Question

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.2 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 34 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 34 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 34 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

It is given that the taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.2 minutes.

Now, the total of 34 jets taxi and takeoff time will have a normal distribution with mean as (34)(8.9) = 302.6 minutes and standard deviation as (34)(2.2) = 74.8 minutes.

So, Mean() = 302.6 , Standard deviation () = 74.8

a) Here, we have to find the probability that for 34 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

So, here, we will find P(x<320)

Thus, standardize x to z =

So, P(x<320) = P[z < {(320-302.6)/74.8}]

= P(z < 0.233) = 0.59095 [from normal probability table]

Thus, P(x<320) = 0.59095

Thus, the required probability is 0.5910(rounded up to four decimal places).

Thus, the required probability that for 34 jets on a given runway total taxi and takeoff time will be less than 320 minutes = 0.5910 .

b) Here, we have to find the probability that for 34 jets on a given runway total taxi and takeoff time will be more than 275 minutes.

So, here we will find P(x>275) .

Thus, standardize x to z =

So, P(x>275) = P[z > {(275-302.6)/74.8}]

= P(z > -0.37) = 0.64431 [from normal probability table]

Thus, P(x>275) = 0.64431

Thus, the required probability is 0.6443(rounded up to four decimal places).

Thus, the required probability that for 34 jets on a given runway total taxi and takeoff time will be more than 275 minutes = 0.6443 .

c) Here, we have to find the probability that for 34 jets on a given runway total taxi and takeoff time will be between 275 and 320 minutes.

Thus, here, we have to find P(275<x<320)

Thus, standardize x to z =

So, P(275<x<320) = P[{(275-302.6)/74.8} < z < {(320-302.6)/74.8}]

= P(-0.37 < z < 0.233) = 0.5910 - 0.35569 = 0.23531 [from normal probability table]

Thus, P(275<x<320) = 0.2353(rounded up to four decimal places).

Thus, the required probability that for 34 jets on a given runway total taxi and takeoff time will be between 275 and 320 minutes = 0.2353 .