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 525 and standard deviation 11 and the second plane's speed is also normally distributed with mean and standard deviation 505 and 11, 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.)
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(b) Determine the probability that the planes are separated by at most 10 km after 2 hr. (Round your answer to four decimal places.)

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Explanation / Answer

here let plance 1st is A and plane 2 is B

hence mean distance after two hours =2*(525-505) =40

and std deviation of mean distance =(112+112)1/2*(2)1/2 =22

hence a) P(A-B>-10) =1-P(A-B<-10) =1-P(Z<(-10-40)/22) =1-P(Z<-2.2727)=1-0.0115=0.9885

b) P(-20<A-B<0) =P(-2.7272<Z<-1.8182)=0.0345-0.0032 =0.0313

please revert

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