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 8.4 minutes and a standard deviation of 3.1 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 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.)

Explanation / Answer

Solution:

Total of 36 jets taxi and takeoff time will have a normal distribution with mean (36)(8.4) = 302.4 minutes and standard deviation 36( 3.1) = 111.6 minutes

a)
= 302.4
= 111.6
standardize x to z = (x - ) /
P(x < 320) = P( z < (320-302.4) / 111.6)
= P(z < 0.1577) = 0.5636
(From Normal probability table)

b)
= 302.4
= 111.6
standardize x to z = (x - ) /
P(x > 275) = P( z > (275-302.4) / 111.6)
= P(z > -0.2455)
= P(Z < 0.2455) = 0.5987
(From Normal probability table)

c)
= 302.4
= 111.6
standardize x to z = (x - ) /
P( 275 < x < 320) = P[( 275 - 302.4) /111.6 < Z < ( 320 - 302.4) / 111.6]
P( -0.2455 < Z < 0.1577) = 0.4013 - 0.5636 = 0.1623
(From Normal probability table)

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