Answer :
Answer:
[tex] \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})[/tex]
In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.
Step-by-step explanation:
For this case we have that the sample size is n =6
The sample man is defined as :
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And we want a normal distribution for the sample mean
[tex] \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})[/tex]
In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.
So for this case we need to satisfy the following condition:
[tex] X_i \sim N(\mu , \sigma), i=1,2,...,n[/tex]
Because if we find the parameters we got:
[tex] E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n}=\mu[/tex]
[tex] Var(\bar X)= \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{n\sigma^2}{n^2}= \frac{\sigma^2}{n}[/tex]
And the deviation would be:
[tex] Sd (\bar X) = \frac{\sigma}{\sqrt{n}}[/tex]
And we satisfy the condition:
[tex] \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})[/tex]