Question

assume that x follows a normal distribution, and we take samples of size 16. can you apply the clt to conclude that (\bar{x}) follows a normal distribution?

Explanation:

Step1: Recall the Central Limit Theorem (CLT)

The CLT states that for a sample of size \(n\) from a population (regardless of the population distribution), the sampling distribution of the sample mean \(\bar{X}\) is approximately normal if \(n\) is large (\(n\geq30\) is a common rule - of - thumb).

Step2: Check the sample size condition

Here, the sample size \(n = 16\). Since \(16<30\), and we are only told that \(X\) follows a normal distribution (not using the CLT assumption about the population distribution). If \(X\) is normally distributed, then \(\bar{X}=\frac{1}{n}\sum_{i = 1}^{n}X_{i}\) (where \(X_{i}\) are i.i.d. copies of \(X\)) is exactly normal. The formula for the mean of \(\bar{X}\) is \(\mu_{\bar{X}}=\mu\) (where \(\mu\) is the mean of \(X\)) and the formula for the variance of \(\bar{X}\) is \(\sigma_{\bar{X}}^{2}=\frac{\sigma^{2}}{n}\) (where \(\sigma^{2}\) is the variance of \(X\)). But if we were trying to use the CLT (which is about non - normal populations), \(n = 16\) is not large enough.

Answer:

No, we cannot apply the CLT (in the sense of using it for non - normal populations) since \(n=16<30\). However, if \(X\) is normally distributed, \(\bar{X}\) is exactly normal (not because of the CLT for non - normal populations, but because of the properties of normal distributions and linear combinations of normal random variables).