# SOLUTION: The length of time a domestic flight must wait between gate departure and takeoff taxi at BWI International Airport is approximately normally distributed with a mean of 22 minutes

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 Click here to see ALL problems on Probability-and-statistics Question 270362: The length of time a domestic flight must wait between gate departure and takeoff taxi at BWI International Airport is approximately normally distributed with a mean of 22 minutes and a standard deviation of 7 minutes. a. what is the probability that a plane must wait between 18 and 24 minutes before taxiing for takeoff after leaving the gate? b. what is the probability a plane must wait between 22 and 30 minutes before taxiing for takeoff after leaving the gate? c. what is the probability a plane waits more than 12 minutes between gate departure takeoff taxi? d. what is the length of time such that only 5% of planes wait this long or longer before gate departure and taxiing down the runway? Answer by stanbon(60771)   (Show Source): You can put this solution on YOUR website!The length of time a domestic flight must wait between gate departure and takeoff taxi at BWI International Airport is approximately normally distributed with a mean of 22 minutes and a standard deviation of 7 minutes. ------------------------------------------------ a. what is the probability that a plane must wait between 18 and 24 minutes before taxiing for takeoff after leaving the gate? z(18) = (18-22)/7 = -4/7 z(24) = (24-22)/7 = 2/7 P(18< x <24) = P(-4/7 < z < 2/7) = 0.3286 --------------------------------------------------- b. what is the probability a plane must wait between 22 and 30 minutes before taxiing for takeoff after leaving the gate? Ans: Same procedure as above. ----- c. what is the probability a plane waits more than 12 minutes between gate departure takeoff taxi? z(12) = (12-22)/7 = -10/7 = -1.4286 P(x > 12) = P(z > -1.4286) Note: z = -1.4286 is a point 1.4286 standard deviations to the left of the mean. You want the area under the normal curve that is to the right of -1.4286. Use Your z-chart or calculator, if you have one. Using a TI calculator I get normalcdf(-1.4286,100) = 0.9234 --- Comment: If these calculations are a mystery I would strongly suggest you use a sketch of the normal curve as you work each problem. That is so you can see where your x-values, z-values, and areas under the curve really are. You will better understand what is going on when you get the answers. ------------ d. what is the length of time such that only 5% of planes wait this long or longer before gate departure and taxiing down the runway? Ans: Find the z-value that corresponds to a left-tail of 5%. invNorm(0.5) = -1.645 Find the corresponding x-value: x = zs + u x = -1.645*7 + 22 x = 10.485 minutes ======================= Cheers, Stan H.