Two airplanes are flying in the same direction in adjacent p

Two airplanes are flying in the same direction in adjacent parallel corridors. At time t=0, the first airplane is 10km 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 plane speed, independent of the first is also normally distributed with mean and standard deviation 500 and 10, respectively.

a.What is 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 10km after 2 hr.

Solution

Note: I am giving u answer of both part together.

To solve this problem, I will need to assume that the planes\' speeds are independent of each other and also do not change with time.

Let X_1 be the first plane\'s speed, and let X_2 be the second plane\'s speed.
E(X_1) = 525; E(X_2) = 505; Var(X_1) = Var(X_2) = 8^2.

Let D be the number of km the first plane is ahead of the second plane, after 2 hours. We need to compute P(-10 <= D <= 10).

Since at t=0, the first plane is 10 km ahead of the second one, we have
D = 10 + 2X_1 - 2X_2, which is also normally distributed.
We need to find D\'s mean and standard deviation.

E(D) = E(10 + 2X_1 - 2X_2) = 10 + 2E(X_1) - 2E(X_2)
= 10 + 2(525) - 2(505)
= 50.

Var(D) = Var(10 + 2X_1 - 2X_2)
= Var(2X_1 - 2X_2), since adding a constant doesn\'t affect variance
= Var(2X_1) + Var(-2X_2) since X_1 and X_2 are independent (and so variances add)
= 2^2 Var(X_1) + (-2)^2 Var(X_2), from the property Var(aX) = a^2 Var(X)
= (2^2)(8^2) + (2^2)(8^2)
= 2*(2*8)^2

St.Dev(D) = sqrt(Var(D)) = sqrt(2*(2*8)^2) = 16sqrt(2).

Finally, the probability that the planes are separated by at most 10 km after 2 hr is
P(-10 <= D <= 10)
= P((-10 - E(D))/St.Dev(D) <= Z <= (10 - E(D))/St.Dev(D))
= P((-10 - 50)/(16sqrt(2)) <= Z <= (10 - 50)/(16sqrt(2)))
= P((-15/8)sqrt(2) <= Z <= (-5/4)sqrt(2))
= P((5/4)sqrt(2) <= Z <= (15/8)sqrt(2)) from the symmetry of the normal curve
= P(Z <= (15/8)sqrt(2)) - P(Z < (5/4)sqrt(2))
= P(Z <= 2.65) - P(Z <= 1.77)
= 0.9960 - 0.9616 from using the normal table
= 0.0344 .