Two airplanes are flying in the same direction in adjacent parallel corridors. A
ID: 3220960 • Letter: T
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 8 and the second plane's speed is also normally distributed with mean and standard deviation 505 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? (Round your answer to four decimal places.) Incorrect: Your answer is incorrect. . (b) Determine the probability that the planes are separated by at most 10 km after 2 hr. (Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question.
Explanation / Answer
ans=
Let X_1 be the first plane's speed, and let X_2 be the second plane's speed.
E(X_1) = 520; 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(520) - 2(505)
= 40.
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 .