Respuesta :
Answer:
[tex] \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})[/tex]
And replacing:
[tex] \mu_{\bar X}= 5[/tex]
And the deviation:
[tex] \sigma_{\bar X}= \frac{0.283}{\sqrt{5}}= 0.126[/tex]
And the distribution is given:
[tex] \bar X \sim N(\mu= 0.08, \sigma= 0.126)[/tex]
Step-by-step explanation:
For this case we have the following info given :
[tex] \mu= 5. \sigma^2 =0.08[/tex]
And the deviation would be [tex] \sigma = \sqrt{0.08}= 0.283[/tex]
For this case we select a sample size of n = 5 and the distirbution for the sample mean would be:
[tex] \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})[/tex]
And replacing:
[tex] \mu_{\bar X}= 5[/tex]
And the deviation:
[tex] \sigma_{\bar X}= \frac{0.283}{\sqrt{5}}= 0.126[/tex]
And the distribution is given:
[tex] \bar X \sim N(\mu= 0.08, \sigma= 0.126)[/tex]
