The random variable \(X\) is said to have a standard normal distribution iff it has density
\[ \phi(x) = \frac{1}{\sqrt{2 \pi}}e^{-\frac12 x^2} \]
Let \(X_1,..,X_n\) be a sequence of independent and identically distributied random variables with mean \(\mu\) and variance \(\sigma^2\). Let \(\bar{X}=\frac1{n}\sum_{i=1}^n X_i\). Then
\[ P(\sqrt{n}\frac{\bar{X}-\mu}{\sigma} \le x) \rightarrow \Phi(x) \]
where \(\Phi(x) = \int_{-\infty}^{x} \phi(t) dt\).
As an illustration we do the following: we generate 100 observations from a uniform distribution on [0, 1] and find it’s sample mean. We know that if \(U\sim U[0, 1]\) we have \(\mu=\frac12\) and \(\sigma^2=\frac1{12}\). So
\[ \begin{aligned} &Z_1 = \sqrt{n}\frac{\bar{X}-\mu}{\sigma} = \\ &\sqrt{100}\frac{\bar{X}-1/2}{\sqrt{1/12}} = \\ &\sqrt{300}\left(2\bar{X}-1\right) \\ \end{aligned} \]
now we repeat this 10000 times:
z <- rep(0, 10000)
for(i in 1:10000) {
x <- runif(100)
z[i] <- sqrt(300)*(2*mean(x)-1)
}Finally we can compare the relative frequencies with the theoretical probabilities for diffeent values of \(x\). For example if \(x=1.5\) we find
round(c(length(z[z<1.5])/10000, pnorm(1.5)), 4)## [1] 0.9365 0.9332and so the difference is 0.9365 - 0.9332 = 0.0033.