Respuesta :
Answer:
[tex]X \sim N(2500,50)[/tex]
Where [tex]\mu=2500[/tex] and [tex]\sigma=50[/tex]
We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And the standard error would be:
[tex]SE = \frac{50}{\sqrt{8}}= 17.678 psi[/tex]
Step-by-step 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".
Solution to the problem
Let X the random variable that represent the compressive strength of concrete of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2500,50)[/tex]
Where [tex]\mu=2500[/tex] and [tex]\sigma=50[/tex]
We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And the standard error would be:
[tex]SE = \frac{50}{\sqrt{8}}= 17.678 psi[/tex]
Using the Central Limit Theorem, it is found that the standard error of the sample mean is of 17.678 psi.
The Central Limit Theorem states that for a sample of size n in a population with standard deviation [tex]\sigma[/tex], the standard error of the sample mean is given by:
[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we are given that [tex]\sigma = 50, n = 8[/tex]. Thus, the standard error of the sample mean is of:
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{8}} = 17.678[/tex]
The standard error of the sample mean is of 17.678 psi.
A similar problem is given at https://brainly.com/question/22934264
