Continuity Corrections and Normal-Approximation Diagnostics

Why Integer Cutoffs Need Care

A normal distribution is continuous, so it assigns probability zero to any single point. An integer-valued sum, by contrast, places probability at individual integers. If a cutoff is used without adjustment, the normal calculation may include too much or too little of the area near the endpoint. A continuity correction addresses this mismatch by moving a cutoff halfway between adjacent integers. It improves the alignment of the event boundaries; it does not guarantee that a normal distribution is a good model for the sum.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(B_1,\ldots,B_n\) be independent Bernoulli random variables with common success probability \(p\), where \(0<p<1\). Each \(B_i\) has state space \(S=\{0,1\}\), with \(\mathbb{P}(B_i=1)=p\) and \(\mathbb{P}(B_i=0)=1-p\); these probabilities sum to \(1\). The sum \(S_n=\sum_{i=1}^n B_i\) is binomial. More generally, the correction below applies to sums supported on consecutive integers when a normal approximation is being used.

Suppose a sum \(S\) has mean \(\mu\) and positive variance \(\sigma^2\). Its normal approximation uses a normal random variable with mean \(\mu\) and variance \(\sigma^2\). Write \(\Phi(x)\) for the standard normal cumulative distribution function, the probability that a standard normal variable is at most \(x\). The corrected rules are:

Here \(a,b,k\) are integers, \(a\le b\), and the strict versus inclusive inequalities matter. If \(\sigma=0\), the sum is constant and these standardized formulas do not apply; compute the event directly. If the integer support has gaps or a lattice spacing other than one, the half-unit rule may not match its support, so the boundary adjustment must reflect the actual lattice rather than be applied mechanically.

Why the Half-Unit Rule Matches the Event

Proof. Since \(S\) takes integer values, an outcome satisfies \(S\le k\) exactly when it satisfies \(S<k+\tfrac12\): the largest integer below \(k+\tfrac12\) is \(k\). Similarly, \(S<k\) is equivalent to \(S<k-\tfrac12\), since the largest integer below \(k\) is \(k-1\), which is also the largest integer below \(k-\tfrac12\). Taking complements gives \(S\ge k\) equivalent to \(S\ge k-\tfrac12\), and \(S>k\) equivalent to \(S>k+\tfrac12\). For \(a\le S\le b\), the same argument gives \(a-\tfrac12<S<b+\tfrac12\). A continuous normal variable assigns probability zero to either endpoint, so its probability on that open interval equals its probability between the displayed boundaries. Standardizing those boundaries gives the formulas. The event identities are exact; replacing the distribution of \(S\) by a normal distribution is the approximation. \(\square\)

There is also a useful geometric interpretation. Each integer \(j\) can be associated with the interval from \(j-\tfrac12\) to \(j+\tfrac12\). The event \(a\le S\le b\) corresponds to combining the intervals for the integers from \(a\) through \(b\), giving the interval from \(a-\tfrac12\) to \(b+\tfrac12\). The normal approximation assigns continuous probability to this combined interval. This explains the adjustment but does not prove that each integer probability is accurately represented by the normal area in its interval.

Worked Binomial Calculations

Diagnostics: Is the Normal Approximation Credible?

A continuity correction fixes a boundary mismatch, not a mismatch in distributional shape. Before relying on the answer, check the conditions and the location of the event. For a binomial sum, the mean is \(np\), the variance is \(np(1-p)\), and the skewness is \((1-2p)/\sqrt{np(1-p)}\). Skewness is the standardized third central moment; its sign indicates the direction of asymmetry.

Proof. For one Bernoulli variable \(B_i\), write \(q=1-p\). Its centered third moment is \(\mathbb{E}[(B_i-p)^3]=p(1-p)^3+q(0-p)^3=pq(q^2-p^2)=pq(q-p)=pq(1-2p)\). For the sum, expand the cube of \(\sum_i(B_i-p)\). Every term involving a factor from one variable to the first power and independent factors from other centered variables has expectation zero, because those other factors have mean zero. The remaining terms are the \(n\) individual third moments, so the sum’s third central moment is \(npq(1-2p)\). Its variance is \(npq\). Dividing the third central moment by the variance to the power \(3/2\) gives \(npq(1-2p)/(npq)^{3/2}=(1-2p)/\sqrt{npq}\), as claimed. \(\square\)

Use these checks as diagnostics, not as a mechanical pass/fail theorem:

• Sample size and expected counts: A large \(n\) alone is not enough for a binomial sum if \(p\) is very small or very close to \(1\). Check both \(np\) and \(n(1-p)\); if either is small, the distribution can remain concentrated near an endpoint.
• Variance: A small variance means the distribution occupies only a narrow range of integer values. Standardization does not create information or smooth out a distribution with very few likely outcomes.
• Skewness: A large absolute skewness signals appreciable asymmetry. In the third example, the formula gives \((1-2/10)/\sqrt{10(1/10)(9/10)}=(4/5)/\sqrt{9/10}\), which is not close to zero.
• Tail location: A central probability can be adequately approximated even when an extreme-tail probability is not. Check how far the corrected boundary lies from the mean in standard-deviation units, and be especially cautious when the requested tail is very small.
• Support and cutoff: Confirm that the sum is on the integer lattice assumed by the half-unit correction, and translate strict and inclusive inequalities before standardizing.

The central limit theorem for independent, identically distributed variables with finite positive variance gives convergence in distribution of the standardized sums to a standard normal as the sample size tends to infinity. That asymptotic result motivates normal approximations; it does not promise a particular finite-sample error, especially for an extreme tail. The corrections in the examples are finite-sample calculations, not consequences of a universal accuracy guarantee. When accuracy matters, compare against an exact calculation, a justified bound, or a more suitable approximation.

A Common Mistake and a Practical Workflow

Common mistake: using the integer cutoff unchanged. In the first example, using \(1\) for the event \(S_4\le1\) gives \(\Phi(-1)\), approximately \(0.1587\). The corrected calculation uses \(1+\tfrac12=\tfrac32\), giving \(\Phi(-\tfrac12)\), approximately \(0.3085\); the exact probability is \(\tfrac5{16}=0.3125\). The error is not in standardization but in placing the boundary at the integer rather than between the included and excluded integers.

A reliable calculation follows a short sequence: identify the event’s exact inequality; convert it to its half-unit boundary; compute the sum’s mean and variance; standardize the corrected boundary; evaluate the normal probability; then assess whether sample size, variance, skewness, and tail location support the approximation. Keep the event identity distinct from the approximation step: the first is exact for integer-valued \(S\), while the second depends on how well the normal distribution represents its probabilities.