Continuity Corrections At Discrete Boundaries

Why Integer Boundaries Need an Adjustment

A count can take only integer values, while a normal random variable can take any real value. When a count distribution is approximated by a normal distribution, the event’s integer boundary therefore needs careful translation. A half-unit adjustment, called the continuity correction, makes the normal interval correspond more closely to the integer values included in the event.

The adjustment is especially important when the question uses strict inequalities, or when an event has both a lower and an upper bound. The first step is always to identify the actual integers included in the event. Only then should you place the normal boundaries.

Definition: Suppose $X$ is an integer-valued count with mean $\mu$ and standard deviation $\sigma$, and a normal approximation is appropriate. Approximate the probability of $L\le X\le U$, for integer bounds $L$ and $U$, by $$ P(L\le X\le U)\approx P(L-0.5<Y<U+0.5), $$ where $Y\sim N(\mu,\sigma^2)$. Since $Y$ is continuous, including or excluding either interval endpoint gives the same probability.

For a binomial count $X\sim\operatorname{Bin}(n,p)$, the normal model uses $\mu=np$ and $\sigma^2=np(1-p)$. For a Poisson count $X\sim\operatorname{Poisson}(\lambda)$, it uses $\mu=\lambda$ and $\sigma^2=\lambda$. These are approximations, not exact replacements for the discrete distributions. The normal approximation must be reasonable for the model and parameters; continuity correction does not make a poor normal approximation reliable.

Translating Strict and Inclusive Bounds

Because $X$ takes integer values, a strict inequality can first be rewritten as an inclusive bound on integers. For example, $X<k$ means $X\le k-1$, and $X>k$ means $X\ge k+1$. After that rewrite, apply the half-unit adjustment to the inclusive boundary.

Discrete event	Equivalent integer event	Continuity-corrected normal event
$X\le k$	At most $k$	$Y<k+0.5$
$X<k$	At most $k-1$	$Y<k-0.5$
$X\ge k$	At least $k$	$Y>k-0.5$
$X>k$	At least $k+1$	$Y>k+0.5$
$L\le X\le U$	$L,L+1,\ldots,U$	$L-0.5<Y<U+0.5$

For a standardized calculation, use $Z=(Y-\mu)/\sigma$, where $Z\sim N(0,1)$. If $\Phi(z)=P(Z\le z)$, then, for example,

P(L\le X\le U)\approx \Phi\!\left(\frac{U+0.5-\mu}{\sigma}\right) - \Phi\!\left(\frac{L-0.5-\mu}{\sigma}\right).

The same principle works for a strict or inclusive endpoint: determine the smallest and largest integer values that satisfy the original event, then expand outward by one-half at each end.

Why the Half-Unit Adjustment Works

The correction has a simple geometric justification. Associate the integer $j$ with the unit-width interval from $j-0.5$ to $j+0.5$. That interval is centered at $j$, so it represents the part of a continuous scale naturally associated with that count. The intervals for consecutive integers meet at their half-unit boundaries and do not overlap except at endpoints.

\bigcup_{j=L}^{U}[j-0.5,j+0.5]=[L-0.5,U+0.5].

If the probability mass at each count is approximated by the normal area in its associated unit-width interval, adding the areas for $j=L,\ldots,U$ gives the normal area from $L-0.5$ through $U+0.5$. This explains the correction. The equality above describes the interval geometry; replacing the discrete masses by normal areas is the approximation.

Worked Applications

Worked Example: A Binomial Count in an Inclusive Range

A hypothetical system independently processes 160 transactions. Each transaction is flagged with probability $0.40$, independently of all others. Let $X$ be the number flagged. Use a normal approximation with continuity correction to estimate $P(58\le X\le72)$.

Method and model. The count has distribution $X\sim\operatorname{Bin}(160,0.40)$. Its mean and variance are

\mu=np=160(0.40)=64,\qquad \sigma^2=np(1-p)=160(0.40)(0.60)=38.4.

Thus $\sigma=\sqrt{38.4}\approx6.1968$. The event includes the integers 58 through 72, so the corrected normal interval runs from 57.5 to 72.5:

P(58\le X\le72)\approx P(57.5<Y<72.5),\qquad Y\sim N(64,38.4).

Standardize both boundaries:

z_{\text{low}}=\frac{57.5-64}{6.1968}\approx-1.0489,\qquad z_{\text{high}}=\frac{72.5-64}{6.1968}\approx1.3717.

Using standard normal probabilities, $\Phi(1.3717)\approx0.9149$ and $\Phi(-1.0489)\approx0.1471$. Therefore,

P(58\le X\le72)\approx0.9149-0.1471=\boxed{0.7678}.

Check and trap avoided. The corrected interval is wider than the interval from 58 to 72 on the continuous scale because it represents all 15 included integers. Using 58 and 72 as the normal boundaries would omit part of the probability associated with the endpoint counts.

Worked Example: A Strict Upper Tail for a Poisson Count

In a hypothetical monitoring period, the number of alerts follows a Poisson distribution with mean 25 per period. Estimate the probability of more than 29 alerts using a normal approximation with continuity correction.

Method and model. Let $X\sim\operatorname{Poisson}(25)$. Then $\mu=25$, $\sigma^2=25$, and $\sigma=5$. The event $X>29$ means $X\ge30$. Its first included integer is 30, so the corrected normal event begins at 29.5:

P(X>29)=P(X\ge30)\approx P(Y>29.5),\qquad Y\sim N(25,25).

Standardize the boundary:

z=\frac{29.5-25}{5}=0.9.

Therefore,

P(X>29)\approx1-\Phi(0.9)=1-0.8159=\boxed{0.1841}.

Check and trap avoided. The event starts at 30, but its continuous lower boundary is 29.5. A common incorrect calculation uses 30 as the normal boundary, giving $1-\Phi((30-25)/5)=1-\Phi(1)=0.1587$. That calculation discards part of the area representing the count 30 and understates the corrected tail probability.

Worked Example: A Mixed Strict and Inclusive Interval

A hypothetical quality audit samples 100 independent components. Each component independently fails inspection with probability $0.20$. Let $X$ be the number that fail. Use a normal approximation with continuity correction to estimate the probability that at least 12 but fewer than 18 fail.

Method and model. Here $X\sim\operatorname{Bin}(100,0.20)$, so

\mu=100(0.20)=20,\qquad \sigma^2=100(0.20)(0.80)=16,\qquad \sigma=4.

The event is $12\le X<18$. Since $X$ is an integer, this is $12\le X\le17$, not $12\le X\le18$. The corrected boundaries are therefore 11.5 and 17.5:

P(12\le X<18)\approx P(11.5<Y<17.5),\qquad Y\sim N(20,16).

The standardized boundaries are

z_{\text{low}}=\frac{11.5-20}{4}=-2.125,\qquad z_{\text{high}}=\frac{17.5-20}{4}=-0.625.

Using $\Phi(-0.625)\approx0.2660$ and $\Phi(-2.125)\approx0.0168$, the estimated probability is

P(12\le X<18)\approx0.2660-0.0168=\boxed{0.2492}.

Check and trap avoided. The strict upper bound excludes 18, so the largest included count is 17 and the upper normal boundary is 17.5. Using 18.5 would incorrectly include the count 18 in the approximation.

A Reliable Endpoint Procedure

A short, consistent procedure prevents most continuity-correction errors. Keep the event in integer terms until both endpoints are settled; do not standardize an endpoint directly from the wording if the wording is strict.

Identify the count and its normal model.
State the distribution, parameters, mean, and variance. Use the normal approximation only when it is appropriate for that count model.

Rewrite strict inequalities.
For an integer-valued count, replace $X<k$ by $X\le k-1$, and $X>k$ by $X\ge k+1$.

Find the included integer endpoints.
For a range, name the smallest and largest counts that satisfy the event. For a one-sided event, identify its last or first included count.

Apply the half-unit adjustment.
Move a lower boundary down by 0.5 and an upper boundary up by 0.5, then standardize and evaluate the normal probability.

Check the event direction.
A lower-tail probability uses a cumulative probability; an upper-tail probability uses one minus that cumulative probability. For an interval, subtract the lower cumulative probability from the upper one.

Limits of the Correction and Common Traps

Continuity correction addresses the mismatch between a discrete count and a continuous approximation. It does not change the count model, correct a wrong mean or variance, or guarantee that the normal approximation is accurate. In particular, do not use the correction as a substitute for checking whether a normal approximation is reasonable for the distribution and parameters in the problem.

Applying the wrong strict-bound rule. $X<k$ ends at $k-1$, while $X\le k$ ends at $k$. Their corrected upper boundaries differ by one full unit.
Correcting only one endpoint of a range. A two-sided event needs both the lower boundary moved down by 0.5 and the upper boundary moved up by 0.5.
Subtracting in the wrong order. For a normal interval from $a$ to $b$, with $a<b$, the probability is $\Phi(b)-\Phi(a)$, not the reverse.
Treating an approximation as an exact probability. The half-unit boundaries reflect the integer lattice, but the area under the normal curve only approximates the count probability.

Key takeaway: First translate the wording into the exact integers included; then place the continuous boundaries halfway between included and excluded integers. This order makes strict and inclusive endpoints consistent in lower tails, upper tails, and intervals.

Check Your Understanding

For each count probability, use a normal approximation with continuity correction and give the result to four decimal places.

A hypothetical batch has $X\sim\operatorname{Bin}(100,0.50)$. Estimate $P(45\le X\le55)$. (Answer: 0.7287)
A hypothetical count has $X\sim\operatorname{Poisson}(36)$. Estimate $P(X<32)$. (Answer: 0.2266)
For a hypothetical audit, $X\sim\operatorname{Bin}(200,0.30)$. Estimate $P(55<X\le68)$. (Answer: 0.6614)
A hypothetical system has $X\sim\operatorname{Poisson}(25)$ alerts in a period. Estimate $P(X>28)$. (Answer: 0.2420)

Discrete event	Equivalent integer event	Continuity-corrected normal event
\(X\le k\)	At most \(k\)	\(Y<k+0.5\)
\(X<k\)	At most \(k-1\)	\(Y<k-0.5\)
\(X\ge k\)	At least \(k\)	\(Y>k-0.5\)
\(X>k\)	At least \(k+1\)	\(Y>k+0.5\)
\(L\le X\le U\)	\(L,L+1,\ldots,U\)	\(L-0.5<Y<U+0.5\)

What You'll Learn