Statistics problem.....help!

2009-07-02 7:32 am
The goal at U.S. airports handling international flights is to clear these flights within 45 minutes. Let's interpret ths to mean that 95% of the
flights are cleared in 45 min, 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 min, what is the mean time to clear a flight?

b) Suppose the standard deviation is 10 min, not the 5 min suggested
in part a. What is the new mean?

c) A customer has 30 min from her flight landed to catch her
limousine. Assuming a standard deviation of 10 min, what is the
likelihood that she will be cleared in time?
更新1:

回king_of_physics: For Q1 & Q2, 95% should be 1.96 not 1.645, right? so Q1 answer= 35.2 and Q2 answer = 25.4 how to calculate Q3? Many thanks!

回答 (2)

2009-07-02 3:55 pm
✔ 最佳答案
Let u be the mean time.

Let X be the random variable of time to clear the flight.


1. X ~ N(u , 52)

P(X <= 45) = 95%

P[z <= (45 - u)/5] = 0.95

(45 - u)/5 = 1.645

Mean time, u = 36.775 min


2. X ~ N(u , 102)

P(X <= 45) = 95%

P[z <= (45 - u)/10] = 0.95

(45 - u)/10 = 1.645

New mean time, u = 28.55 min


3. The mean is now 28.55 min

X ~ N(28.55 , 102)

P(X <= 58.55)

= P[z <= (58.55 - 28.55)/10]

= P(z <= 3)

= 0.5 + 0.4987

= 0.9987



2009-07-03 20:21:47 補充:
No, 95% should be 1.645, not 1.96. It is because this is a one-tailed test, not two tailed. If it is two-tailed test, then it is 1.96.

So, question 3 is correct.
參考: Physics king
2009-07-03 9:41 pm
十卜king_of_physics


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