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A random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength μ = 5500 psi and the standard deviation is σ = 100 psi. Find the probability that the sample mean comprehensive strength exceeds 4985 psi. (Express your result to four significant digits.)

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Answer:


P( \bar X >4985)

And the z score is given by:


z = (\bar X -\mu)/((\sigma)/(√(n)))

And using this formula we got:


z = (4985-5500)/((100)/(√(9)))= -15.45)

And using the complement rule and the normal tandard distribution or excel we got:


P(z>-15.45) = 1-P(z<-15.45) \approx 1

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the comprehensive strength of a population, and for this case we know the following info:

Where
\mu=5500 and
\sigma=100

We select a sample size of n=9. Assuming that the distribution for X is normal then we can use the distribution for the sample mean
\bar X is given by:


\bar X \sim N(\mu, (\sigma)/(√(n)))

And we want this probability:


P( \bar X >4985)

And the z score is given by:


z = (\bar X -\mu)/((\sigma)/(√(n)))

And using this formula we got:


z = (4985-5500)/((100)/(√(9)))= -15.45)

And using the complement rule and the normal tandard distribution or excel we got:


P(z>-15.45) = 1-P(z<-15.45) \approx 1

User Terence Hill
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