STAT 400 Lecture AL1 Answers for 3.2 (Part 2) Spring 2015 Dalpiaz Gamma Distribution: θ 1 α 1αθαxexxf, 0 x < E ( X ) = Var ( X ) = 2 xexxf λ 1 αααλ, 0 x < E ( X ) = /, Var ( X ) = /2 If T has a Gamma ( , = 1/ ) distribution, where is an integer, then F T ( t ) = P ( T t ) = P ( X t ), P ( T > t ) = P ( X t – 1 ), where X t has a Poisson ( t = θt ) distribution. 1. Alex is told that he needs to take bus #5 to the train station. He misunderstands the directions and decides to wait for the fifth bus. Suppose that the buses arrive to the bus stop according to Poisson process with the average rate of one bus per 20 minutes. X t = number of buses in t hours. Poisson ( t ) T k = arrival time of the k th bus. Gamma, = k. one bus per 20 minutes = 3. a) Find the probability that Alex would have to wait longer than 1 hour for the fifth bus to arrive. P ( T 5 > 1 ) = P ( X 1 4 ) = P ( Poisson ( 3 ) 4 ) = 0.815. OR P ( T 5 > 1 ) = 13 15 5 53 dttte = 13 4 5 43 ! dttte = …b) Find the probability that the fifth bus arrives during the second hour. P ( 1 < T 5 < 2 ) = P ( T 5 > 1 ) – P ( T 5 > 2 ) = P ( X 1 4 ) – P ( X 2 4 ) = P ( Poisson ( 3 ) 4 ) – P ( Poisson ( 6 ) 4 ) = 0.815 – 0.285 = 0.530. OR P ( 1 < T 5 < 2 ) = 213 15 5 53 dttte = 213 4 5 43 ! dttte = … c) Find the probability that the fifth bus arrives during the third hour. P ( 2 < T 5 < 3 ) = P ( T 5 > 2 ) – P ( T 5 > 3 ) = P ( X 2 4 ) – P ( X 3 4 ) = P ( Poisson ( 6 ) 4 ) – P ( Poisson ( 9 ) 4 ) = 0.285 – 0.055 = 0.230. OR P ( 2 < T 5 < 3 ) = 323 15 5 53 dttte = 323 4 5 43 ! dttte = … 1.3. Traffic accidents at a particular intersection follow Poisson distribution with an average rate of 1.4 per week. 0.2 per day. a) What is the probability that the next accident will not occur for three days? Exponential, = 0.2. P ( T > 3 ) = e – 0.2 3 = e – 0.6 0.5488. OR Poisson, 3 days t = 0.2 3 = 0.6. P ( X = 0 ) = !06.06.00 e 0.5488.b) What is the probability that the next accident will occur during the third day? (That is, the time until the next accident is more than two days, but less than three days.) Exponential, = 0.2. P ( 2 < T < 3 ) = P ( T > 2 ) – P ( T > 3 ) = e – 0.2 2 – e – 0.2 3 = e – 0.4 – e – 0.6 0.1215. OR Poisson, 1 day t = 0.2 1 = 0.2. P ( X = 0 ) = !02.02.00 e 0.81873. 1st day 2nd day 3rd day No accident No accident Accident(s) 0.81873 0.81873 0.18127 0.1215. c) What is the probability that the second accident will occur before the end of the third day? P ( the second accident will occur before the end of the third day ) = P ( at least 2 accidents will occur in three days ) = !! 16.006.016.016.00 ee 0.1219. OR 30 2.0 1 2 12.0 dxxex = 03 2.012.012.0 2.0 22.0 2xxeex 0.1219.d) What is the probability that the fourth accident will occur after the end of the seventh day? Gamma, = 4, = 0.2. P ( T 4 > 7 days ) = 7 2.0 3 4 ! 32.0 dxxex = … OR Gamma, = 4, = 1.4. P ( T 4 > 1 week ) = 1 4.1 3 4 ! 34.1 dxxex = … OR P ( the fourth accident will occur after the end of the seventh day ) = P ( at most 3 accidents will occur in seven days ) = !!!! 34.124.114.104.14.134.124.114.10 eeee 0.9463. e) What is the probability that the third accident will occur during the fourth day? P ( the third accident will occur during the fourth day ) = P ( the third accident will occur after the end of the third day ) – P ( the third accident will occur after the end of the fourth day ) = P ( at most two accidents will occur in three days ) – P ( at most two accidents will occur in four days ) = !!! 26.016.006.06.026.016.00 eee – !!! 28.018.008.08.028.018.00 eee 0.0243. OR P ( 3 < T 3 < 4 ) = 43 2.0 2 3 22.0 dxxex = … 0.0243.1.7. During a radio trivia contest, the radio station receives phone calls according to Poisson process with the average rate of five calls per minute. X t = number of phone calls in t minutes. Poisson ( t ) T k = time of the k th phone call. Gamma, = k. five calls per minute = 5. a) Find the probability that we would have to wait less than two minutes for the ninth phone call. P ( T 9 < 2 ) = P ( X 2 9 ) = 1 – P ( X 2 8 ) = 1 – P ( Poisson ( 10 ) 8 ) = 1 – 0.333 = 0.667. OR P ( T 9 < 2 ) = 205 19 9 95 dttte = 205 8 9 85 ! dttte = … b) Find the probability that the ninth phone call would arrive during the third minute. P ( 2 < T 9 < 3 ) = P ( T 9 > 2 ) – P ( T 9 > 3 ) = P ( X 2 8 ) – P ( X 3 8 ) …
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