Central limit theorem helps us understand the features of 'population of the means' that have been generated from the mean of infinite numbers of random population sample of size. Each of them drawn from the given 'parent population'. Normal distribution curve is handy in computing the probabilities related to continuous variables which are normally distributed. Central Limit theorem asserts that the sampling distributions of large sample size are approximately normal, even if the population distribution itself is not normal. This answers the question about sample means and allows normal distribution to be used to find the probabilities of individual values in the samples.

The statement of Central Limit theorem for Sampling distribution of sample means is as follows:

As the sample size n increases without limit, the shape of the distribution of sample means taken with replacement from a population with mean μ and standard deviation σ will approach a normal distribution.The mean and standard deviation of this distribution are denoted by $\mu _{\overline{x}}$ and $\sigma _{\overline{x}}$ and are correspondingly equal to μ and $\frac{\sigma }{\sqrt{n}}$.
The standard deviation of this sampling distribution $\sigma _{\overline{x}}$ is called the standard error of the sampling distribution.
Sampling distribution of sample means for large samples is approximately normal and approaches normal distribution even when the population from which the samples are drawn is not normally distributed.
Central limit theorem also states about the normal shape of the sampling distributions of sample proportions, difference of means and difference of proportions giving the formulas for mean and standard deviations of such distributions
Following is a list of formulas to describe the sampling distributions of various sample statistics and the conditions that apply given by the corresponding versions of Central Limit theorem.
Sample Statistic Mean of the
Sampling Distribution

Standard error of the
Sampling Distribution
Mean $\mu _{\overline{x}}$ = μ $\sigma _{\overline{x}}$ = $\frac{\sigma }{\sqrt{n}}$ Population is normally distributed
and the population standard deviation is known.
Mean $\mu _{\overline{x}}$ = μ $\sigma _{\overline{x}}$ =$\frac{s}{\sqrt{n}}$ Population unknown. Sample size n ≥ 30
and s is the standard deviation of the
Proportion $\mu _{p}$ = p $\sigma _{p}$ = $\sqrt{\frac{p(1-p)}{n}}$ np ≥ 5 and n(1-p) ≥ 5 where p is the
attributed population proportion

Formulas describing the sampling distributions of difference of means for both the cases of dependent and independent samples also exist.

Solved Examples

Question 1: If repeated samples of size 50 are taken from an infinite population, the distribution of sample means:
A. Will be normal if the population is normal
B. Will always be normal, because the sample mean is always normal
C. Will always be normal as the samples are taken from infinite population.
D. Will be approximately normal as stated by the Central Limit Theorem.
Option 4 is the right answer. As the sample size = 50 > 30 is large the distribution of sample means is approximately normal irrespective of the Population shape as stated by Central limit Theorem.

Question 2: Samples of sizes of 49 are drawn from a population of mean = 36 and standard deviation = 15. Find the probability that P(33 < x < 36).
As the sample size is large, the distribution of sample means is approximately normal as per Central Limit Theorem with $\mu _{\overline{x}}$ = 36 and  $\sigma _{\overline{x}}$ = $\frac{s}{\sqrt{n}}$ = $\frac{15}{\sqrt{49}}$ = $\frac{15}{7}$

   The z scores for the two x values are calculated using the formula, z = $\frac{\overline{x}-\mu _{\overline{x}}}{\sigma _{\overline{x}}}$

   z1 = $\frac{33-36}{\frac{15}{7}}$ = $-1.4$       and z2 = 0

   P(33 < x < 36) = P(-1.4 < z < 0) = 0.4192.