Question

In a manufacturing process, a machine produces bolts that have an average length of 5 inches with a variance of .08. If we randomly select five bolts from this process, what is the standard deviation of the sampling distribution of the sample mean

Answer 1

Answer:

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

And replacing:

$\mu_{\bar X}= 5$

And the deviation:

$\sigma_{\bar X}= \frac{0.283}{\sqrt{5}}= 0.126$

And the distribution is given:

$\bar X \sim N(\mu= 0.08, \sigma= 0.126)$

Step-by-step explanation:

For this case we have the following info given :

$\mu= 5. \sigma^2 =0.08$

And the deviation would be $\sigma = \sqrt{0.08}= 0.283$

For this case we select a sample size of n = 5 and the distirbution for the sample mean would be:

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

And replacing:

$\mu_{\bar X}= 5$

And the deviation:

$\sigma_{\bar X}= \frac{0.283}{\sqrt{5}}= 0.126$

And the distribution is given:

$\bar X \sim N(\mu= 0.08, \sigma= 0.126)$