Question

# 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 9 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 30 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 30 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 30 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.)

Given, = 9, = 3.5

Let X be the total taxi and takeoff time.

Then

X ~ N( n , n2)

= N(30 * 9 , 30 *3.52)

= N(270 , 367.5)

For normal distribution,

P( X < x) = P( Z < x - mean / standard deviation)

a)

P( X < 320) = P( Z < 320 - 270 / sqrt(367.5) )

= P( Z < 2.6082)

= 0.9954

b)

P( X > 275) = P( Z > 275 - 270 / sqrt(367.5) )

= P (Z > 0.2608)

= 1 - P( Z < 0.2608)

= 1 - 0.6029

= 0.3971

c)

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

= P(Z < 320 - 270 / sqrt(367.5) ) - P( Z < 275 - 270 / sqrt(367.5) )

= P( Z < 2.6082) - P( Z < 0.2608)

= 0.9954 - 0.6029

= 0.3926