Answer:
a) 0.9207 = 92.07% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes
b) 0.8907 = 89.07% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes
c) 0.8114 = 81.14% 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:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For sum of n normally distributed variables, the mean is [tex]\mu*n[/tex] and the standard deviation is [tex]s = \sigma\sqrt{n}[/tex].
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this question, we have that:
For a single commercial jet:
[tex]\mu = 8, \sigma = 2.8[/tex]
For the 37 jets combined:
[tex]\mu = 37*8 = 296, s = 2.8\sqrt{37} = 17.03[/tex]
(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?
This is the pvalue of Z when X = 320. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{320 - 296}{17.03}[/tex]
[tex]Z = 1.41[/tex]
[tex]Z = 1.41[/tex] has a pvalue of 0.9207
0.9207 = 92.07% 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?
This is 1 subtracted by the pvalue of Z when [tex]X = 275[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{275 - 296}{17.03}[/tex]
[tex]Z = -1.23[/tex]
[tex]Z = -1.23[/tex] has a pvalue of 0.1093
1 - 0.1093 = 0.8907
0.8907 = 89.07% 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?
This is the pvalue of Z when [tex]X = 320[/tex] subtracted by the pvalue of | when [tex]X = 275[/tex]
From a and b
0.9207 - 0.1093 = 0.8114
0.8114 = 81.14% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes
