Question

If a random sample of size nequals=6 is taken from a​ population, what is required in order to say that the sampling distribution of x overbarx is approximately​ normal?

Answer 1

Answer:

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

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 :

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And we want a normal distribution for the sample mean

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

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:

$X_i \sim N(\mu , \sigma), i=1,2,...,n$

Because if we find the parameters we got:

$E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n}=\mu$

$Var(\bar X)= \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{n\sigma^2}{n^2}= \frac{\sigma^2}{n}$

And the deviation would be:

$Sd (\bar X) = \frac{\sigma}{\sqrt{n}}$

And we satisfy the condition:

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