If x1,x2,x3...xn is a random sample without replacement from a finite population of Y1,Y2,Y3,.....YN of N values, with population mean M and standard deviation sigma or variance sigma^2, then the random sample x1,x2,x3....xn has the sample mean = (sum of all n xi 's)/n = xbar, say. Sorry for notational gimmicks done!
The distribution of sample mean, xbar can be proved to have the mean M and the variance = {sigma^2/n}{N-n)/(n-1)}.
Therefore,
variance of the sample mean = variance (xbar) = (sigma^2/n)*{(N-n)/(N-1)} (1)
The standard deviation of sample mean= (sigma/n)sqrt{(N-n)/(N-1)} (2)
Thus the variance of the sample mean of sample of size n , drawn from a finite population of size N and the population variance, sigma ^2 are connected by the relation at (1) and the standard deviation by (2)
The factor sqrt{(N-n)/(N-1)} involving population size,N and sample size,n is called the finite population correction factor, or in a abbreviation, FPCF.
For n=1, the sample is equivalent to without replacement an this is equivalent to the entire popilation itself. fpc=1 and variance (xbar)=sigma^2.
For large N, the FPCF = (N-n)/(N-1) is nearly 1. But when n is comparatively greater than 5% of the population size, N, we use FPCF to detrmine the variance of the sample mean.
The distribution of sample mean helps us to make the probabilistic statemnts about the sample mean in the theory of Inference. It also makes us determine the confidence limits of the domain of the population parameters like population mean. It also helps us to decide whether our sample is drawn from a population of with particular mean and variance.
For accademic interest you can refer to any text books on Exact Samplig theory and topics like distribution of sample mean and how sample mean tends to Normal Distribution despite parent population not following the Normal Distribution under the influence of Central Limit Theorem.
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