This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 John’s 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 – ex/0.5 ) = ex/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/ ν = 2500, 1 √...
View
Full
Document
 '11
 MTODOROVSKA

Click to edit the document details