###### Discrete Uniform Random Variables, a B ...

Probability distributions have different shapes and characteristics. As such, we describe a random... **Read More**

The central limit theorem asserts that when we have simple random samples each of size *n* from a population with a mean μ and variance σ^{2}, the sample mean X approximately has a normal distribution with mean μ and variance σ^{2}/n as *n*(sample size) becomes large.

Suppose we have a sequence of independent and identically distributed variables X_{1}, X_{2}, X_{3} …, X_{n} with mean finite mean μ and non-zero variance σ^{2}, then the distribution of \( \frac {(X – \mu)}{\left(\frac {\sigma}{\sqrt n}\right)}\) approaches the standard normal distribution as *n* approaches infinity, i.e., \(n \rightarrow \infty \).

Remember that \(X=\left( \frac { 1 }{ n } \right) \ast \sum { { X }_{ i } } \).

The central limit theorem provides very useful normal approximations to some common distributions including the binomial and Poisson distributions.

Note that while X is approximately normally distributed with mean μ and variance σ^{2}/n, ΣX_{i} is approximately normally distributed with mean nμ and variance nσ^{2}. In fact, we can show that both the mean and the variance are exact and it is only the shape of the curve that is approximate.

Although the widely accepted value is n ≥ 30, this answer might be too simple. The truth is that the value of n depends on the shape of the population involved, i.e., the distribution of X_{i} and its skewness.

In a non-normal but fairly symmetric distribution, n = 10 can be considered large enough. With a very skewed distribution, the value of n can be 50 or even more.

Suppose we have a set of independent and identically distributed variables X_{i}, i = 1…, n such that:

\(\sum X_i \sim \text{binomial} (n,\theta)\).Applying the CLT, for large n:

\(X \sim N \left( \mu, \frac {\sigma^2}{n} \right)\) and \(\sum X_i \sim N(n \mu, n \sigma^2)\)

Additionally, note that the binomial distribution is a sequence of Bernoulli variables such that the mean is θ and the variance θ(1- θ).

Therefore, applying the CLT:

\(\sum X_i \sim N(n\theta, n \theta (1 – \theta))\) which is the normal approximation to the binomial and n = 10 is considered large enough for CLT application.

Under Poisson, \(\mu = \sigma^2=\lambda\)

Therefore, applying CLT:

$$ \sum X_i \sim N(n\lambda, n \lambda) $$

The central limit theorem is a very useful tool, especially in the construction of confidence intervals or testing of hypotheses. As long as n is “sufficiently large,” just about any non-normal distribution can be approximated as normal.

*Reading 10 LOS 10e:*

*Explain the central limit theorem and its importance.*