Question

The compressive strength of concrete is normally distributed with mu = 2500 psi and sigma = 50 psi. A random sample of n = 8 specimens is collected. What is the standard error of the sample mean? Round your final answer to three decimal places (e.g. 12.345). The standard error of the sample mean is [ ]psi.

Answer 1

Answer:

$X \sim N(2500,50)$  

Where $\mu=2500$ and $\sigma=50$

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 $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$SE = \frac{50}{\sqrt{8}}= 17.678 psi$

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:

$X \sim N(2500,50)$  

Where $\mu=2500$ and $\sigma=50$

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 $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$SE = \frac{50}{\sqrt{8}}= 17.678 psi$