Answer:
[tex]\bar X \sim N( \mu, \frac{\sigma}{\sqrt{n}})[/tex]
And for the parameters we have:
[tex]\mu_{\bar X}= 36[/tex]
[tex]\sigma_{\bar X}=\frac{8}{\sqrt{17}}=1.94[/tex]
And the best option would be:
B) 36 and 1.94
Step-by-step explanation:
From this case we have the following info given:
[tex] n = 17[/tex] represent the sample size
[tex] N=200[/tex] represent the population size
[tex] \mu = 36[/tex] represent the mean
[tex]\sigma = 8[/tex]
For this case the distribution for the sample mean would be approximately as:
[tex]\bar X \sim N( \mu, \frac{\sigma}{\sqrt{n}})[/tex]
And for the parameters we have:
[tex]\mu_{\bar X}= 36[/tex]
[tex]\sigma_{\bar X}=\frac{8}{\sqrt{17}}=1.94[/tex]
And the best option would be:
B) 36 and 1.94
