To explain:Whether the given statement is correct or not

Given information:
The given statement is,
"if $\bar{x}$ is the mean of a large (n > 30) simple random from a population with mean \mu and standard deviation $\sigma$ , then $\bar{x}$ is approximately normal with ${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}$.
The central limit theorem is one of the important concepts of the large sample theory.
According to the central limit theorem, for a sufficiently large sample size, the sampling distributions of the mean tend to be normal distribution, irrespective of the distribution of the population.Generally, a sample size more than 30 is considered a large sample.
The sampling distribution of the sample mean bar x for the large samples follows normal distribution with mean and variance $\frac{{\sigma }^2}{n}$ , where mu is the population mean, n is the sample size and ${\sigma }^2$ is the population variance, that is,
$\bar{x}simN(\mu ,\frac{\sigma }{\sqrt{n}})$
$\bar{x}simN(\mu ,{\sigma }_{\bar{x}})$
Thus, the provided statement is true.