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Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is 13.4 years with a standard deviation of 1.7 years. Suppose that 40 airplanes were randomly selected from the fleets of these 10 carriers and were inspected for cracks in these airplanes that are considered too large for flying. What is the probability that the average age of these 40 airplanes is at least 14 years old? Round your answer to 4 decimal places. Remember to round your z-value to two decimal places.

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JeanaShupp

Answer: 0.0129

Step-by-step explanation:

Given : Mean : [tex]\mu=13.4\text{ years}[/tex]

Standard deviation : [tex]\sigma=1.7\text{ years}[/tex]

Sample size : [tex]n=40[/tex]

Let X be the random variable that represents the age of of the fleets.

Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x = 14

[tex]z=\dfrac{14-13.4}{\dfrac{1.7}{\sqrt{40}}}\approx2.23[/tex]

By using standard normal distribution table , the probability that the average age of these 40 airplanes is at least 14 years old will be

[tex]P(x\geq14)=P(z\geq2.23)=1-P(z<2.23)\\\\=1-0.9871262=0.0128738\approx0.0129[/tex]

Hence, the the probability that the average age of these 40 airplanes is at least 14 years old =0.0129

z scores of some value of some normal distribution is that value in standard normal distribution transformation of the specified normal distribution.

The probability that the average age of selected 40 airplanes is at least 14 years is 0.0129

We can use the z score for sample with size n and use that to get the p value(probability of getting at least that z score) from the z tables.

What is the z score for sample mean from a normal distribution?

Suppose that the population was normally distributed with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex]

Then, if a sample of size n is selected from it, then the z score for that sample's mean being [tex]\overline{x}[/tex] is given as

[tex]z = \dfrac{\overline{x} - \mu}{\sigma / \sqrt{n}}[/tex]

Suppose that we have X = random variable tracking the age of the airplane from fleet of those 10 largest US commercial passenger carriers.

Then, by the given data, we have:

[tex]X \sim N(\mu, \sigma)\\\mu = 13.4\\\sigma = 1.7\\or\\X \sim (13.4, 1.7)[/tex]

Since the sample taken was of size n = 40, and the sample mean is assumed to be [tex]\overline{x} = 14[/tex]

Thus, its z score is calculated as

[tex]z = \dfrac{\overline{x} - \mu}{\sigma / \sqrt{n}} = \dfrac{14 - 13.4}{1.7 / \sqrt{40}} \approx 2.23[/tex]

From the z tables, we get the p value as 0.9871 which tells that

[tex]P(Z \leq 2.23) = 0.9871[/tex]

The probability of the average age of these 40 airplanes to be "at least" 14 is equal to  [tex]P(Z \geq 2.23)[/tex] = 1 - 0.9871 = 0.0129 approximately.

Thus,

The probability that the average age of selected 40 airplanes is at least 14 years is 0.9871

Learn more about z score for sample mean here:

https://brainly.com/question/21118324

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