The taxi and takeoff time for commercial jets is a random variable x with a mean of 8 minutes and a standard deviation of 2.8 minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway.
A button hyperlink to the SALT program that reads: Use SALT.
(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Round your answer to four decimal places.)
Incorrect: Your answer is incorrect.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? (Round your answer to four decimal places.)
Incorrect: Your answer is incorrect.

(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? (Round your answer to four decimal places.)
Incorrect: Your answer is incorrect.

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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

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