EGR 252 Walpole Chapter 5 Some Discrete Probability

EGR 252 Walpole Chapter 5 Some Discrete Probability Distributions Part 2

Multinomial Experiments q What if there are more than 2 possible outcomes? (e. g. , acceptable, scrap, rework) q That is, suppose we have: ü n independent trials ü k outcomes that are § mutually exclusive (e. g. , ♠, ♣, ♥, ♦) § exhaustive (i. e. , ∑all kpi = 1) q Then f(x 1, x 2, …, xk; p 1, p 2, …, pk, n) = JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 2

Multinomial Examples q Example 5. 7 refer to page 150 q Exercise 5. 22, refer to page 152 a) Convert ratio 8: 4: 4 to probabilities (8/16, 4/16) or 0. 5, 0. 25 b) Identify x, p pairs. x 1 = _______ p 1 = 0. 50 x 2 = _______ p 2 = 0. 25 x 3 = _______ p 3 = 0. 25 c) Enter values into multinomial equation (page 149)and solve. f ( 5, 2, 1; 0. 50, 0. 25, 8). 25, 8 = 8!/(5!2!1!)* (0. 5)5(0. 25)2(0. 25)1 = 0. 082031 JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 3

Binomial vs. Hypergeometric Distribution q Replacement and Independence ü Binomial (assumes sampling “with replacement”) and hypergeometric (sampling “without replacement”) ü Binomial assumes independence, while hypergeometric does not. q Hypergeometric: The probability associated with getting x successes in the sample (given k successes in the lot. ) JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 4

Hypergeometric Example q Example from Complete Business Statistics, 4 th ed (Mc. Graw-Hill) ü 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? q This example follows a hypergeometric distribution: ü A random sample of size n is selected without replacement from N items. ü k of the N items may be classified as “successes” and N-k are “failures. ” q The probability associated with getting x successes in the sample (given k successes in the lot. ) JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 5

Solution: Hypergeometric Example q In our example, k = number of “successes” = 2 N = the lot size = 10 n = number in sample = 5 x = number found = 1 or 2 P(X > 1) = 0. 556 + 0. 222 = 0. 778 JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 6

Expectations: Hypergeometric Distribution q The mean and variance of the hypergeometric distribution are given by q 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 σ2 σ = (5/9)(5*2/10)(1 -2/10) = 0. 444 = 0. 667 JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 7

Additional problems … 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? σ2 = (14/19)(6*3/20)(1 -3/20) = 0. 5637 3. What is the probability that the lot will be rejected? P(X>2) = 1 – [P(0)+P(1)] JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 8

Binomial Approximation q Note, if N >> n, then we can approximate the hypergeometric with the binomial distribution. q Example: Automobiles arrive in a dealership in lots of 100. 5 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 inspected will not meet safety standards? q Recall: P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0) Hypergeometric distribution 1 - h(0; 100, 5, 20) = 0. 6807 Binomial distribution 1 - b(0; 5, 0. 2) 1 - 0. 3277 = 0. 6723 (See also example 5. 12, pg. 155 -6) JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 9

Negative Binomial Distribution b* q A binomial experiment in which trials are repeated until a fixed number of successes occur. q 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 Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 10

Negative Binomial Equation q This example follows a negative binomial distribution: ü Repeated independent trials. ü Probability of success = p and probability of failure = q = 1 -p. ü Random variable, X, is the number of the trial on which the kth success occurs. q The probability associated with the kth success occurring on trial x is given by, Where, k = “success number” x = trial number on which k occurs p = probability of success (error) q=1–p JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 11

Example: Negative Binomial Distribution q What is the probability that the 10 th error will occur on the 25 th trial? 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 Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 12

Geometric Distribution q Continuing with our example in which p = probability of success (error) = 0. 3 q What is the probability that the 1 st bit received in error will occur on the 5 th trial? q This is an example of the geometric distribution, which is a special case of the negative binomial in which k = 1. ü The probability associated with the 1 st success occurring on trial x is ü P = (0. 3)(0. 7)4 = 0. 072 JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 13

Additional problems … A worn machine tool produces 1% defective parts. If we assume that parts produced are independent: 1. What is the probability that the 2 nd defective part will be the 6 th one produced? 2. What is the probability that the 1 st defective part will be seen before 3 are produced? 3. How many parts can we expect to produce before we see the 1 st defective part? Negative binomial or geometric? Expected value = ? JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 14

Poisson Process q The number of occurrences in a given interval or region with the following properties: ü “memoryless” ie number in one interval is independent of the number in a different interval ü 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. ü P(X>1) in a very short interval is negligible JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 15

Poisson Process Situations q Number of bits transmitted per minute. q Number of calls to customer service in an hour. q Number of bacteria present in a given sample. q Number of hurricanes per year in a given region. JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 16

Poisson Distribution Probabilities q 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 Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 17

Service Call Example - Poisson Process q 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. q λ = 2. 7 and t = 1 minute JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 18

Our Example: λ = 2. 7 and t = 1 minute q What is the probability that fewer than 2 calls will be received in any given minute? q The probability that fewer than 2 calls will be received in any given minute is P(X < 2) = P(X = 0) + P(X = 1) q The mean and variance are both λt, so μ = λt = 2. 7 (1) = 2. 7 q Note: Table A. 2, pp. 732 -734, gives Σt p(x; μ) JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 19

Service Call Example (Part 2) q 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? ü μ = λt = (2. 7) (3)= 8. 1 ü 8. 1 is not in the table; we must use basic equation q Suppose λt = 8; see table with μ = 8 and r = 6 P(X > 6) = 1 – P(X < 6) = 1 - 0. 3134 = 0. 6866 JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 20

Poisson Distribution JMB Ch. 5 Part 2 rev 2019 EGR 252 JMB 9 th ed. Slide 21