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(75887)   (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.
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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
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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.
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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
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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.
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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
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Cheers,
Stan H.
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