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 8.2 minutes and a standard deviation of 2.9 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 33 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 33 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 33 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 33 jets taxi and takeoff time will have a normal distribution with mean (33)(8.2) = 270.6 minutes and standard deviation 33( 2.9) = 95.7 minutes

a)
= 270.6
= 95.7
standardize x to z = (x - ) /
P(x < 320) = P( z < (320-270.6) / 95.7)
= P(z < 0.5161) = 0.6985
(From Normal probability table)

b)
= 270.6
= 95.7
standardize x to z = (x - ) /
P(x > 275) = P( z > (275-270.6) / 95.7)
= P(z > 0.0459) = 0.5199
(From Normal probability table)

c)
= 270.6
= 95.7
standardize x to z = (x - ) /
P( 275 < x < 320) = P[( 275 - 270.6) / 95.7 < Z < ( 320 - 270.6) / 95.7]
P( 0.0459< Z < 0.5161) =0.6985 - 0.5199 = 0.1786
(From Normal probability table)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.