A goal of U.S. airports handing international flights is to clear these flights

Question

A goal of U.S. airports handing international flights is to clear these flights within 45 minutes. Let's interpret this to mean that 95% of the fights are cleared minutes, so 5% of the flights take longer to clear. Let's also assume that the distribution is approximately normal. (a) If the standard deviation of the time to clear an international flight is 5 minutes, what is the mean time to clear a flight? Mean (b) Suppose the standard deviation is 10 minutes, not the 5 minutes suggested in part (a) What is the new mean? New Mean (c) A customer has 30 minutes from the time her flight lands to catch her. Assuming a standard deviation of 10 minutes, what is the likelihood that she get her luggage in time? Probability

Explanation / Answer

a) Here, x bar = 45 , sd= 5

At 95% confidence interval z value = 1.64 for one tail

z = (x - mean ) / sd

mean = x - z*sd

mean = 45 - 1.64 * 5

= 36.8

b)

Here, x bar = 45 , sd=10

At 95% confidence interval z value = 1.64 for one tail

z = (x - mean ) / sd

mean = x - z*sd

mean = 45 - 1.64 * 10

= 28.6

c) P(x < 30)

By normal distribution formula,

z = ( x - mean ) / sd

= ( 30 - 28.6 ) / 10

= 0.14

Now, we need to find P( z < 0.14)

p(x <30 ) = P(z <0.14) = .5557

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