Binomial to Normal Distribution

Given \(n\) as the number of Bernoulli trials and \(p\) as the probability of success, we have:

\[ P(\mathbf{X} = 1) = p \\ P(\mathbf{X} = 0) = q = 1 - p \]

If \(n\) is large and \(n \cdot p \cdot q\) is not too small, the distribution approaches a Normal Distribution:

\[ \mathbf{X} \sim \text{Binomial}(n, p) \rightarrow \mathbf{X} \sim \mathcal{N}(np, \sqrt{npq}) \]

where \(np = \mu\) (mean), and \(npq = \sigma^2\) (variance). The standard deviation is \(\sigma\), and we can compute the probability using the cumulative distribution function (CDF) of the normal distribution.

The probability density function (PDF) of the normal distribution is:

\[ f(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \]

To calculate the probability of a specific random value \(\mathbf{X}\) over the interval \([a, b]\) using the CDF, we have:

\[ P(a \leq \mathbf{X} \leq b) = \int_a^b f(x) \, dx = \int_a^b \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \, dx \]

It is easier to compute this for the standard normal distribution \(\mathcal{N}(\mu = 0, \sigma = 1)\), also called the z-distribution. To standardize \(\mathbf{X}\), a random variable with any \(\mu\) and \(\sigma\), to mean 0 and standard deviation 1, we compute the z-score \(Z\) corresponding to \(\mathbf{X}\):

\[ Z = \frac{\mathbf{X} - \mu}{\sigma} \]

This equation shifts the mean of \(\mathbf{X}\) to 0 and scales the distribution to have \(\sigma = 1\). The standard normal PDF is:

\[ \phi(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \]

and the CDF of the standard normal distribution is:

\[ \Phi(z) = \int_{-\infty}^z \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \, dz \]

You can compute \(P(Z \leq \text{z-score})\) using the z-table.