Why a Continuous Curve Can Miss a Discrete Event
A count can take only whole-number values. For example, the number of defective items in a shipment might be 0, 1, 2, and so on. A normal random variable, by contrast, can take any real value, and its probability at any single point is zero. This difference matters when a probability is determined by a boundary such as “at least 56” or “no more than 44.”
The central limit theorem can make a normal distribution a useful approximation for a count, but it does not make the count continuous. If a normal approximation is evaluated at the same integer cutoff as the count event, its area may start or stop in the wrong place. A continuity correction adjusts the boundary so that the area under the continuous curve better represents the probability of the relevant integer values.
Where the Boundary Mismatch Comes From
Let \(X\) be an integer-valued count, and suppose a normal random variable \(Y\) is being used to approximate it. The discrete cumulative probability changes in jumps: moving from \(P(X\le k-1)\) to \(P(X\le k)\) adds exactly \(P(X=k)\). A continuous normal cumulative distribution has no such jump, because \(P(Y=k)=0\).
This difference explains why using an integer cutoff directly can miss some of the probability associated with the boundary count. For instance, the event \(X\ge k\) includes the entire probability at \(k\), while the uncorrected normal event \(Y\ge k\) starts at \(k\) and assigns no probability to the point \(k\). The correction moves the normal boundary to account for the width associated with an integer count. The exact geometric derivation of the half-unit adjustment is a separate step; here the important point is why matching the integer cutoff without adjustment can misstate the event probability.
For the straightforward events used below, the standard half-unit adjustment is to place the normal boundary halfway between adjacent integer counts. Thus, an event of the form \(X\ge k\) is approximated using \(Y\ge k-0.5\), and \(X\le k\) using \(Y\le k+0.5\). An interval including both integer endpoints, \(a\le X\le b\), is approximated using \(a-0.5\le Y\le b+0.5\).
Before standardizing, calculate the count’s mean and variance from its probability model. For \(X\sim\operatorname{Binomial}(n,p)\), these are \(np\) and \(np(1-p)\). For \(X\sim\operatorname{Poisson}(\lambda)\), both are \(\lambda\). For a sum of independent Bernoulli variables with success probabilities \(p_i\), the mean is \(\sum_i p_i\) and the variance is \(\sum_i p_i(1-p_i)\). In each case, standardize a boundary \(c\) using \(z=(c-\mu)/\sigma\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation.
Worked Examples
Worked Example: An Upper-Tail Binomial Count
A hypothetical quality audit checks 100 components. Assume the components are independent and each has probability \(0.5\) of failing the audit. Let \(X\) be the number that fail. Approximate \(P(X\ge56)\) with and without a continuity correction, and compare the results. There is no interest or mortality assumption in this probability model; \(X\sim\operatorname{Binomial}(100,0.5)\).
Method. First find the binomial mean and variance, then standardize the two candidate boundaries. The normal approximation is being used to illustrate the boundary effect; the correction does not itself establish that the approximation is accurate.
The mean and variance are \( \mu=np=100(0.5)=50 \) and \( \sigma^2=np(1-p)=100(0.5)(0.5)=25 \). Therefore, \( \sigma=\sqrt{25}=5 \).
Without a correction, the event is approximated by \(Y\ge56\), where \(Y\sim N(50,25)\). Its standardized boundary is \( z=(56-50)/5=1.2 \). Thus, \( P(Y\ge56)=1-\Phi(1.2)\approx1-0.8849=0.1151 \).
The integer event \(X\ge56\) includes the count 56. Using the half-unit boundary gives \(Y\ge55.5\), so \( z=(55.5-50)/5=1.1 \). The corrected approximation is \( P(Y\ge55.5)=1-\Phi(1.1)\approx1-0.8643=0.1357 \).
The uncorrected and corrected approximations differ by \(0.1357-0.1151=0.0206\), or 2.06 percentage points. The correction gives a larger value because it includes normal area representing the boundary count and its associated half-unit interval. Check: the upper-tail probability decreases as its lower boundary moves upward; since 55.5 is below 56, the corrected probability should be larger.
Worked Example: A Lower-Tail Binomial Count
A hypothetical system performs 100 independent tests, each with probability \(0.5\) of producing a flagged result. Let \(X\) be the number of flagged results. Approximate \(P(X\le44)\) both ways. There is no interest or mortality assumption; \(X\sim\operatorname{Binomial}(100,0.5)\).
As in the first example, \(\mu=100(0.5)=50\), \(\sigma^2=100(0.5)(0.5)=25\), and \(\sigma=5\). Without a correction, standardize 44: \( z=(44-50)/5=-1.2 \). Therefore, \( P(Y\le44)=\Phi(-1.2)\approx0.1151 \).
The event \(X\le44\) includes the integer count 44. The half-unit boundary is 44.5. Standardizing gives \( z=(44.5-50)/5=-1.1 \), and the corrected approximation is \( P(Y\le44.5)=\Phi(-1.1)\approx0.1357 \).
The corrected result is larger because the upper boundary moved from 44 to 44.5. This is the appropriate direction: increasing the cutoff of a lower-tail event includes more area. Check: by symmetry of the normal distribution centered at 50, this result matches the corrected upper-tail probability \(P(Y\ge55.5)\) in the first example.
Worked Example: An Interval for a Sum of Unequal Bernoulli Counts
A hypothetical monitoring network has 80 sensors, each independently triggering with probability \(0.3\), and 120 other independent sensors, each triggering with probability \(0.5\). Let \(S\) be the total number of triggers. Approximate \(P(78\le S\le90)\) with and without a continuity correction. There is no interest or mortality assumption; \(S\) is a sum of independent Bernoulli variables with the stated probabilities.
Method. Add the means and variances of the independent indicators. Then compare the normal area over the integer endpoints with the area extended halfway beyond each endpoint.
The mean is \( \mu=80(0.3)+120(0.5)=24+60=84 \). The variance is \( \sigma^2=80(0.3)(0.7)+120(0.5)(0.5)=16.8+30=46.8 \), so \( \sigma=\sqrt{46.8}\approx6.8411 \).
Without a correction, standardize the endpoints 78 and 90: \( z_{\mathrm{low}}=(78-84)/6.8411\approx-0.8771 \) and \( z_{\mathrm{high}}=(90-84)/6.8411\approx0.8771 \). Thus the uncorrected probability is approximately \( \Phi(0.8771)-\Phi(-0.8771)\approx0.8098-0.1902=0.6196 \).
The interval includes every integer from 78 through 90, so the corrected normal interval is from 77.5 through 90.5. Its standardized endpoints are \( (77.5-84)/6.8411\approx-0.9507 \) and \( (90.5-84)/6.8411\approx0.9507 \). The corrected probability is \( \Phi(0.9507)-\Phi(-0.9507)\approx0.8291-0.1709=0.6582 \).
The corrected approximation exceeds the uncorrected one because it extends the interval at both ends. Check: the model is centered at 84, and both intervals are symmetric about 84, so the lower and upper normal areas outside each interval are equal.
Common Mistakes and Limits
- Using the same integer cutoff without checking the event. For \(X\ge k\), the count \(k\) is included; for \(X\le k\), it is included on the other side of the boundary. The correction must reflect which integer counts belong to the event.
- Moving the boundary in the wrong direction. For an upper-tail event, lowering the boundary should increase the probability. For a lower-tail event, raising the boundary should increase it. Use these checks to catch sign errors.
- Confusing a continuity correction with an exact calculation. The corrected result remains a normal approximation. It is not the exact binomial or Poisson probability.
- Assuming a correction fixes model shape. It adjusts the event boundary, not skewness or other departures from a normal shape. A poor normal approximation can remain poor after correction.
A useful justification check is to compare the corrected and uncorrected answers. If a correction expands the interval or moves a one-sided cutoff outward to include additional counts, the corrected probability should move in the direction implied by that change. A result moving the other way usually signals an incorrectly standardized boundary or tail.
Check Your Understanding
For each problem, use a normal approximation with a continuity correction. Give probabilities to four decimal places.
- A hypothetical count has distribution \(X\sim\operatorname{Binomial}(100,0.5)\). Approximate \(P(X\ge57)\). (Answer: 0.0968)
- A hypothetical count has distribution \(X\sim\operatorname{Binomial}(100,0.5)\). Approximate \(P(X\le43)\). (Answer: 0.0968)
- A hypothetical event count has a Poisson distribution with parameter \(\lambda=81\). Approximate \(P(X\ge91)\). Assume for this calculation that the normal approximation is being used. (Answer: 0.1456)
- A hypothetical group has 60 independent trials with success probability \(0.4\) and 40 independent trials with success probability \(0.6\). Let \(S\) be the total number of successes. Approximate \(P(45\le S\le52)\). (Answer: 0.5833)
- A hypothetical count has distribution \(X\sim\operatorname{Binomial}(240,0.3)\). Approximate \(P(65\le X\le80)\). (Answer: 0.7390)