The taxi and takeoff time for commercial jets is a random variable x with a mean

Question

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.1 minutes and a standard deviation of 2.7 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 35 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 35 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 35 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal

Explanation / Answer

a) Mu = 8.1
Sigma = 2.7
Z = ((X – mu) / sigma) = (((320/35) – 8.1) / 2.7) = 0.386
So the probability that for 35 jets on a given runway, total taxi and takeoff time will be less than 320 minutes is
P (Z < 0.386) = 0.6480

(b)
Here,
Mu = 8.1
Sigma = 2.7
Z = ((X – mu) / sigma) = (((275/35) – 8.1) / 2.7) = -0.0899
So the probability that for 39 jets on a given runway, total taxi and takeoff time will be more than 275 minutes is
P (Z > -0.0899) =1- 0.5319=0.4681

c) z1 = ((275/35-8.1)/2.7 =-0.0899
= P ( Z <-00899) From Standard Normal Table
=0.4681
Z2 = ((X – mu) / sigma) = (((320/35) – 8.1) / 2.7) = 0.386

P (Z < 0.386) = 0.6480 From Standard Normal Table

P(Z1<X>Z2)=0.6480-0.4681= 0.1799

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