STA3007_0506_t08_ssol

STA3007_0506_t08_ssol - (5 + 5 t ) dt = [5 t + 5 t 2 / 2] 1...

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STA 3007 Applied Probability 2005 Tutorial 8 Suggested Solution 1. Poisson Process i. (a) Pr { X (1) = 2 } = Pr { X (1) - X (0) = 2 } = [(2)(1)] 2 e - (2)(1) 2! = 0 . 2706 (b) Pr { X (1) = 2 and X (3) = 6 } = Pr { X (1) - X (0) = 2 and X (3) - X (1) = 4 } = Pr { X (1) - X (0) = 2 } Pr { X (3) - X (1) = 4 } = 2 e - 2 [(2)(2)] 4 e - (2)(2) 4! = 0 . 05288 (c) Pr { X (1) = 2 | X (3) = 6 } = Pr { X (1) = 2 , X (3) = 6 } /Pr { X (3) = 6 } = 0 . 05288 / [(3)(2)] 6 e - (3)(2) 6! = 0 . 329218 (d) Pr { X (3) = 6 | X (1) = 2 } = Pr { X (3) = 6 , X (1) = 2 } /Pr { X (1) = 2 } = 0 . 05288 / 0 . 2706 = 0 . 1954 ii. (a) Pr { X (1) 2 } = Pr { X (1) = 0 } + Pr { X (1) = 1 } + Pr { X (1) = 2 } = 2 0 e - 2 0! + 2 1 e - 2 1! + 2 2 e - 2 2! = 5 e - 2 = 0 . 6767 (b) Pr { X (1) = 2 and X (2) = 3 } = Pr { X (1) = 2 , X (2) - X (1) = 1 } = Pr { X (1) = 2 } Pr { X (2) - X (1) = 1 } = 2 2 e - 2 2! 2 1 e - 2 1! = 4 e - 4 = 0 . 07326 (c) Pr { X (1) 2 | X (1) 1 } = Pr { X (1) 2 | X (1) 1 } /Pr { X (1) 1 } = Pr { X (1) 2 } /Pr { X (1) 1 } = [1 - 2 0 e - 2 0! - 2 1 e - 2 1! ] / [1 - 2 0 e - 2 0! ] = 1 - 3 e - 2 1 - e - 2 = 0 . 6870 1
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iii. A good model assume that arrivals constitute a nonhomogenous Poisson process with intensity function λ ( t ) λ ( t ) = 5 + 5 t , 0 t 3 λ ( t ) = 20 , 3 t 5 λ ( t ) = 20 - 2( t - 5) , 5 t 9 The expected number of arrivals in the perios 0830-0930 = R 1 . 5 0 . 5
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Unformatted text preview: (5 + 5 t ) dt = [5 t + 5 t 2 / 2] 1 . 5 . 5 = 10 The probability that no customers in this period: = 10 e-10 / 0! = e-10 iv. Pr { X (1) = 2 | X (3) = 5 } = C 5 2 ( 1 2 ) 2 ( 2 3 ) 3 = 0 . 3292 v. Pr { X (1) = 5 | X (2) = 12 } = C 12 5 ( 1 2 ) 5 ( 1 2 ) 7 = 0 . 1934 vi. Pr { W n ≤ t } = Pr { at least n events in the interval (0 , t ] } = Pr { X ( t ) ≥ n } Pr { W X 1 1 ≤ t, W X 2 1 ≤ t, W X 3 1 ≤ t, . .., W X n 1 ≤ t } = Pr { W X 1 1 ≤ t } Pr { W X 2 1 ≤ t } · · · Pr { W X n 1 ≤ t } = Pr { X 1 ( t ) ≥ 1 } Pr { X 2 ( t ) ≥ 1 } · · · Pr { X n ( t ) ≥ 1 } = (1-Pr { X ( t ) = 0 } ) n = (1-e-λt ( λt ) 0! ) n = (1-e-λt ) n 2...
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STA3007_0506_t08_ssol - (5 + 5 t ) dt = [5 t + 5 t 2 / 2] 1...

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