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 530 and standard deviation 8 and the second plane's speed is also normally distributed with mean and standard deviation 510 and 8, respectively. (a)What is the probability that after 2 hr 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 hr.

Answer 1

First we consider,

A = first plane and

B = scond plane

The values provided in the above problem is as

A~ N(530, 82)

B~ N(510, 82)

a) distance = speed * time

distanceA = 2A + 10

distanceB = 2B

we have to find P(distanceA - distanceB > 0)

Let random variable X = distance A - distance B

X = 2A + 10 - 2B

We have to find Mean of X by following way

mean X = mean(distanceA) - mean(distanceB)

mean X = mean(2*530) + mean(10)) - mean(2*510)

mean X = 50............(1)

We have to find Variance of X by following way

var X = var(distanceA) - var(distanceB)

var X = 22 * 82 + 22 * 82

var X = 512.............(2)

Here X ~ N(50, 512)

We have to find P(X >0)

That is

b) 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

mean Y = 2*530 - 2*510 = 40...........(3)

var Y = 512............(4)

Y~N(40,512)

Hence we have to find P(Y <= 0)

That is

Summary : -

a)

b)