****~** Can Anyone Help Me Solve This Statistics Problem?? **~**?**

a.An airline knows that the duration of the flight from Pittsburgh to Minneapolis is uniformly distributed between 120 and 140 minutes. The flight departs at 1:00 p.m. The airline wants the probability that the flight will be late not to exceed .25. What scheduled arrival time should the airline announce? b.If the flight is early, it may have to wait until its arrival gate is available for deplaning. Assume that the arrival gate becomes available between 3:00 p.m. and 3:10 p.m, and that the time the arrival gate becomes available and the arrival time of the flight are statistically independent and jointly uniformly distributed. What is the probability that the flight arriving from Pittsburgh will have to wait for its gate? c. Assume the arrival gate becomes available between 3:00 p.m. and the scheduled arrival time, and (as in part b) the time the arrival gate becomes available and the arrival time are statistically independent and jointly uniformly distributed. (For example, if the scheduled arrival time is 3:07, the arrival gate becomes available between 3:00 and 3:07.) Suppose every minute a flight is late costs the airline $1 thousand dollars, and every minute a flight has to wait for the arrival gate to become available costs $2 thousand dollars. What scheduled arrival time minimizes expected cost?

Mathematics - 1 Answers

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a) The flight arrives between 3:00 and 3:20 and the arrival time is 3:00 + X, where X ~ UNIF(0,20) If the scheduled time is 3:15, then the plane is late if X > 15 and this happens with probability 0,25 b) Here Y = the time at which the gate comes available ~ UNIF(0,10) and (X,Y) ~ UNIF (on the square [0,10]x[0,20]) This means that the density of (X,Y) is constant on the square and the constant = 1/10 * 1/20 = 1/200 We have to find P(X < = Y) Making a picture of the square and of the event {X <= Y} shows that P(X < = Y) = (100/2)/200 = 1/4