plugin-HW_09_solutions

plugin-HW_09_solutions - HW 9 Ang and Tang: 3.33, 3.51,...

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Unformatted text preview: HW 9 Ang and Tang: 3.33, 3.51, 3.57, 4.15, 4.17,4.20 Due Thurs 11/17/2011 3.33 Let J 1 and J 2 denote the events that Johns scheduled connection time is 1 and 2 hours, respectively, where P(J 1 ) = 0.3 and P(J 2 ) = 0.7. Also, let X be the delay time of the flight in hours. Note that P( X > x ) = 1 P( X x ) = 1 F ( x ) = 1 (1 e-x/0.5 ) = e-x/0.5 (a) Let M denote the event that John misses his connection, i.e. the flight delay time exceeded his scheduled time for connection. Using the theorem of total probability, P(M) = P(M | J 1 )P(J 1 ) + P(M | J 2 )P(J 2 ) = P( X > 1) 0.3 + P( X > 2) 0.7 = e 1/0.5 0.3 + e 2/0.5 0.7 = 0.135 0.3 + 0.0183 0.7 2245 0.053 (b) Regardless of whether John has a connection time of 1 hour and Mike has 2, or the opposite, for them to both miss their connections the flight must experience a delay of more than two hours , and the probability of such an event is P( X > 2) = e 2/0.5 2245 0.018 (c) Since Mary has already waited for 30 minutes, the flight will have a delay time of at least 0.5 hours when it arrives. Hence the desired probability is P( X > 1 | X > 0.5) = P( X > 1 and X > 0.5) / P( X > 0.5) = P( X > 1) / P( X > 0.5) = e 1/0.5 / e 0.5/0.5 = e 2 / e 1 = 1/e 2245 0.368 3.51 The capacity C is gamma distributed with a mean of 2,500 tone and a c.o.v. of 35% k/ = 2500, 1...
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plugin-HW_09_solutions - HW 9 Ang and Tang: 3.33, 3.51,...

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