 0.9827 0.9500 0.7869 0.5000 8 0.9992 0.9848 0.9050 0.6964 9 0.9999 0.9963 0.9662 0.8491 10 0.9993 0.9907 0.9408 © 2002 Steven E. Guffey, PhD, CIH

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**Example Disease -continued**

IMSE 213 Prob&Statistics Example Disease -continued 4/1/2017 The prob that a patient recovers from a rare blood disease is B) What is the probability that, of 15, P(3 < X < 8) 8 15 0.4 3 Trials Success 0.1 0.2 0.25 0.3 0.4 15 2 0.8159 0.3980 0.2361 0.1268 0.0271 3 0.9444 0.6482 0.4613 0.2969 0.0905 4 0.9873 0.8358 0.6865 0.5155 0.2173 6 0.9997 0.9819 0.9434 0.8689 0.6098 8 1.0000 0.9992 0.9958 0.9848 0.9050 © 2002 Steven E. Guffey, PhD, CIH

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IMSE 213 Prob&Statistics Example Disease C 4/1/2017 The probability that a patient recovers from a blood disease = 0.4. What is the probability that: C) exactly 5 of 15 survive? P(X = 5) = b(5;15,0.4) = = Trials Success 0.1 0.2 0.25 0.3 0.4 15 2 0.8159 0.3980 0.2361 0.1268 0.0271 3 0.9444 0.6482 0.4613 0.2969 0.0905 4 0.9873 0.8358 0.6865 0.5155 0.2173 5 0.9978 0.9389 0.8516 0.7216 0.4032 © 2002 Steven E. Guffey, PhD, CIH

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**Example Chain Retailer**

IMSE 213 Prob&Statistics Example Chain Retailer 4/1/2017 A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. Problem (a) The inspector of the retailer randomly picks 20 items for a shipment. What is the probability that there will be at least one defective item among the 20? Solution (a): Denote X = number of defectives devices among the 20 The distribution of the X follows: b(x; 20, 0.03) P( X > 1) = 1 – P(X < 1) = 1- b(0;20,0.03) = 1 – (1)(0.03)x(1-0.03)n-x = 1 – (1)(0.03)0(1-0.03)20-0 = 1 – (1)(1)(0.97)20 = 1 – = © 2002 Steven E. Guffey, PhD, CIH

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**Example Chain Retailer - continued**

IMSE 213 Prob&Statistics Example Chain Retailer - continued 4/1/2017 A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. From part a, P(X > 1) = Problem (b) Suppose the retailer receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be 3 shipments containing at least one defective device? Solution (b): Each shipment contains at least 1 defective or not From part a, P(X > 1) = Let Y = number of shipments with at least one defective item Assuming independence from shipment to shipment, the Y follows binomial distribution with p = b( y ; 10, ) P(Y = 3) = 10! (0.4562)3( )10-3 = ! 7! © 2002 Steven E. Guffey, PhD, CIH

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IMSE 213 Prob&Statistics 4/1/2017 Areas of Application Industrial engineers are interested in the proportion of defectives. Quality control measures and sampling schemes are based on the binomial distribution. Also used extensively for medical (survive, die) and military applications (hit, miss). © 2002 Steven E. Guffey, PhD, CIH

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IMSE 213 Prob&Statistics 4/1/2017 Mean and Variance The mean and variance of the binomial distribution [b(x;n,p)] are: = n p 2 = n p q © 2002 Steven E. Guffey, PhD, CIH

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**Example Binomial Mean and Variance**

IMSE 213 Prob&Statistics Example Binomial Mean and Variance 4/1/2017 Find the mean and variance of the binomial random variable of Example 5.5 Solution: Since was a binomial experiment with n = 15 and p = 0.4, then: = n p = 15(0.4) = 6 2 = n p q = 15(0.4)(0.6) = 3.6 = = 1.897 © 2002 Steven E. Guffey, PhD, CIH

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**Example Binomial Impure Wells**

IMSE 213 Prob&Statistics 4/1/2017 Example Binomial Impure Wells An impurity exists in perhaps 30% of wells. Ten wells tested at random. a) What is the probability that exactly 3 wells have the impurity, assuming the rate really is 30%: = = b) What is the probability that more than 3 wells are impure? P(X > 3|p=30%) = 1 – P(X < 3) = 1 – = © 2002 Steven E. Guffey, PhD, CIH

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**Example Impure Wells - continued**

IMSE 213 Prob&Statistics 4/1/2017 Example Impure Wells - continued How likely is that 6 of 10 are contaminated if p = 0.3 ? Solution: = = © 2002 Steven E. Guffey, PhD, CIH

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**Multinomial Experiments – not on exam**

IMSE 213 Prob&Statistics Multinomial Experiments – not on exam 4/1/2017 More than 2 possible outcomes Small, medium, large If given trial can result in any one of k outcomes, E1, E2, …, Ek with probabilities p1, p2, …, pk occurs in n independent trials. © 2002 Steven E. Guffey, PhD, CIH

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**Example Multinomials – not on exam**

IMSE 213 Prob&Statistics Example Multinomials – not on exam 4/1/2017 Arrivals and departures from an airport 3 runways, each with a probability of being used for any random plane: R1: p1 = 2/9 R2: p2 = 1/6 R3: p3 = 11/18 What is the probability that 6 planes will land such that: R1: 2 planes R2: 1 plane R3: 3 planes © 2002 Steven E. Guffey, PhD, CIH

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**Homework: 5A Do problemson pp. 150-151 Up to but not including 17**

IMSE 213 Prob&Statistics 4/1/2017 Homework: 5A Do problemson pp Up to but not including 17 Artificial intelligence is no match for natural stupidity. "This 'telephone' has too many shortcomings to be seriously considered as a means of communication. The device is inherently of no value to us." --Western Union internal memo, 1876. "The wireless music box has no imaginable commercial value. Who would pay for a message sent to nobody in particular?" --David Sarnoff's associates in response to his urging for investment in the radio in the 1920s. "We don't like their sound, and guitar music is on the way out." -- Decca Recording Co. rejecting the Beatles, 1962. © 2002 Steven E. Guffey, PhD, CIH

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**Hypergeometric Distribution**

IMSE 213 Prob&Statistics 4/1/2017 Hypergeometric Distribution Involves the way sampling is done Binomial requires independence (e.g., replace card after dealing) Hypergeometric does NOT require independence. Therefore, highly useful when testing destroys item. Applies, also, when sample is a significant fraction of the population Used in acceptance sampling, electronics testing, and quality assurance. Start here 26 Feb 04 © 2002 Steven E. Guffey, PhD, CIH

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**Hypergeometric Distribution**

IMSE 213 Prob&Statistics 4/1/2017 Hypergeometric Distribution Hypergeometric Experiment Random sample of size n selected without replacement from N items k of the N items may be classified as successes and N-k are classified as failure Hypergeometric random variable: number, X, of successes in hypergeometric experiment. Hypergeometric distribution: Prob distr. of a hypergeometric variable h( x; N, n, k ) © 2002 Steven E. Guffey, PhD, CIH

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**Hypergeometric Distribution in Acceptance Sampling**

IMSE 213 Prob&Statistics 4/1/2017 Hypergeometric Distribution in Acceptance Sampling Acceptance sampling: parts are sampled to determine whether the entire lot should be sampled. © 2002 Steven E. Guffey, PhD, CIH

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**Hypergeometric Distribution**

IMSE 213 Prob&Statistics Hypergeometric Distribution 4/1/2017 P(X= x) = x = assumed number of successes in sample k = number of success in population to be sampled n = size of random sample N = size of population from which sample is taken Proportion of successes in population = k/N Proporton of successes in sample = x/n © 2002 Steven E. Guffey, PhD, CIH

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**Hypergeometric Distribution**

IMSE 213 Prob&Statistics Hypergeometric Distribution 4/1/2017 © 2002 Steven E. Guffey, PhD, CIH

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**Example Shipment Sampling**

IMSE 213 Prob&Statistics 4/1/2017 Suppose our plan is to sample 3 out of a lot of 10 in each shipment. We will accept the lot if none of the 3 are bad. How often will we accept the lot when 2 of 10 are actually bad? Solution: find P(accept | 2 of 10 in shipment are bad) i.e., NONE of the 3 are unacceptable) x = assumed number of “successes” (defectives) in sample = N = size of population from which sample is taken = 10 n = size of random sample = 3 k = number of success in population to be sampled = 2 bad good 2 10- 2 3- 0.467 10 all 3 © 2002 Steven E. Guffey, PhD, CIH

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**0.3011 Example Lots of 40 3! (40- 3)! 1!(3-1)! (5 -1)! (40- 3-5+1)! 1**

IMSE 213 Prob&Statistics Example Lots of 40 4/1/2017 Lots of 40 components should be rejected if they contain 3 or more defectives. However, we sample 5 at random from each lot and reject if even 1 defective is found. What is the prob that exactly 1 sample is bad if there are 3 defectives in the whole lot? 3! (40- 3)! 1!(3-1)! (5 -1)! (40- 3-5+1)! 1 40 5 3 0.3011 40 5!(40-5)! © 2002 Steven E. Guffey, PhD, CIH

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**Mean and Variance of Hypergeometric**

IMSE 213 Prob&Statistics Mean and Variance of Hypergeometric 4/1/2017 The mean and variance of a hypergeometric distribution, h(x; N, n, k) are: © 2002 Steven E. Guffey, PhD, CIH

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**Example: Hypergeometric Mean and Variance**

IMSE 213 Prob&Statistics Example: Hypergeometric Mean and Variance 4/1/2017 Find the mean and variance for the data given n = 5 N = 40 k = 3 x = 1 nk N 40 5 x 3 3 1 0.3113 1 © 2002 Steven E. Guffey, PhD, CIH

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**Relationship to Binomial Distribution**

IMSE 213 Prob&Statistics Relationship to Binomial Distribution 4/1/2017 If n/N < 0.05 then doesn’t matter much if replace or not. k/N plays the role of the binomial value, p Variance differs by a factor of (N-n)/N-1 negligible as n/N becomes very small Therefore, binomial is “large population edition” of the hypergeometric distribution. n/N < 0.05 1 © 2002 Steven E. Guffey, PhD, CIH

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IMSE 213 Prob&Statistics Example Tires 4/1/2017 Among 5000 tires shipped, 1000 are blemished. If one selects 10 at random, what is prob that exactly 3 are blemished? Solution: n/N = 10/5000 << 0.05 Therefore, can use the binomial distribution h(3:5000,10,1000) ~ b(3;10,0.2) = – = Compares well to h(3;5000,10,1000) = © 2002 Steven E. Guffey, PhD, CIH

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**Multivariate Hypergeometric Distribution – not on exam**

IMSE 213 Prob&Statistics Multivariate Hypergeometric Distribution – not on exam 4/1/2017 If N items can partitioned in the k cells A1, A2, …. Ak with a1, a2,…ak elements, then the prob distr of the random variables X1, X2,…,Xk, representing the number of elements selected from A1, A2, …. Ak in a random sample of size n, is: © 2002 Steven E. Guffey, PhD, CIH

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**Example Multivariate Hypergeometric – not on exam**

IMSE 213 Prob&Statistics 4/1/2017 A group of 10 individuals are used for a biological case study with the following blood types: Type O = 3 people Type A = 4 people Type B = 3 people What is the prob that a random sample of 5 will contain 1 Type O, 2 Type A, 2 Type B ? Solution: X1 = 1 x2 = 2 x3 = 2 A1 = 3 a2 = 4 a3 = 3 N = 10 n = 5 © 2002 Steven E. Guffey, PhD, CIH

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*** Set up and substitute; do not calculate**

IMSE 213 Prob&Statistics 4/1/2017 HW: 5B p , problems: 29, 30, 31, 33, 35, 41, 43, 47*, 49* * Set up and substitute; do not calculate © 2002 Steven E. Guffey, PhD, CIH

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**Negative Binomial and Geometric Distributions**

IMSE 213 Prob&Statistics Negative Binomial and Geometric Distributions 4/1/2017 Consider case where repeat experiments until a fixed number of successes occur. Called negative binomial experiments. Example: Suppose testing drug we hope will provide some relief to 60% of patients, and we want to know the probability that the 5th patient to experience relief will be the 7th patient to receive the drug. Let k= number of success Prob for a specific order = pk-1 qx-k p = pk qx-k P = 0.65(1-0.6)2 for each of several orders of successes and failures For the kth success to occur on the xth trial, there must have been (k-1) successes and (x-k) failures among the first (x-1) trials since the xth trial must be a success. © 2002 Steven E. Guffey, PhD, CIH

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**Negative Binomial Random Variable**

IMSE 213 Prob&Statistics 4/1/2017 Negative Binomial Random Variable Probability distribution of X, the number of trials on which the kth success occurs, is: Negative binomial random (X) Number of trials to produce k successes in a negative binomial experiment © 2002 Steven E. Guffey, PhD, CIH

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IMSE 213 Prob&Statistics Example NBA-a 4/1/2017 NBA team must win 4 of 7 games to be Champion. Suppose Team A has prob. of beating Team B of 55%. A) what is the probabilty A will win in 6 games? Solution: x = 6 k = 4 p = 0.55 6 4 6-4 6 4 0.55 0.55 0.55 3! 2! © 2002 Steven E. Guffey, PhD, CIH

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IMSE 213 Prob&Statistics Example NBA-b 4/1/2017 NBA team must win 4 of 7 games to be Champion. Suppose Team A has probability of beating Team B of 55% each game that is played. B) what is the probability A will win the series? Solution: (Hint: can win in 4, 5, 6, or 7 games) k = 4 p = 0.55 © 2002 Steven E. Guffey, PhD, CIH

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**Geometric Distribution**

IMSE 213 Prob&Statistics Geometric Distribution 4/1/2017 Special case of negative binomial with k=1 Example: toss coin until first occurrence of a “head” If repeated trials can result in success with probability, p, and failure with q =1-p, then the probability distribution of the random variable, X, the number of the trial on which the first success occurs is: g(x,p) = p qx-1, x = 1,2,3,…. © 2002 Steven E. Guffey, PhD, CIH

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**Example: First Defective**

IMSE 213 Prob&Statistics 4/1/2017 Example: First Defective In a certain mfg process, on the average, 1/100 items is defective. What is the probability that the 5th item inspected is the 1st defective? p = k=1 x = 5 Solution: g(x,p) = p q x-1 g(5; 0.01) = = (0.01) (1-0.01)4 = © 2002 Steven E. Guffey, PhD, CIH

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**Example: Telephone Exchanges**

IMSE 213 Prob&Statistics Example: Telephone Exchanges 4/1/2017 When many calls are coming in at once, telephone exchanges can be overwhelmed, making callers have to call repeatedly to get a connection. p = 0.05 for probability of connection What is probability that 5 attempts are necessary? Solution: x = 5 P(X=x) = g(x,p) = g(5; 0.05) = p q x-1 = 0.05(1-0.05)5-1 = 0.041 © 2002 Steven E. Guffey, PhD, CIH

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**Example: Particles in Wafers**

IMSE 213 Prob&Statistics Example: Particles in Wafers 4/1/2017 The probability that a wafer contains a large particle of contamination is If it is assumed that the wafers are independent, what is the probability that exactly 125 wafers need to be analyzed before a large particle is detected? Let X denote the number of samples analyzed until a large particle is detected p = 0.01 P(X=125) = g( ; ) 125 0.01 = p q x-1 125-1 = ( ) = 0.0029 0.01 © 2002 Steven E. Guffey, PhD, CIH

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**Mean and Variance of Geometric**

IMSE 213 Prob&Statistics 4/1/2017 Mean and Variance of Geometric Mean and variance of a random variable following the geometric distribution are: © 2002 Steven E. Guffey, PhD, CIH

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**Poisson Distribution and the Poisson Process**

IMSE 213 Prob&Statistics 4/1/2017 Poisson Distribution and the Poisson Process “Poisson Experiments” yield numerical values of a random variable, X, the number of outcomes occurring during a given time interval or in a specified region. Example: number of calls received per hour Example: number of field mice per acre Derived from Poisson process Used for quality control, quality assurance, and acceptance sampling. © 2002 Steven E. Guffey, PhD, CIH

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**Properties of Poisson Process**

IMSE 213 Prob&Statistics Properties of Poisson Process 4/1/2017 Properties Number of outcomes independent of number occurring in other times or spaces Probability that a single outcome will occur during a short period or small space is proportional to the size of the region and independent of outcomes outside that region The probability that more than one outcome will occur in a short time interval or small place is negligible. Can sample only a small fraction of the total volume (“law of rare events”). X = Poisson random variable Follows Poisson distribution Mean = µ = t t = specific time or region of interest = single unit P(x; t ) © 2002 Steven E. Guffey, PhD, CIH

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**Derivation from Binomial Distribution**

Wikipedia: This limit is sometimes known as the law of rare events, since each of the individual Bernouilli event rarely triggers. The name may be misleading because the total count of success events in a Poisson process need not be rare if the parameter λ is not small. For example, the number of telephone calls to a busy switchboard in one hour follows a Poisson distribution with the events appearing frequent to the operator, but they are rare from the point of the average member of the population who is very unlikely to make a call to that switchboard in that hour. The proof may proceed as follows. First, recall from from calculus that: and the definition of the Binomial distribution: If the binomial probability can be defined such that p = λ / n, we can evaluate the limit of P as n goes large: The F term can be written as: and then note that, since k is fixed, this is a rational function of n with limit 1.

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**Poisson Distribution lt = 2 lt = 5**

IMSE 213 Prob&Statistics 4/1/2017 Poisson Distribution The probability distribution of a Poisson random variable, X, representing the number of outcomes occurring in a given time interval or specified region denoted by t, is: lt = 2 lt = 5 © 2002 Steven E. Guffey, PhD, CIH

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**Example: Flaws in Copper Wire**

IMSE 213 Prob&Statistics Example: Flaws in Copper Wire 4/1/2017 For thin copper wire the number of flaws follows a Poisson distribution with a mean of 2.3 flaws per millimeter. Determine the probability of exactly 2 flaws in 1 millimeter of wire. l = t = lt = 2.3/mm * 1 mm = 2.3 2.3 flaws/mm 1 mm x = 2 in 1 mm P(4 flaws in 2 mm of wire) = lt = 2.3/mm * 2 mm = 4.6 P(6 flaws in 3 mm of wire) = lt = 2.3/mm * 3 mm = 6.9 © 2002 Steven E. Guffey, PhD, CIH

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**Example: Flaws in Copper Wire - B**

IMSE 213 Prob&Statistics Example: Flaws in Copper Wire - B 4/1/2017 For thin copper wire, the number of flaws follows a Poisson distribution with a mean of 2.3 flaws per millimeter. Determine the probability that there are < 2 flaws in 1 millimeter of wire. X ≤ 2 flaws lt = 2.3 flaws = = 0.596 © 2002 Steven E. Guffey, PhD, CIH

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**Example: Particle Counter**

IMSE 213 Prob&Statistics Example: Particle Counter 4/1/2017 In a lab experiment, average number of particles in 1 ms is 4. What is the probability that 6 particles enter the counter in a given milli-second? Solution: Using the Poisson distribution with: x = t = 6 4 © 2002 Steven E. Guffey, PhD, CIH

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**Example: Particle Counter (Using Table A.2)**

IMSE 213 Prob&Statistics Example: Particle Counter (Using Table A.2) 4/1/2017 In lab experiment, the average number of particles counted in 1 ms is 4. What is the prob that exactly 6 particles enter the counter in a given ms? Poisson distribution with: x = 6 t = 4 0.7851 = x p(x,lt) P(x,lt) 0.0183 1 0.0733 0.0916 2 0.1465 0.2381 3 0.1954 0.4335 4 0.6288 5 0.1563 0.7851 6 0.1042 0.8893 © 2002 Steven E. Guffey, PhD, CIH

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IMSE 213 Prob&Statistics Example Tankers 4/1/2017 On the average, 10 tankers arrive each day. We can handle 15 tankers per day. What is the probability that on a given day some tankers will be turned away? Solution: Let X = number tankers arriving Using Table A.2, we have: © 2002 Steven E. Guffey, PhD, CIH

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**Mean and Variance of Poisson**

IMSE 213 Prob&Statistics 4/1/2017 Mean and Variance of Poisson The mean and variance of the Poisson distribution p(x, t) both have the value t. © 2002 Steven E. Guffey, PhD, CIH

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**Poisson as a limiting form of the binomial**

IMSE 213 Prob&Statistics 4/1/2017 Poisson as a limiting form of the binomial If n is large and p approaches 0, conditions are nearer to the continuous space or time region of the Poisson process. Note that if p approaches 1, we can just exchange the definition of success and failure and use the Poisson distribution. Accurate if: n ≥ 20 and p ≤ 0.05, or n ≥ 100 and np ≤ 10 © 2002 Steven E. Guffey, PhD, CIH

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**Then: t = np and b(x;n,p) = p(x, t) = p(x;µ)**

IMSE 213 Prob&Statistics 4/1/2017 Theorem Let X be a binomial random variable with probability distribution, b(x;n,p). When: n ∞ p 0 µ = np remains constant, Since binomial: µ = np Poisson: µ = t Then: t = np and b(x;n,p) = p(x, t) = p(x;µ) © 2002 Steven E. Guffey, PhD, CIH

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**Since p 0 and n is large, use Poisson distribution. **

IMSE 213 Prob&Statistics Example: Accidents 4/1/2017 Accidents occur infrequently in an particular factory. Assume accidents are independent and p = on any given day. What is the probability of an accident on one day of a period of 400 days? Solution: Since p 0 and n is large, use Poisson distribution. t = np = 400(0.005) = 2 X = 1 © 2002 Steven E. Guffey, PhD, CIH

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**3 < Example : Accidents - B**

IMSE 213 Prob&Statistics Example : Accidents - B 4/1/2017 Accidents occur infrequently in an particular factory. Assume accidents are independent and p = What is the prob that there are at most 3 days out of 400 with an accident? Solution: Since p 0 and n is large, use Poisson distribution. t = np = 400(0.005) = 2 < 3 © 2002 Steven E. Guffey, PhD, CIH

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**Example: Glass Making (8000)(0.001) = 8**

IMSE 213 Prob&Statistics Example: Glass Making 4/1/2017 In glass-making, one or more defects occur on the avg on 1/1000 pieces. What is the prob that 8000 pieces will contain < 7 with defects? Solution: Is a binomial experiment, but since n=8000 and p=0.0001, can approximate with the Poisson distr: µ = (8000)(0.001) = 8 x p(x,lt) P(x,lt) 0.0003 1 0.0027 0.0030 2 0.0107 0.0138 3 0.0286 0.0424 4 0.0573 0.0996 5 0.0916 0.1912 6 0.1221 0.3134 © 2002 Steven E. Guffey, PhD, CIH

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**Odd problems on p. 165-167, omitting nos. 63 and 67**

IMSE 213 Prob&Statistics 4/1/2017 HW: 5C Odd problems on p , omitting nos. 63 and 67 © 2002 Steven E. Guffey, PhD, CIH

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**The End IMSE 213 Prob&Statistics 4/1/2017**

© 2002 Steven E. Guffey, PhD, CIH

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