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UIUC STAT 400 - 400Ex2_6ans

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STAT 400 Lecture AL1 Examples for 2.6 Spring 2015 Dalpiaz Poisson Distribution: X = the number of occurrences of a particular event in an interval of time or space. P( X = x ) = !λλ xx e −⋅, x = 0, 1, 2, 3, … . E( X ) = λ, Var( X ) = λ. Table III ( pp. 580 – 582 ) gives P( X ≤ x ) 1. Traffic accidents at a particular intersection follow Poisson distribution with an average rate of 1.4 per week. a) What is the probability that the next week is accident-free? 1 week ⇒ λ = 1.4. P ( X = 0 ) = !04.14.10 e −⋅ ≈ 0.2466. b) What is the probability that there will be exactly 3 accidents next week? 1 week ⇒ λ = 1.4. P ( X = 3 ) = !34.14.13 e −⋅ ≈ 0.1128. c) What is the probability that there will be at most 2 accidents next week? 1 week ⇒ λ = 1.4. P ( X ≤ 2 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) = !!! 24.114.104.14.124.114.10 eee −−−⋅⋅⋅++ ≈ 0.2466 + 0.3452 + 0.2417 = 0.8335.d) What is the probability that there will be at least 2 accidents during the next two weeks? 2 weeks ⇒ λ = 2.8. P ( X ≥ 2 ) = 1 – [ P ( X = 0 ) + P ( X = 1 ) ] = +−−−⋅⋅!! 18.208.218.218.20 ee ≈ 1 – [ 0.0608 + 0.1703 ] = 0.7689. e) What is the probability that there will be exactly 5 accidents during the next four weeks? 4 weeks ⇒ λ = 5.6. P ( X = 5 ) = !56.56.55 e −⋅ ≈ 0.1697. f) What is the probability that there will be exactly 2 accidents tomorrow? 1 day ⇒ λ = 0.2. P ( X = 2 ) = !22.02.02 e −⋅ ≈ 0.0164. g) What is the probability that the next accident will not occur for three days? 3 days ⇒ λ = 0.6. P ( X = 0 ) = !06.06.00 e −⋅ ≈ 0.5488. h) What is the probability that there will be exactly three accident-free weeks during the next eight weeks? “Success” = an accident-free week 1 week ⇒ λ = 1.4. p = P ( “Success” ) = P ( X = 0 ) = !04.14.10 e −⋅ ≈ 0.2466. P ( exactly 3 accident-free weeks in 8 weeks ) = 8 C 3 ⋅ 0.2466 3 ⋅ 0.7534 5 ≈ 0.20384. ( Binomial distribution )i) What is the probability that there will be exactly five accident-free days during the next week? “Success” = an accident-free day 1 day ⇒ λ = 0.2. p = P ( “Success” ) = P ( X = 0 ) = !02.02.00 e −⋅ ≈ 0.81873. P ( exactly 5 accident-free days in 7 days ) = 7 C 5 ⋅ 0.81873 5 ⋅ 0.18127 2 ≈ 0.25385. ( Binomial distribution ) When n is large ( n ≥ 20 ) and p is small ( p ≤ 0.05 ) and n ⋅ p ≤ 5, Binomial probabilities can be approximated by Poisson probabilities. For this, set λ = n ⋅ p. 2. Suppose the defective rate at a particular factory is 1%. Suppose 50 parts were selected from the daily output of parts. Let X denote the number of defective parts in the sample. a) Find the probability that the sample contains exactly 2 defective parts. P ( X = 2 ) = ( ) ( )48299.001.0250 ⋅⋅ ≈ 0.075618. b) Use Poisson approximation to find the probability that the sample contains exactly 2 defective parts. λ = n ⋅ p = 0.5. P ( X = 2 ) = !25.05.02 e−⋅ ≈ 0.075816.c) Find the probability that the sample contains at most 1 defective part. P ( X ≤ 1 ) = P ( X = 0 ) + P ( X = 1 ) = ( )( )() ()49150099.001.015099.001.0050 ⋅⋅⋅⋅+ ≈ 0.910565. d) Use Poisson approximation to find the probability that the sample contains at most 1defective part. P ( X ≤ 1 ) = P ( X = 0 ) + P ( X = 1 ) = !! 15.005.05.015.00 ee−−⋅⋅+ ≈


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UIUC STAT 400 - 400Ex2_6ans

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