Question

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.2 minutes and a standard deviation of 2.8 minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway. (a) What is the probability that for 40 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Round your answer to four decimal places.) (b) What is the probability that for 40 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? (Round your answer to four decimal places.) (c) What is the probability that for 40 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.)

Answer 1

a)
Here, μ = 8.2*40 = 328,
σ = 2.8*sqrt(40) = 17.7088 and x = 320. We need to compute P(X <= 320). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (320 - 328)/17.7088 = -0.45

Therefore,
P(X <= 320) = P(z <= (320 - 328)/17.7088)
= P(z <= -0.45)
= 0.3264

b)
P(X >= 275) = P(z <= (275 - 328)/17.7088)
= P(z >= -2.99)
= 1 - 0.0014 = 0.9986

c)
P(275 <= X <= 320) = P((320 - 328)/17.7088) <= z <= (320 - 328)/17.7088)
= P(-2.99 <= z <= -0.45) = P(z <= -0.45) - P(z <= -2.99)
= 0.3264 - 0.0014
= 0.3250