A Bernoulli random variable (sometimes called a binary random variable) is any random variable with only two possible outcomes. ## Notation $X \sim Bern(p)$ ## PMF $P(X = x) = p^x(1-p)^{1-x} \ I_{(0,1)}(x)$ ## CDF $ F(x) = P(X <= x) = \begin{cases} 0 & x <= 0 \\ 1-p & 0 < x < 1 \\ 1 & x >= 1 \end{cases} $ ## Expectation $E(X) = p$ ## Variance $V(X) = p(1-p)$ ## Approximation of the Bernoulli Distribution with the Normal Distribution The Bernoulli distribution can be approximated by the Normal Distribution when $n$ is sufficiently large. This is especially useful because at large $n$ the calculations for the Bernoulli distribution become incredibly computationally intensive. For large $n$, $X$ can be approximated by a normal random variable with $\mu=np$ and $\sigma^2=np(1-p)$. > [!Example] Example > In a given day, there are approximately 1,000 visitors to a website. Of these, 25% register for a service. Estimate the probability that between 200 and 225 people will register for a service tomorrow. >First, we calculate $\mu$ and $\sigma$ from our data. >$\mu = np = 1000*0.25 = 250$ >$\sigma^2 = np(1-p) = 1000 * 0.25 * (1 - 0.25) = 187.5$ >Next we transform to the standard normal distribution. > >$P(200 <= X <= 225) = P(\frac{199.5 - 250}{\sqrt{187.5}} <= Z <= \frac{225.5 - 250}{\sqrt{187.5}}$ >Why use 199.5 and 225.5 in place of 200 and 225? This is called the "continuity correction". Because we are transforming a discrete variable to the continuous normal distribution, we want to split the difference between the two possible discrete values 199 and 200. > >Finally, we will look up the values in a look up table to arrive at the probability. >$P(-3.688 <= Z <= -1.789) = \Phi(-1.789) - \Phi(-3.688) = 0.367$ >In this case, because the probability is what we were asked to evaluate, we do not need to transform back to the original scale. The probability is simply the probability.