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08 Discrete Probability Distributions Part 2

22 poisson distribution properties and assumptions of

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22 Poisson distribution Properties and assumptions of the poisson distribution Can take any non-negative integer value (x=0,1,2,3, …… .) : the average number of outcomes of interest per unit time segment or unit space segment. The probability that an outcome of interest occurs in a given (time) segment is the same for all segments of interest of the same length. What happens in one segment is independent from what happens in any other segment.
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23 Poisson distribution Probability mass function: Expectation (Mean) x = number of successes in segment of interest  = average rate of occurrence per unit segment (e.g., #of calls per hour) t = size of the segment of interest (e.g., a three-hour interval) e = base of the natural logarithm system (2.71828...) Variance Standard Deviation ! ) ( ) ( x e t x X P t x λ λ - = = λt μ = λt σ 2 = = σ t λ
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24 =Poisson( x , t, 0/FALSE) =Poisson( x , t, 1/TRUE) POISSON(1,2,1) = P(X≤1) when X~Poisson( λ t=2) = 0.4060. Poisson distribution in Excel ! ) ( ) ( ) ( x e t x X P x P t x λ λ - = = = = - = = x i t i i e t x X P x F 0 ! ) ( ) ( ) ( λ λ
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25 Surface-finish defects in an LCD board occur according to a Poisson distribution at a mean rate of 0.01 defects per in2. 1. Find the probability that a randomly inspected surface of 1 in2 has no defect. X ~ Poisson ( λ =0.01, t=1) Need P(X=0) = 0.010e-0.01/0! = 0.9905 =POISSON(0,0.01,0) Poisson random variable example: Surface defects on a LCD screen.
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26 2. Find the probability that a randomly selected unit of a 21 inch LCD monitor with an actual screen size of 250 in2 will contain at least one defect? 3. What is the expected number of defects on the 21 inch monitor above? X ~ Poisson ( λ =0.01, t=250) Need P(X≥1) = 1-P(X=0) = 1-e-0.01*250 = 0.9179 = 1-POISSON(0,2.5,0) E[X] = λ t = 0.01*250 = 2.50 Poisson random variable example: Surface defects on a LCD screen
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27 Lab accidents at Drexel occurs at an average rate of three accidents per month. 1. What is the probability that there is no accident in 2 months? 2. What is the probability that there will be less than 2 accidents in two months? X ~ Poisson ( λ =3, t=2) P(X=0) = poisson(0, 6, 0) = 0.002479 P(X < 2) = P(X ≤ 1) = poisson(1, 6, 1) = 0.01735 = P(X=0) + P(X=1) = poisson(1, 6, 0) + poisson(0, 6, 0) = 0.00248 + 0.01487 = 0.01735 Poisson random variable example: Sci-fi insurance
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28 3. In a 2-month period during which the university had Dr. Tim “the toolman” Taylor, 16 (reported) accidents occurred. Does this number seem highly improbable if the average rate of accidents per month is 3? Does this indicate that an increase in the mean accident rate might be necessary?
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