Respuesta :
Answer:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And replacing we got:
[tex]\mu_{\bar X}= 2.2[/tex]
[tex]\sigma_{\bar X}= \frac{6}{\sqrt{100}}= 0.6[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] \mu = 2.2[/tex] represent the mean
[tex]\sigma = 6[/tex] represent the deviation
We select a sample size of n=100. This sample is >30 so then we can use the central limit theorem. And we want to find the distribution for the sample mean and we know that the distribution is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And replacing we got:
[tex]\mu_{\bar X}= 2.2[/tex]
[tex]\sigma_{\bar X}= \frac{6}{\sqrt{100}}= 0.6[/tex]
