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Unformatted text preview: Stat 333  Tutorial 3 Answers/Solutions 1. Mean = μ p Variance = μ p + μ 2 (1 p ) p 2 2. (a) P A = μ 1 μ 1 + μ 2 (b) P B = 1 μ 2 μ 1 + μ 2 2 (c) The total time spent in the system can be written as T = S 1 + S 2 + W A + W B where • S 1 is your service time from server 1, • S 2 is your service time from server 2, • W A is the service time for A from server 2 if A is still in service when you move over to server 2, and • W B is the service time for B from server 2 if B is still in the system when you move over to server 2. We can write W A = ( X 2 ,A , if A is still in service when you move over to server 2 , otherwise where X 2 ,A is the service time for A from server 2. Therefore, E ( W A ) = E ( X 2 ,A ) · ( Probablity from part a) ) = 1 μ 2 · P A Additionally, W B = ( X 2 ,B , if B is still in the system when you move over to server 2 , otherwise where X 2 ,B is the service time for B from server 2. Therefore, E ( W B ) = E ( X 2 ,B ) · ( Probablity from part b)...
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This note was uploaded on 01/21/2012 for the course STAT 333 taught by Professor Chisholm during the Fall '08 term at Waterloo.
 Fall '08
 Chisholm
 Probability, Variance

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