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IMSE 213: Probability and Statistics for Engineers Instructor: Steven E. Guffey, PhD, CIH © 2002-2011.

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Presentation on theme: "IMSE 213: Probability and Statistics for Engineers Instructor: Steven E. Guffey, PhD, CIH © 2002-2011."— Presentation transcript:

1 IMSE 213: Probability and Statistics for Engineers Instructor: Steven E. Guffey, PhD, CIH ©

2 2 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

3 3 Theorem 5.1: Mean and Variance of Discrete Uniform Distribution f(i) = n/N = 1/k

4 4 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

5 5 Binomial Distribution 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 = p x q n-x –Total number of sample points that have n trials and x successes and n-x failures is

6 Graph of Binomial 6 f(x)

7 7 Example 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: 243/4

8 8 Sum of Probabilites Must Equal Unity As with all other distributions, the sum of all probabilities must equal unity:

9 9 Table A.1: Binomial Cumulative Distribution Proportion, p Trials Success

10 TrialsSuccess Example Disease 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) x = What is included?

11 11 Example Disease -continued The prob that a patient recovers from a rare blood disease is 0.4. B) What is the probability that, of 15, P(3 < X < 8) TrialsSuccess

12 12 Example Disease C The probability that a patient recovers from a blood disease = 0.4. What is the probability that: –C) exactly 5 of 15 survive? TrialsSuccess = = P(X = 5) =b(5;15,0.4)

13 13 Example Chain Retailer 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 – =

14 14 Example Chain Retailer - continued 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!

15 15 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).

16 16 Mean and Variance The mean and variance of the binomial distribution [b(x;n,p)] are: = n p 2 = n p q

17 17 Example Binomial Mean and Variance 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 pn p = 15(0.4) = 6 2 = = = = n p qn p q = 15(0.4)(0.6) = 3.6

18 18 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 – =

19 19 Example Impure Wells - continued How likely is that 6 of 10 are contaminated if p = 0.3 ? Solution: = =

20 20 Multinomial Experiments – not on exam More than 2 possible outcomes –Small, medium, large If given trial can result in any one of k outcomes, E 1, E 2, …, E k with probabilities p 1, p 2, …, p k occurs in n independent trials.

21 21 Example Multinomials – not on exam Arrivals and departures from an airport 3 runways, each with a probability of being used for any random plane: –R 1 : p 1 = 2/9R 2 : p 2 = 1/6R 3 : p 3 = 11/18 What is the probability that 6 planes will land such that: –R 1 : 2 planesR 2 : 1 planeR 3 : 3 planes

22 22 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, "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.

23 23 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.

24 24 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 )

25 25 Hypergeometric Distribution in Acceptance Sampling Acceptance sampling: parts are sampled to determine whether the entire lot should be sampled.

26 26 Hypergeometric Distribution 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 P(X= x) = Proportion of successes in population = k/N Proporton of successes in sample = x/n

27 27 Hypergeometric Distribution

28 28 Example Shipment Sampling 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) badgood all x=assumed number of successes (defectives) 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 = 10- 0

29 29 Example Lots of 40 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? 1!(3-1)! 3! 5 (40- -1)! 40 5!(40-5)! )! (5 ( )!

30 30 Mean and Variance of Hypergeometric The mean and variance of a hypergeometric distribution, h(x; N, n, k) are:

31 31 Example: Hypergeometric Mean and Variance Find the mean and variance for the data given n = 5N = 40k = 3x = x nk N

32 32 Relationship to Binomial Distribution If n/N < 0.05 then doesnt 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. 1 n/N < 0.05

33 33 Example Tires 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) =

34 34 Multivariate Hypergeometric Distribution – not on exam If N items can partitioned in the k cells A 1, A 2, …. A k with a 1, a 2,…a k elements, then the prob distr of the random variables X 1, X 2,…,X k, representing the number of elements selected from A 1, A 2, …. A k in a random sample of size n, is:

35 35 Example Multivariate Hypergeometric – not on exam 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: –X 1 = 1x 2 = 2x 3 = 2 –A 1 = 3a 2 = 4a 3 = 3N = 10n = 5

36 36 HW: 5B p , problems: 29, 30, 31, 33, 35, 41, 43, 47*, 49* * Set up and substitute; do not calculate

37 37 Negative Binomial and Geometric Distributions 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 5 th patient to experience relief will be the 7 th patient to receive the drug. Let k= number of success Prob for a specific order = p k-1 q x-k p = p k q x-k P = (1-0.6) 2 for each of several orders of successes and failures For the k th success to occur on the x th trial, there must have been (k-1) successes and (x-k) failures among the first (x-1) trials since the x th trial must be a success.

38 38 Negative Binomial Random Variable Negative binomial random (X) –Number of trials to produce k successes in a negative binomial experiment Probability distribution of X, the number of trials on which the k th success occurs, is:

39 39 Example NBA-a 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 = ! !

40 40 Example NBA-b 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

41 41 Geometric Distribution 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 q x-1,x = 1,2,3,….

42 42 Example: First Defective In a certain mfg process, on the average, 1/100 items is defective. What is the probability that the 5 th item inspected is the 1 st defective? p = 0.01k=1 x = 5 Solution: g(x,p) = p q x-1 g(5; 0.01) == (0.01)(1-0.01) 4 =

43 43 Example: Telephone Exchanges 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) = p q x-1 = g(5; 0.05) = 0.05(1-0.05) 5-1 = 0.041

44 44 Example: Particles in Wafers 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 q x-1 P(X=125) = g( ; ) = ( ) =

45 45 Mean and Variance of Geometric Mean and variance of a random variable following the geometric distribution are:

46 46 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.

47 47 Properties of Poisson Process 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 )

48 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: 48 If the binomial probability can be defined such that p = λ / n, we can evaluate the limit of P as n goes large: and then note that, since k is fixed, this is a rational function of n with limit 1. The F term can be written as: and the definition of the Binomial distribution:

49 49 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: t = 2 t = 5

50 50 Example: Flaws in Copper Wire 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. t = x = 2 in 1 mm P(4 flaws in 2 mm of wire) = P(6 flaws in 3 mm of wire) = t = 2.3/mm * 2 mm = 4.6 t = 2.3/mm * 3 mm = 6.9 t = 2.3 flaws/mm1 mm 2.3/mm * 1 mm = 2.3

51 51 Example: Flaws in Copper Wire - B 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. t = 2.3 flaws X 2 flaws = = 0.596

52 52 Example: Particle Counter 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 = 46

53 53 Example: Particle Counter (Using Table A.2) 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 x p(x, t)P(x, t) =

54 54 Example Tankers 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:

55 55 Mean and Variance of Poisson The mean and variance of the Poisson distribution p(x, t) both have the value t.

56 56 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

57 57 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)

58 58 Example: Accidents 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

59 Example : Accidents - B 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 < 59

60 60 Example: Glass Making 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? x p(x, t)P(x, t) Solution: –Is a binomial experiment, but since n=8000 and p=0.0001, can approximate with the Poisson distr: µ = (8000)(0.001) = 8

61 61 HW: 5C Odd problems on p , omitting nos. 63 and 67

62 62


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