Answer:
a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that
.
(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?
We are working with a sample mean of 37 jets. So we have that:
![s = \frac{3.3}{\sqrt{37}} = 0.5425](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B3.3%7D%7B%5Csqrt%7B37%7D%7D%20%3D%200.5425)
Total time of 320 minutes for 37 jets, so
![X = \frac{320}{37} = 8.65](https://tex.z-dn.net/?f=X%20%3D%20%5Cfrac%7B320%7D%7B37%7D%20%3D%208.65)
This probability is the pvalue of Z when
. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{8.65 - 8.3}{0.5425}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B8.65%20-%208.3%7D%7B0.5425%7D)
![Z = 0.65](https://tex.z-dn.net/?f=Z%20%3D%200.65)
has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.
(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?
Total time of 275 minutes for 37 jets, so
![X = \frac{275}{37} = 7.43](https://tex.z-dn.net/?f=X%20%3D%20%5Cfrac%7B275%7D%7B37%7D%20%3D%207.43)
This probability is subtracted by the pvalue of Z when ![X = 7.43](https://tex.z-dn.net/?f=X%20%3D%207.43)
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{7.43 - 8.3}{0.5425}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B7.43%20-%208.3%7D%7B0.5425%7D)
![Z = -1.60](https://tex.z-dn.net/?f=Z%20%3D%20-1.60)
has a pvalue of 0.0548.
There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.
(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?
Total time of 320 minutes for 37 jets, so
![X = \frac{320}{37} = 8.65](https://tex.z-dn.net/?f=X%20%3D%20%5Cfrac%7B320%7D%7B37%7D%20%3D%208.65)
Total time of 275 minutes for 37 jets, so
![X = \frac{275}{37} = 7.43](https://tex.z-dn.net/?f=X%20%3D%20%5Cfrac%7B275%7D%7B37%7D%20%3D%207.43)
This probability is the pvalue of Z when
subtracted by the pvalue of Z when
.
So:
From a), we have that for
, we have
, that has a pvalue of 0.7422.
From b), we have that for
, we have
, that has a pvalue of 0.0548.
So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.