Continuity Correction¶

Method¶

When approximating $\mathbb{P}(X\leq k)$ for an integer-valued random variable $X$ by a normal distribution with mean $\mu$ and standard deviation $\sigma>0$$0$, use

(and similarly for $\geq$). The half-unit shift is an approximation rule, not an exact identity.

Here $\Phi$ is the CDF of the Standard Normal Distribution. Subtracting $\mu$ and dividing by $\sigma$ converts the shifted boundary to its standard-normal scale.

Interpretation¶

An integer-valued probability is a sum of bar heights, and the bar for the value $k$ has width $1$ and covers $[k-\tfrac{1}{2},k+\tfrac{1}{2}]$$\frac{1}{2}]$. A normal area approximates those bars best when it covers the same half-unit cells around the included integers.

The figure shows $X\sim\operatorname{Bin}(100,0.5)$ with the $\mathcal{N}(50,25)$ density. The highlighted bars $45,\ldots,55$ together cover $[44.5,55.5]$, so the normal area is taken between these endpoints: $\mathbb{P}(45\le X\le55)\approx\Phi(1.1)-\Phi(-1.1)\approx0.7287$$0.7287$, which agrees with the exact value to four decimals. Without the correction, $\Phi(1)-\Phi(-1)\approx0.6827$ leaves out half of each end bar.

The direction of the shift depends on which integers the event includes. $\mathbb{P}(X\le k)$ includes the bar at $k$, so the area ends at $k+\tfrac{1}{2}$$\frac{1}{2}$; $\mathbb{P}(X<k)=\mathbb{P}(X\le k-1)$ ends at $k-\tfrac{1}{2}$$\frac{1}{2}$; $\mathbb{P}(X\ge k)$ starts at $k-\tfrac{1}{2}$$\frac{1}{2}$. Write the event as a range of integers first, then extend it by half a unit at both ends. For a single value, the uncorrected normal answer to $\mathbb{P}(X=k)$ would be $0$, while the corrected one uses the width of the bar, as in the first Example.

The correction matters most when $\sigma$ is small, because half a unit is then a large fraction of a standard deviation. It is used with the normal approximation to the binomial and, more generally, with the central limit theorem for integer-valued sums such as totals of die rolls.

Key Results¶

• $\mathbb{P}(X\ge k)\approx1-\Phi((k-1/2-\mu)/\sigma)$$\mu )/\sigma )$.
• For integers $a\le b$$b$, $\mathbb{P}(a\le X\le b)\approx\Phi((b+1/2-\mu)/\sigma)-\Phi((a-1/2-\mu)/\sigma)$$\mu )/\sigma )$.

Examples¶

• For $X\sim\operatorname{Bin}(100,0.5)$$\mathrm{Bin}(100,0.5)$, $\mathbb{P}(X=50)\approx\Phi(0.1)-\Phi(-0.1)\approx0.0797$$0.0797$.
• For the same $X$, $\mathbb{P}(45\le X\le55)\approx\Phi(1.1)-\Phi(-1.1)\approx0.7287$$0.7287$.

Prerequisites¶

• Standard Normal Distribution

Related Concepts¶

• Central Limit Theorem
• Normal Approximation to the Binomial

Sources¶

• Nicky van Foreest · Probability Distributions · Week 8

Aliases: correction for continuity Symbols: mu, phi, sigma

Last updated: 20 October 2026 03:02