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 and standard deviation 10 and the second planes speed is also normally distributed with mean and standard deviation 500 and 10, respectively. What is the probability that after 2 hr of flying, the second plane has not caught up to the first plane?

Answer 1

I worked out the distance for both of these:

Distance for plane A: 2(520) + 10 = 1050
Distance for plane B: 2(500) = 1000

Standard deviation for both is 10, but I'm not sure whether I should multiply these by 2 as well. Should it be:

\[\sigma = \sqrt{2^210^2 + 2^210^2} = 28.28\] I'm not sure how to set up the probability from here.

Answer 2

Should my probability be: \[\phi(1.77)\] The z-table gives me 0.9616 = ~96.2%