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Central Limit Theorem – Statement & Examples

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The central pillar theorem is one of the pillars of modern statistics. The central theorem gives us information about the sampling distribution of the sample mean.

It states that as the sample size drawn from a population becomes larger and larger, then the sample mean follows the normal distribution.

The central theorem is important because it often happens that we do not know the distribution of the parent population.

But the central limit theorem enables us to make statements about the probability of the sample mean because we are assured that the sample mean follows the normal distribution.

We can also calculate confidence intervals for the population mean as a consequence of the central limit theorem.

This theorem was first stated by Laplace in 1812 and rigorous proof under general conditions was given by Liapounov in 1901.

Mathematical Statement of the theorem:

If Xi(i=1,2,…,n)are independent and identically distributed random variables with mean µ and variance σ2, then the sample mean is a random variable has probability distribution asymptotically approaching the normal distribution with mean µ and variance σ2/n as the sample size n tends to infinity.

Central limit theorem with coins
Demonstrating Central limit theorem with coins

Generally speaking, the sample size is considered large if n>30 and in this case the central limit theorem can be applied.

Examples of Calculating Probability using Central Limit Theorem:

1. Suppose that we consider a random sample of 100 people drawn from a population with a mean height of 170cm and a variance of 10cm. Find the probability that the sample mean is greater than 170cm.

Solution: Let x̄ denote the average sample height.

By the central limit theorem, we know that x̄ follows a normal distribution with a mean of 170 cm and variance 100/102=1.

We convert the normal distribution to the standard Z score by subtracting the mean and dividing it by the standard deviation.

So Z =  x̄ -170

Hence,

P(x̄>170)

=P( x̄-170>0)

=P(Z>0)

=0.5 since we know that the standard normal distribution is symmetric with mode 0.

2. Suppose that we consider a random sample of 100 people drawn from a population with a mean weight of 60 kg and a variance of 10kg. Find the probability that the sample mean lies between 59 and 61 kgs.

Solution: Let x̄ denote the average sample weight

By the central limit theorem, we know that x̄ follows the normal distribution with mean 60kgs and variance 100/102=1.

We convert the normal distribution to the standard Z score by subtracting the mean and dividing it by the standard deviation.

So Z =  x̄ -60

Hence,

P(59<x̄<61)

= P(59-60<x̄<61-60)

= P(-1<Z<1)

= 0.67 or 67% as we see by looking at the Z table (or by applying the empirical rule).

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