**Can someone explain to me probability density function. how would i apply it to this question?**

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? i know that this is 3:15 but this is where i'm having troubnle 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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Let X be the time the flight arrives and Y the time the gate is available, both in minutes after 3pm. Then X is uniform over [0, 20] and Y is uniform over [0, 10]. Since they are independent the joint pdf will be the product of the pdfs of X and of Y, which is to say 1/200 on the region {(x, y): x âˆˆ [0, 20] and y âˆˆ [0, 10]} and 0 elsewhere. For (b) we want P(X > Y) = âˆ«(0 to 10) âˆ«(0 to x) 1/200 dy dx (note that if X>10, the plane does not have to wait) = âˆ«(0 to 10) x/200 dx = [x^2 / 400] [0 to 10] = 1/4. For (c), we need to modify this slightly: if the scheduled flight time is s minutes after 3pm, then Y is uniform on [0, s]. So the pdf is 1/(20s) where x âˆˆ [0, 20] and y âˆˆ[0, s], and 0 elsewhere. We can assume s âˆˆ [0, 20] since moving it outside this interval is guaranteed to increase the cost. The plane is late if X > s (cost $1000/min), and the plane has to wait for its gate if Y > X (cost $2000/min). Note that since Y is confined to [0, s], these events are mutually exclusive. So the cost of being late is âˆ«(s to 20) 1000(x-s) . 1/20 dx = [50 (x^2 / 2 - xs)] [s to 20] = 50 ((200 - 20s)) - (s^2 / 2 - s^2)) = 50 (200 - 20s + s^2 / 2) = 10000 - 1000s + 25s^2 and the cost of waiting for the gate is âˆ«(0 to s) âˆ«(x to s) 2000(y-x) . 1/(20s) dy dx = (100/s) âˆ«(0 to s) [y^2 / 2 - xy] [x to s] dx = (100/s) âˆ«(0 to s) (s^2 / 2 - sx + x^2 / 2) dx = (50/s) âˆ«(0 to s) (s^2 - 2sx + x^2) dx = (50/s) [s^2 x - sx^2 + x^3 / 3] [0 to s] = (50/s) (s^3 / 3) = 50s^2 / 3. So the total cost is 10000 - 1000s + 125s^2 / 3 which has a minimum at s = 1000 / (250/3) = 12. So the scheduled arrival time should be 3:12 pm.