Normal approximation to the binomial

When $n$ is large, the stair-step shape of a binomial distribution smooths out into a normal curve with mean $\mu = np$ and standard deviation $\sigma = \sqrt{np(1-p)}$ (mu and sigma). The usual check is that $np$ and $n(1-p)$ are both at least 10, the same Large Counts idea that sits behind inference for proportions. With $n = 100$ and $p = 0.4$, $np = 40$ and $n(1-p) = 60$ both clear 10, so the count is roughly normal with mean $40$ and standard deviation $\sqrt{24} \approx 4.899$, putting about 95% of counts within $1.96$ standard deviations of the mean, between $30.4$ and $49.6$. Calculators return exact binomial probabilities now, so this approximation matters most as the reason $\hat{p}$ (p-hat, the sample proportion) has a normal sampling distribution.