Questions A, B and C

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 3.5minutes. 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 36 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 36 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 36 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

1.) You are asked to find the following:

a.) p(sum(x)) < 320 min

b.) p(sum(x)) > 275 min

c.) 275 < sum(x) < 320

1.) You are given the following information in the problem:

1. The population of taxi-takeoff times is normally distributed with mu = 8.2 and sigma = 3.5
2. N airplanes are involved, i.e., N = 36

2.) Very important point: The sum of statistically independent (i.e., outcomes don't depend on each other) normally distributed random variables is, itself, a normally distributed random variable with mu = N*mu, and sigma = sigma/sqrt(N). So, your new mean = (36)*(8.2), and your new sigma = 3.5/sqrt(36)

3.) Next, go to work with your Z tables or excel functions:

a.) Convert 320 and 275 to z values, Z = (x - new mu)/(new sigma), e.g., (320 - (36)*(8.2))/(3.5/sqrt(36))

b.)

1. p(Z <= z) that you found for 320 = p(sum(x)) < 320
2. p(Z > z), calculate 1 - p(Z <= z) for the Z you found for 275. this will give you p(sum(x)) > 275

c.) The difference of these two probabilities will give you: 275 < p(sum(x)) < 320

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