Poisson as an approximation, and the normal approximation

Quick reminder of the three distributions:

• $X \sim \text{B}(n, p)$, binomial: the number of successes in $n$ independent trials, each with probability $p$ of success. Mean $np$, variance $np(1 - p)$.
• $X \sim \text{Po}(\lambda)$, Poisson: the number of random events in a fixed interval when they happen at an average rate of $\lambda$ (the Greek letter "lambda"). Mean $\lambda$ and variance $\lambda$.
• $Y \sim \text{N}(\mu, \sigma^2)$, normal: a continuous, bell-shaped distribution with mean $\mu$ ("mu") and standard deviation $\sigma$ ("sigma"). $\Phi(z)$ ("phi") is the table value $\text{P}(Z < z)$ for the standard normal $Z \sim \text{N}(0, 1)$.

Why approximate? Exact binomial or Poisson sums can be huge. $\text{P}(X \leqslant 40)$ for $X \sim \text{B}(1000, 0.003)$ needs 41 messy terms. A simpler distribution with the same shape gives a very close answer in a line or two.

Picture 200 light bulbs, each with a $2\%$ chance of being faulty. Lots of trials, and a success (a fault) in any one is rare. "Rare events scattered among many opportunities" is exactly what a Poisson describes. So the binomial count of faults behaves almost like a Poisson with the same mean:

$\lambda = np = 200 \times 0.02 = 4.$

Why must $np$ be small, not just $n$ large? A Poisson always has variance equal to its mean. The binomial's variance is $np(1 - p)$, which is close to $np$ only when $p$ is small (so $1 - p \approx 1$). With $n = 1000$ and $p = 0.4$, the mean is 400 but the binomial variance is only 240: a Poisson would have the wrong spread. That's why the rule is about $np < 5$ too.