Normal approximations and continuity correction

What you’ll learn

• Choose the matching Normal mean and variance.
• Apply continuity corrections to count events.
• Assess whether a Normal approximation is plausible.

Understand the idea

A Normal approximation replaces a discrete count by a continuous variable with the same mean and variance. For a binomial variable the variance is $np(1-p)$; for a Poisson variable it is $\lambda$. Approximation improves when the count distribution is sufficiently spread and not strongly skewed; both expected binomial successes and failures should be reasonably large.

Represent each integer count by a cell extending half a unit on each side. Thus $X\leq k$ corresponds to a continuous boundary at $k+0.5$, while $X\geq k$ starts at $k-0.5$. Translate a strict inequality into its included integer values before choosing the boundary.

After the correction, standardise with the approximating standard deviation. State that the final probability is approximate and retain enough working precision. A continuity correction improves the alignment of boundaries but cannot rescue an unsuitable distributional shape.

Key formulae

$$B(n,p)\approx N\!\left(np,np(1-p)\right)$$

$$\operatorname{Po}(\lambda)\approx N(\lambda,\lambda)$$

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

$$P(X\geq k)\approx P(Y>k-0.5)$$

Worked example

Use a Normal approximation to estimate $P(X\leq45)$ when $X\sim B(100,0.4)$.

1. Use $Y\sim N(40,24)$ since $np=40$ and $np(1-p)=24$.
2. The event includes 45, so the continuous upper boundary is $45.5$.
3. Standardise: $z=(45.5-40)/\sqrt{24}=1.1227$.
4. The estimate is $\Phi(1.1227)\approx0.8692$.

Answer: $0.8692$ approximately.