Question

In: Statistics and Probability

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

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.5 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 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Round your answer to four decimal places.) Incorrect: Your answer is incorrect.

(b) What is the probability that for 37 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 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.)

Solutions

Expert Solution

We have modelled together the take off and taxi times for the jets.

We have been asked in the questions about the probability of the total time of a number of jets. Hence, we need the distribution of the sum of the times. Using the central limit theorem we have

We have 37 jets therefore n = 37

z-score = z-score = (x- E(X)) / (S(X)

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Round your answer to four decimal places.)

P(< 320 ) = P( Z < 0.61)

.....................using normal distribution tables

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? (Round your answer to four decimal places.)

P(> 275) = P( Z > -1.51)

= P( Z < 1.51)

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.)

P( 275 < <320) = P( < 320) - P( < 275)

= P(Z < 0.61) - P(Z < -1.51)

= P( Z < 0.61) - [1 - P( Z < 1.51)]

Using the probabilities from above we have


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