Normal Approximations of Binomial Distributions

When $n$ is large, computing binomial probabilities by summing individual terms becomes impractical. Fortunately, the binomial distribution can be approximated by a normal distribution with the same mean and variance.

If $X \sim \text{Bin}(n, p)$ with $n$ sufficiently large, then

$$X \;\approx\; N\big(np,\; np(1-p)\big).$$

The approximation is reliable when both

$$np \geq 10 \qquad \text{and} \qquad n(1-p) \geq 10.$$

Because $X$ is discrete and the approximating $Y \sim N(np,\, np(1-p))$ is continuous, we apply a continuity correction: each integer value $k$ of $X$ corresponds to the interval $[k - 0.5,\, k + 0.5]$ under the normal density. For a left-tail probability,

$$P(X \leq k) \;\approx\; P(Y \leq k + 0.5).$$

For example, let $X \sim \text{Bin}(100, 0.5)$. Then $\mu = np = 50$ and $\sigma^2 = np(1-p) = 25$, so $\sigma = 5$. To estimate $P(X \leq 48){:}$

$$\begin{align*} P(X \leq 48) &\approx P(Y \leq 48.5) \\[3pt] &= P\!\left(Z \leq \dfrac{48.5 - 50}{5}\right) \\[3pt] &= P(Z \leq -0.3) \\[3pt] &\approx 0.3821. \end{align*}$$