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 3.4 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 38 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 38 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 38 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

Total of 38 jets taxi and takeoff time will have a normal distribution with mean (38)(8.4) = 319.2 minutes and standard deviation 38( 3.4) = 129.2 minutes

a)
= 319.2
= 129.2
standardize x to z = (x - ) /
P(x < 320) = P( z < (320-319.2) / 129.2)
= P(z < 0.0061) = 0.504
(From Normal probability table)

b)
= 319.2
= 129.2
standardize x to z = (x - ) /
P(x > 275) = P( z > (275-319.2) / 129.2)
= P(z > -0.3421) = 0.6331
(From Normal probability table)

c)
= 319.2
= 129.2
standardize x to z = (x - ) /
P( 275 < x < 320) = P[( 275 - 319.2) / 129.2 < Z < ( 320 - 319.2) / 129.2]
P( -0.3421 < Z < 0.0061 ) = 0.6331 - 0.504 = 0.1291
(From Normal probability table)

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