Question

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 2.8 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 35 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 35 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 35 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.) 7. 2/2 points

Answer 1

MEAN= 8.3 MINUTES AND S.D= 2.8 MINUTES

a) Approximately normal since

Xbar= 320/35= 9.14

s.e = 2.8/sqrt(35)=2.8/5.916=0.47

P ( Xbar<9.14 )=P ( Xbar−μ<9.14−8.3 )=P ((Xbar−μ)/s.e<(9.14−8.3)0.47)

Since (x−μ)/s.e=Z and (9.14−8.3)/0.47=1.79 we have:

P (X<9.14)=P (Z<1.79)

Use the standard normal table to conclude that:

P (Z<1.79)=0.9633

b) Xbar= 275/35=7.86

S.E= 2.8/sqrt(35)=0.47

P ( Xbar>7.86 )=P ( Xbar−μ>7.86−8.3 )=P ( Xbar−μ/s.e>(7.86−8.3)/0.47)
(7.86−8.3)/0.47=−0.94 we have:
P ( Xbar>7.86 )=P ( Z>−0.94 )
Use the standard normal table to conclude that:
P (Z>−0.94)=0.8264

c) P ( 7.86<<9.14 )=P ( 7.86−8.3< −μ<9.14−8.3 )=P ((7.86−8.3)/0.47<(−μ)/σ<(9.14−8.3)/0.47)

Since Z=(xbar−μ)/S.E ,

(7.86−8.3)/0.47=−0.94 and (9.14−8.3)/0.47=1.79 we have:

P ( 7.86<Xbar<9.14 )=P ( −0.94<Z<1.79 )

Use the standard normal table to conclude that:

P ( −0.94<Z<1.79 )=0.7897