Deriving the Half-Unit Continuity Correction

Why a Half Unit Appears

A discrete count takes integer values, while a normal random variable varies continuously along the real line. A continuous random variable has probability zero at any one exact point, so a normal approximation cannot represent the probability of a particular count by assigning area to that point alone. The continuity correction supplies a geometric bridge: it gives each integer count the interval of real values closest to that integer.

The adjustment is not an arbitrary rule to memorize. It follows from the spacing of the possible count values. Consecutive integers are one unit apart, and the boundary between the values \(k\) and \(k+1\) is halfway between them. Thus the interval associated with \(k\) extends half a unit in each direction. This tutorial derives that construction and shows how it supports a normal approximation.

Definition: For an integer-valued count \(K\), the unit-width cell centered at the integer \(k\) is the interval from \(k-\tfrac12\) to \(k+\tfrac12\). When approximating count probabilities with a continuous normal distribution, this cell is the region used to represent the count \(k\). The interval endpoints can be included or excluded without changing a normal probability, because a continuous distribution assigns probability zero to a single endpoint.

Deriving the Cell From Nearest-Integer Assignment

Imagine assigning every real number to the nearest integer. A real value is closer to \(k\) than to \(k-1\) when it lies above the midpoint \(k-\tfrac12\); it is closer to \(k\) than to \(k+1\) when it lies below \(k+\tfrac12\). Therefore, the values assigned to \(k\) occupy a width-one interval centered on \(k\). Adjacent cells meet at their midpoint boundaries, so the cells cover the real line without gaps or overlapping interiors.

\text{cell for } k = \left[k-\frac12,\;k+\frac12\right)

The half-unit is exactly half the distance between consecutive integer counts. It is not half the count, half the standard deviation, or a correction to the mean. Its role is to locate the boundaries of the region representing a count. The choice of which cell contains a midpoint is immaterial for a continuous normal approximation: a boundary point has area zero.

Suppose a count \(K\) has mean \(\mu\) and variance \(\sigma^2\), and a suitable normal approximation uses \(Y\sim N(\mu,\sigma^2)\). The approximation represents the probability at \(K=k\) by the normal area over the cell for \(k\):

P(K=k) \ \approx\ P\left(k-\frac12<Y<k+\frac12\right) = \Phi\left(\frac{k+\frac12-\mu}{\sigma}\right) - \Phi\left(\frac{k-\frac12-\mu}{\sigma}\right),

where \(\Phi\) is the standard normal cumulative distribution function. The normal model is an approximation to the count distribution, not an exact replacement. Its use still depends on having a reasonable normal approximation for the count in the first place.

Why the Area Represents a Count Probability

The cell construction matches the support of the count to the continuous scale. Each possible integer receives one non-overlapping interval, and the center of that interval is the integer itself. Under the normal approximation, the area in that interval is therefore assigned to the corresponding count. This gives a local approximation to the probability mass at \(k\), rather than treating the count as if it could take every real value.

The unit width matters. If an integer count is approximated by a density value at \(k\), that density is not itself a probability: density has units of probability per unit of the variable. A cell with width one gives the density area a region to occupy. The full normal-area calculation integrates across the cell, which is what the CDF difference expresses.

Justification: The midpoint between two adjacent integer values \(k\) and \(k+1\) is \(\frac{k+(k+1)}{2}=k+\frac12\). The midpoint between \(k-1\) and \(k\) is \(k-\frac12\). A real value between these two midpoints is closer to \(k\) than to either neighboring integer. Those boundaries are one unit apart, so the nearest-integer region for \(k\) has width one and is centered at \(k\). Assigning the normal area in that region to \(P(K=k)\) yields the half-unit continuity correction.

Worked Applications

Each example assumes a normal approximation has been selected for the stated count. The calculations focus on the reason for the half-unit boundaries and on evaluating the resulting area.

Worked Example: A Binomial Count Near Its Mean

A hypothetical quality-control line inspects 100 independent components. Each component has probability \(0.50\) of failing a specified test, independently of the others. Let \(K\) be the number of failures. Use a normal approximation with a continuity correction to estimate the probability of exactly 57 failures.

Method. The count is binomial with \(n=100\) and \(p=0.50\). Its mean and variance are

\mu=np=100(0.50)=50, \qquad \sigma^2=np(1-p)=100(0.50)(0.50)=25, \qquad \sigma=5.

The count 57 is represented by the real-number cell from \(56.5\) to \(57.5\). Standardizing the two boundaries gives

z_{\mathrm{lower}}=\frac{56.5-50}{5}=1.3, \qquad z_{\mathrm{upper}}=\frac{57.5-50}{5}=1.5.

Thus the normal area for that cell is

P(K=57) \ \approx\ \Phi(1.5)-\Phi(1.3) = 0.9331928-0.9031995 = 0.0299933 \ \approx\ 0.0300.

The estimated probability is 0.0300, to four decimal places. The cell has width one and is centered at 57, as required. A common mistake is to use the interval from 57 to 58; that shifts the region upward and does not represent the values nearest to 57. This approximation is a probability, so it must lie between zero and one; the result passes that basic check.

Worked Example: A Poisson Count With a Large Mean

Suppose the number \(K\) of service requests arriving in a fixed operating period follows a Poisson distribution with parameter \(\lambda=100\). Use the normal approximation with a continuity correction to estimate the probability of exactly 110 requests. The Poisson model assumes a fixed observation period and a constant event rate, with independent event counts in disjoint subintervals.

Method. For a Poisson count, the mean and variance both equal \(\lambda\). Therefore,

\mu=100, \qquad \sigma^2=100, \qquad \sigma=10.

The count 110 is assigned the cell from \(109.5\) to \(110.5\), not a cell beginning at 110. The standardized boundaries are

z_{\mathrm{lower}}=\frac{109.5-100}{10}=0.95, \qquad z_{\mathrm{upper}}=\frac{110.5-100}{10}=1.05.

The resulting approximation is

P(K=110) \ \approx\ \Phi(1.05)-\Phi(0.95) = 0.8531409-0.8289439 = 0.0241971 \ \approx\ 0.0242.

The estimated probability is 0.0242, to four decimal places. The result is positive because the upper boundary has the larger standardized value and hence the larger CDF. Using \([110,111)\) instead would assign the count 110 the cell centered at 110.5; this is the wrong cell. The normal approximation itself is not exact, even though the boundaries follow directly from the integer spacing.

Worked Example: Checking the Boundary on Both Sides

A hypothetical laboratory records the number of positive results among 64 independent specimens. Each specimen has probability \(0.50\) of a positive result. Let \(K\) be the number of positives. Use the normal approximation with a continuity correction to estimate the probability of exactly 33 positives, and compare it with the result from using the incorrectly shifted interval \([33,34)\).

Method. The binomial parameters are \(n=64\) and \(p=0.50\), so

\mu=np=64(0.50)=32, \qquad \sigma^2=np(1-p)=64(0.50)(0.50)=16, \qquad \sigma=4.

The correct cell centered at 33 is \([32.5,33.5)\). After standardization,

z_{\mathrm{lower}}=\frac{32.5-32}{4}=0.125, \qquad z_{\mathrm{upper}}=\frac{33.5-32}{4}=0.375.

Therefore,

P(K=33) \ \approx\ \Phi(0.375)-\Phi(0.125) = 0.6461698-0.5497382 = 0.0964316 \ \approx\ 0.0964.

The incorrect interval \([33,34)\) would instead give

\Phi\left(\frac{34-32}{4}\right) - \Phi\left(\frac{33-32}{4}\right) = \Phi(0.5)-\Phi(0.25) = 0.6914625-0.5987063 = 0.0927561 \ \approx\ 0.0928.

That is a different area because the incorrect interval is centered at \(33.5\), not at 33. The corrected estimate is 0.0964, to four decimal places. This comparison shows why the correction must be applied at both boundaries: the lower boundary moves down by \(0.5\), and the upper boundary moves up by \(0.5\).

Common Mistakes and Scope

Moving only one boundary. A cell centered at \(k\) has lower endpoint \(k-\tfrac12\) and upper endpoint \(k+\tfrac12\). Moving only one endpoint changes both the center and the width.
Using a full-unit shift. The gap between consecutive counts is one, so the dividing point is half a unit from each count. A shift of one assigns part of a neighboring count’s cell to \(k\).
Calling the correction an adjustment to the mean. The mean and standard deviation come from the count model. The half-unit changes the cell boundaries before standardization; it does not change those model parameters.
Treating the normal area as exact. The cell construction explains how to form the approximation. It does not make a discrete distribution continuous or guarantee that the normal approximation is accurate.

The half-unit is specific to counts whose possible values are spaced one unit apart. If a variable can take values only at intervals of two, the nearest-value cell has width two and extends one unit on either side of its center. The general geometric principle is to use half the spacing on each side; the familiar half-unit follows when the spacing is one.

Key takeaway: For an integer count \(k\), the nearest-integer cell is centered at \(k\) and has endpoints \(k-\tfrac12\) and \(k+\tfrac12\). A normal approximation assigns the area over that cell to the probability of the count. The correction comes from the midpoint geometry of consecutive integers.

Check Your Understanding

For each question, focus on how the count’s spacing determines the cell before calculating or interpreting a normal area.

A count can take every nonnegative integer value. Write the unit-width cell assigned to the count 26 under nearest-integer assignment. (Answer: \([25.5,26.5)\).)
A count is approximated by \(N(40,25)\), where the second parameter is the variance. Use a continuity correction to approximate the probability that the count equals 44. (Answer: \(\Phi(0.9)-\Phi(0.7)\approx 0.0579\).)
A Poisson count has parameter \(\lambda=100\). Using its normal approximation, state the standardized boundaries for approximating the probability of exactly 110. (Answer: \(0.95\) and \(1.05\).)
A binomial count has \(n=64\) and \(p=0.50\). Using a normal approximation, give the continuity-corrected interval and approximate probability for exactly 33. (Answer: \([32.5,33.5)\), with probability approximately \(0.0964\).)
A variable takes values \(0,2,4,6,\ldots\), so adjacent possible values are two units apart. Under nearest-value assignment, what interval represents the value 10? (Answer: \([9,11)\), a width-two cell centered at 10.)

What You'll Learn