Question 1076003
 Suppose that long-distance domestic flight arrival times follow normal distributions that all have standard deviation 10 minutes. 
a. You book a flight from Honolulu that is scheduled to arrive in Denver at 3:30 pm.
i. What is the probability that the flight actually arrives in Denver 
before 3:00 pm?
z(3) = (3-3.5)/10 = -30min/10min = -3
P(x < 3:00) = P(z< -3) = normalcdf(-100,-3) = 0.0013
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ii. What is the probability that the flight actually arrives in Denver 
after 3:40 pm?
z(3.40) = (3:40-3:30)/10 = 1
P(x > 3.40) = P(z > 1) = normalcdf(1,100) = 0.1587
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b. You have a connecting flight that is scheduled to depart from Denver to Chicago at 4:20 pm. It ordinarily takes 15 minutes for those seated in economy to completely deplane and another 10 minutes for you to walk (or run) from your arrival gate to your departure gate. By airline policy, the airplane door will close 10 minutes prior to departure time. What is the probability that the boarding door will close (and you will miss your connecting flight) by the time you get to the gate?
 Time from 3:30 to 4:10 = 40 min
Time to get there = 25 min
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Landing time for you to miss connection:: > 3:30+15 min > 3:45
z(3:45) = (3:45-3:30)/10 = 15/10 = 1.5
P(x > 3:45) = P(z > 1.5) = normalcdf(1.5,100) = 0.0668
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Cheers,
Stan H.
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