Continuity Corrections for Discrete Sums

Why a Discrete Sum Needs a Boundary Adjustment

The central limit theorem can approximate the distribution of a sum by a continuous normal distribution. But a discrete sum takes isolated values, while a normal random variable assigns probability to intervals. A continuity correction bridges that difference: for an integer-valued sum with unit spacing, move each event boundary by one-half before calculating the normal probability.

The adjustment is small in size, but it can have a substantial effect when the standard deviation is not large. The key is to translate the event into the exact integers it includes first. Then place the continuous boundaries halfway between the last included integer and the nearest excluded integer.

Definition: Let \(S\) be an integer-valued random variable whose possible values in the region of interest are spaced one unit apart. Suppose a normal approximation \(Y\sim N(\mu,\sigma^2)\), with \(\sigma>0\), is appropriate for \(S\). A continuity correction replaces an integer event for \(S\) with an interval event for \(Y\) whose endpoints lie halfway between adjacent integer values.

For example, the event \(a\le S\le b\), where \(a\) and \(b\) are integers, includes the integers from \(a\) through \(b\). Its corrected continuous interval is approximately \(a-0.5<Y<b+0.5\). The use of strict inequalities at the continuous endpoints is immaterial because a continuous normal variable has zero probability at any single point.

The Boundary Rules

The most reliable approach is to identify the smallest and largest integers included in the event. If these are \(L\) and \(U\), respectively, the corrected interval is \(L-0.5<Y<U+0.5\). This rule handles inclusive, strict, and mixed inequalities without memorizing separate formulas.

Discrete event	Integers included	Corrected normal event
\(a\le S\le b\)	\(a,\ldots,b\)	\(a-0.5<Y<b+0.5\)
\(a<S\le b\)	\(a+1,\ldots,b\)	\(a+0.5<Y<b+0.5\)
\(a\le S<b\)	\(a,\ldots,b-1\)	\(a-0.5<Y<b-0.5\)
\(S\ge c\)	\(c,c+1,\ldots\)	\(Y>c-0.5\)
\(S>c\)	\(c+1,c+2,\ldots\)	\(Y>c+0.5\)
\(S\le c\)	\(\ldots,c-1,c\)	\(Y<c+0.5\)
\(S<c\)	\(\ldots,c-2,c-1\)	\(Y<c-0.5\)

If \(S\) is a sum of independent random variables and a normal approximation is appropriate, use the sum’s mean and variance. In particular, for independent \(X_1,\ldots,X_n\) with finite means and variances, the sum has mean \(\mu=\sum_{j=1}^n E[X_j]\) and variance \(\sigma^2=\sum_{j=1}^n\operatorname{Var}(X_j)\). The continuity correction changes the event boundaries, not these parameters.

Write \(\Phi(z)\) for the cumulative distribution function of a standard normal random variable. After correcting the boundaries, standardize using \(Z=(Y-\mu)/\sigma\). For instance, an inclusive interval becomes

P(a\le S\le b)\ \approx\ \Phi\!\left(\frac{b+0.5-\mu}{\sigma}\right) -\Phi\!\left(\frac{a-0.5-\mu}{\sigma}\right).

The same logic gives the one-sided forms \(P(S\ge c)\approx 1-\Phi((c-0.5-\mu)/\sigma)\) and \(P(S\le c)\approx\Phi((c+0.5-\mu)/\sigma)\). These are approximations, not exact equalities.

Why the Half-Unit Adjustment Works

The adjustment can be justified by matching each integer value to the unit-width interval around it. For an integer \(k\), associate the interval from \(k-0.5\) to \(k+0.5\). These intervals meet at their endpoints and cover the real line without overlapping interiors. If a normal curve is a reasonable smooth approximation to the probabilities of the integer values, the probability assigned to the interval around \(k\) approximates the probability \(P(S=k)\).

P(S=k)\ \approx\ P(k-0.5<Y<k+0.5).

For the event \(a\le S\le b\), add the probabilities for \(k=a,\ldots,b\). The corresponding intervals join to form the single interval from \(a-0.5\) to \(b+0.5\). Thus

P(a\le S\le b) =\sum_{k=a}^{b}P(S=k) \ \approx\ P(a-0.5<Y<b+0.5).

This argument explains both half-unit shifts. It relies on a smooth normal approximation being suitable for the sum and on unit spacing in the relevant part of its support. It does not make a poor normal approximation accurate by itself.

Worked Examples

Worked Example: An Inclusive Binomial Interval

A hypothetical quality-control system independently flags each inspected unit with probability \(0.40\). Let \(S\) be the number flagged among \(100\) units. Use a normal approximation with a continuity correction to estimate \(P(35\le S\le45)\).

Model and method. Here \(S\sim\operatorname{Binomial}(100,0.40)\). A binomial variable is a sum of independent Bernoulli variables, so its mean and variance are

\mu=np=100(0.40)=40,\qquad \sigma^2=np(1-p)=100(0.40)(0.60)=24,\qquad \sigma=\sqrt{24}\approx4.8990.

The event includes the integers \(35,36,\ldots,45\). Its corrected normal boundaries are \(34.5\) and \(45.5\). Standardizing gives

z_{\mathrm{lower}}=\frac{34.5-40}{\sqrt{24}}\approx-1.1227, \qquad z_{\mathrm{upper}}=\frac{45.5-40}{\sqrt{24}}\approx1.1227.

Therefore,

P(35\le S\le45) \ \approx\ \Phi(1.1227)-\Phi(-1.1227) \ \approx\ 0.8692-0.1308 \ =\ 0.7384.

The endpoints are symmetric about the mean, so their standardized values are equal in magnitude and opposite in sign. The interval probability is \(\Phi(z)-\Phi(-z)=2\Phi(z)-1\), which here is \(0.7384\); symmetry does not make this probability equal to \(0.5\). The result is reported to four decimal places. Check: the corrected interval is centered at \(40\), and its probability is greater than \(0.5\), as expected for this relatively broad central interval. The method avoids treating the integer endpoints as though they were continuous cutoffs.

Worked Example: An Upper-Tail Event

In a hypothetical screening process, each of \(80\) records independently has probability \(0.25\) of requiring a manual review. Let \(S\) be the number requiring review. Approximate the probability that at least \(25\) records require review.

Model and method. The count has distribution \(S\sim\operatorname{Binomial}(80,0.25)\), with

\mu=80(0.25)=20,\qquad \sigma^2=80(0.25)(0.75)=15,\qquad \sigma=\sqrt{15}\approx3.8730.

“At least \(25\)” means \(S\ge25\). The first included integer is \(25\), so the corrected lower boundary is \(24.5\), not \(25.5\). Thus

z=\frac{24.5-20}{\sqrt{15}}\approx1.1619, \qquad P(S\ge25)\approx P(Y>24.5)=1-\Phi(1.1619)\approx1-0.8774=0.1226.

The approximation is \(0.1226\), to four decimal places. Check: the cutoff lies above the mean of \(20\), so the upper-tail probability is below \(0.5\). The correction moves the cutoff toward the included integer \(25\), rather than past it.

Worked Example: A Lower Tail for Nonidentical Indicators

A hypothetical monitoring system records \(20\) independent sensors of type A, each of which signals with probability \(0.20\), and \(30\) independent sensors of type B, each of which signals with probability \(0.50\). Let \(S\) be the total number of signals. Approximate the probability of at most \(34\) signals.

Model and method. The \(50\) indicators are independent but do not all have the same probability. Add their means and variances. A Bernoulli variable with success probability \(p\) has mean \(p\) and variance \(p(1-p)\). Therefore,

\mu=20(0.20)+30(0.50)=4+15=19,

\sigma^2=20(0.20)(0.80)+30(0.50)(0.50)=3.2+7.5=10.7, \qquad \sigma=\sqrt{10.7}\approx3.2711.

The event \(S\le34\) includes \(34\) as its largest integer. Apply the lower-tail correction at \(34.5\):

z=\frac{34.5-19}{\sqrt{10.7}}\approx4.738.

Consequently, \(P(S\le34)\approx\Phi(4.738)\), which is approximately \(1.0000\) to four decimal places. Check: the corrected threshold is more than four standard deviations above the mean, so a probability very close to \(1\) is consistent with the model. This method uses the independent-sum mean and variance even though the indicators have different probabilities.

Worked Example: Strict and Mixed Inequalities

A hypothetical batch test independently detects a defect in each item with probability \(0.50\). For \(64\) items, let \(S\) be the number of defects detected. Approximate \(P(28<S<37)\).

Model and method. Here \(S\sim\operatorname{Binomial}(64,0.50)\), so

\mu=64(0.50)=32,\qquad \sigma^2=64(0.50)(0.50)=16,\qquad \sigma=4.

The strict inequalities include \(29,30,\ldots,36\), not \(28\) and \(37\). The corrected continuous boundaries are therefore \(28.5\) and \(36.5\). After standardizing,

z_{\mathrm{lower}}=\frac{28.5-32}{4}=-0.875, \qquad z_{\mathrm{upper}}=\frac{36.5-32}{4}=1.125.

Thus

P(28<S<37) \ \approx\ \Phi(1.125)-\Phi(-0.875) \ \approx\ 0.8697-0.1908 \ =\ 0.6789.

The approximation is \(0.6789\), to four decimal places. Check: the event is not symmetric about \(32\), so the standardized endpoints are not opposites; subtracting the two CDF values gives the interval probability. The method avoids mistakenly including \(28\) or \(37\).

Common Mistakes and Limits of the Rule

Using the integer endpoints unchanged. For an inclusive event \(35\le S\le45\), using \(35\) and \(45\) as continuous endpoints omits part of the probability represented by the endpoint integers. In the first worked example, that uncorrected calculation would be \(\Phi(1.0206)-\Phi(-1.0206)\approx0.6926\), rather than the corrected approximation \(0.7384\).
Moving a tail boundary in the wrong direction. For \(S\ge25\), the cutoff is \(24.5\); for \(S>25\), the first included integer is \(26\), so the cutoff is \(25.5\). Read the inequality before shifting.
Applying a correction to the mean or standard deviation. The half-unit adjustment changes only the event boundaries. The normal approximation still uses the sum’s actual mean and variance.
Assuming every integer-valued variable has unit spacing. If the possible values are separated by a lattice spacing \(h\), the corresponding half-cell adjustment is \(h/2\), with boundaries placed between adjacent possible values. A unit half-correction is not appropriate when the support has gaps.
Expecting the correction to fix every approximation. The normal approximation may still be inaccurate, especially when the distribution is strongly skewed or the event is in a far tail. A continuity correction improves the boundary match; it does not replace the conditions needed for a useful normal approximation.

Key takeaway: First list the integer values included in the event. Move the lower boundary down by \(0.5\) and the upper boundary up by \(0.5\) for a unit-spaced inclusive interval; for a tail, place the boundary halfway between the last excluded and first included integer. Then standardize with the sum’s mean and standard deviation.

Check Your Understanding

For each problem, identify the included integers before applying the continuity correction. The answers use normal approximations and are rounded to four decimal places.

A hypothetical count is \(S\sim\operatorname{Binomial}(50,0.40)\). Approximate \(P(18\le S\le23)\). (Answer: \(0.6086\))
A hypothetical event count has distribution \(S\sim\operatorname{Poisson}(36)\). Using a normal approximation with mean \(36\) and variance \(36\), estimate \(P(S\ge43)\). (Answer: \(0.1393\))
A system has \(20\) independent indicators with probability \(0.30\) and \(30\) independent indicators with probability \(0.60\). If \(S\) is their total, approximate \(P(S\le20)\). (Answer: \(0.1500\))
For \(S\sim\operatorname{Binomial}(120,0.50)\), approximate \(P(54<S\le68)\). (Answer: \(0.7819\))
For \(S\sim\operatorname{Binomial}(90,0.20)\), approximate \(P(S<13)\). (Answer: \(0.0736\))

What You'll Learn