The Central Limit Theorem is an important theorem used in mathematical statistics used to make inferences about populations based on limited amounts of information.

The principle is that if you have n random variables, Y1, Y2,…,Yn each with mean (expected value) u; and each with some variance s^2, then

U = sqrt(n)*((Y - u)/s^2), where Y is the average of the realised value of these n random variables

will converge to the standard normal distribution as n approaches infinity. The standard normal distribution is with mean 0 and variance 1, usually denoted Z. Note that the CLT can be applied to any random sample Y1, Y2,…,Yn so long as n is large (say >30) and as long as the mean and variance of Y are known and finite.

There are two other important ways to think about the CLT.

1. Y is approximately normally distributed with mean u and variance s^2/n. This makes sense because as n gets lager and larger, the variance will get smaller and smaller making Y, a better and better estimate of u. That is, Y ~` N(u , s^2/n ).

2. Alternatively, Y1+ Y+,…,+Yn are approximately normally distributed with mean nu and variance ns^2. That is Y1+ Y+,…,+Yn ~` N(nu , ns^2).

An Example. Suppose the test scores of all high school students in a certain state have mean 60 and variance 64. A random sample of 100 students from a large high school had mean 58. Is there any evidence to suggest that the high school is inferior?

Let Y denote the mean of the random sample of n=100 scores from a population with mean u = 60 and variance s^2 = 64. We want to find the probability that this sample mean is less than or equal to 58. If this probability is small, then there is reason to suggest that the school is inferior. We know from the CLT that Y is approximately normally distributed with mean u and variance s^2/n from (1).

So, we want: P(Y less than 58) = P({Y - u}/sqrt(s^2/n) less than {58 - u}/sqrt(s^2/n)), this has standardised Y

This expression is now in the context of the CLT and so we can replace the left hand side by Z, where Z has the standard normal distribution. Ie Z ~ N(0,1).

So, = P(Z less than {58 – 60}/sqrt(64/100))
= P(Z less than -2.5)

Since Z is a continuous random variable, then we can ignore the ‘=’ sign and just consider values of Z less than –2.5. At this point we consult our standard normal distribution tables and look up a value of –2.5, to find that the probability of Z being less than –2.5 is just 0.0062.

That is, we can say that the probability that this school obtained an average score of 58 given that it has the same abilities as the rest of the state is approximately 0.0062, or 0.62%. Hence, there is reason to suggest that the school is inferior.