Question

5. Two airplanes are flying in the same direction in adjacent parallel corridors. At time t = 0, the first airplane is 15 km behind the second one. Suppose the speed of the first plane (km/hr) is normally distributed with mean 510 and standard deviation 14 and the second plane's speed is also normally distributed with mean and standard deviation 498 and 14, respectively. (a) (2) What is the probability that after 1 hour of flying, the first plane has not caught up to the second plane? (b) [2] Determine the probability that the planes are apart by at least 10 km after one hour of flying (c) [2] What is the probability that after 2 hours of flying, the first plane has caught up to the second plane? (d) [2] Determine the probability that the planes are apart by at most 10 km after 2 hours of flying

Answer 1

Solution : la 510, -14, 0214, liza c.498, n=14 let the number of im the Gost plane is ahead of te D be the number of second plane, after D = 1x, 1X2 + 15 1 hour soille : 19.80 elp = El X.-X2 +15) = E(XI) - E(X2) +15 = 510-498+15=27 Do (V4*, -X2 +15) = 1; (%) +V1X2J - 1144142 @ PID>0) - PC kartta > 07256) EP12)-1,36) = 1- P124-):36) 21-0.0869 use a table) -10.9131 ) Ano ® P[112,103 =1-P1-10405.00) 71-P1-1017.s blagoste 101973) 71-191-1.87525 -0.86] 21 § Plz<-0.861-P126-1.87) 3 1- $0.1949 -0.03.07 a table) = -0.1642 (tuse - 10.8358/ Ave in 2 hours. . Now, the distance covered By first plane : 2x by second plane : 2x2

Now, D; 2*, -2X2 +15 IDEl 2 X,-2*2+15) - E (281) -E 1242 ) + E (15) = 2€1) - 2 ElX 2) +15=20,-242 +15 • 5 24510 - 2* 498 +15 39 To: julix, -142+15) - j22vLX.) +2?lx2 14 x 142 +40149 39.60 Now, Plodo) = pl por ello > 033.7.) -P123 -0.93) al-P124-0.98) i 1-0.1635 luse 2 table) = [0.8365 Ans @ P1101310) = pl-10 DL10) 5P1937.7%. < Docente e 1930 S ; Pl_124 <22-0.73) i plZc-0.73) - P12 < -1:24) : 0.2327-0.1075 [use 2 table] - 10.1252 / Ane