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 545 and standard deviation 12 and the second plane's speed is also normally distributed with mean and standard deviation 520 and 12, respectively.

(a) What is the probability that after 2 hr of flying, the second plane has not caught up to the first plane? (Round your answer to four decimal places.)


(b) Determine the probability that the planes are separated by at most 10 km after 2 hr. (Round your answer to four decimal places.)

Explanation / Answer

Ans:

Measure position as x, with x=0 at the position of the second (lagging) plane at time 0. Let X be the position of the first (leading) plane after two hours, and let X be the position of the second plane after two hours.

X = 2V+10 where V N(545,12), so X N(1100,24), and
X = 2V where V N(520,12), so X N(1040,24).

It should have been in the problem statement, but in order to solve this problem, we need to know how V and V are jointly distributed. The question designer probably expects us to assume that they are independent, so we will assume that.
Then
X-X N(60,(24²+24²)) = N(60,242).

(a) P(X-X>0)

z=(0-60)/242=-1.77

P(z>-1.77)=0.9616

(b) P(-10<X-X<10)

z=(10-60)/242=1.47

P(-1.47<=z<=1.47)=P(z<=1.47)-P(z<=-1.47)=0.9293-0.0708=0.8584

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