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 8.6 minutes and a standard deviation of 3 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 32 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 32 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 32 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

mean=8.6
dev=3 --> var=3² = 9

The mean and variance for the total time of 32 jets are

mean=32*8.6 = 275.2

variance = 32*9 = 288 --> dev = 288 = 16.9705

Normal (275.2 , 16.9705)

a)

P(X<320)

Standarize with Z=(X-mean)/dev

X=320 --> Z=(320-275.2) / 16.9705 = 2.63

P(X<320) = P(Z<2.63) = using the tables = 0.9957

b)

P(X>275)

Standarize with Z=(X-mean)/dev

X=275 --> Z=(275-275.2) / 16.9705 = -0.0117

P(X>275) = P(Z>-0.0117) = 0.504 using the tables

c)

P(275<X<320) =

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

P(X<320) = 0.9957 (see point a)

P(X<275) =1P ( Z<0.0117 )=10.504= 0.496

P(X<320) - P(X<275) = 0.9957 - 0.496 = 0.4997

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