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andrew-mc [135]
2 years ago
10

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 s

tandard 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.
Mathematics
1 answer:
Sliva [168]2 years ago
6 0

Answer: 0.0129

Step-by-step explanation:

Given : Mean : \mu=13.4\text{ years}

Standard deviation : \sigma=1.7\text{ years}

Sample size : n=40

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

Z-score : z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x = 14

z=\dfrac{14-13.4}{\dfrac{1.7}{\sqrt{40}}}\approx2.23

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

P(x\geq14)=P(z\geq2.23)=1-P(z

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

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Answer:

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Step-by-step explanation:

<em>"One side of a rectangle is 3 feet shorter than twice the other side find the sides if the area is 209 feet squared"</em>

Ok, so this is a tricky one- <em> 209 feet squared, </em>not 209 square feet, therefore the area is 209^2, or 43,681.

Next, let's define our variables-

<em>x= the "other side"</em>

<em>z= 3 feet shorter than twice side</em>

We can now make these (useful) equations

z=2x-3

43681= (2x-3)+x

We will focus on the latter for now-

Simplify

43681= (2x-3)+x

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/3

14561.3333...=x

z=2x-3

z=2*(14561.3333)-3

z=29119.66666....

29119.66666...+14561.3333...=43681

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