Two airplanes are flying in the same direction in adjacent parallel corridors. A

Question

Two airplanes are flying in the same direction in adjacent parallel corridors. At time t = 0, the first airplane is 10 km ahead of the second one. Suppose the speed of the first plane (km/hr) is normally distributed with mean 520 km and standard deviation 10 km, and the second plane's speed is also normally distributed with mean and standard deviation 500 km and 10 km, respectively (a) Find the probability that after 2 hours of flying, the second plane has not caught up to the first plane. (b) Determine the probability that the planes are separated by at most 10 km after 2 hours.

Explanation / Answer

A~N(520, 102 )

B~N(500, 102 )

distance = speed * time

distanceA = 2A + 10

distanceB = 2B

You want to find P(distanceA - distanceB > 0)

Let random variable X = distance A - distance B

X = 2A + 10 + 2B

meanX = mean(distanceA) - mean(distanceB)

meanX = mean(2*520) + mean(10)) - mean(2*500)

meanX = 50

varX = var(distanceA) - var(distanceB)

varX = 22 * 102 + 22 * 102

varX = 800

X~N(50,800)

Hence, P(X>0) = 96.15%

For the second part:

You want to find P(distanceA - distanceB <= 10)

Let the random variable Y be such that distanceA - distanceB <= 10

Y = 2A - 2B

meanY = 2*520 - 2*500 = 40

varY = 800

Y~N(40,800)

Hence, P(Y<=0) = 7.86%

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