Why Poisson Count Probabilities Need Corrected Endpoints
A Poisson random variable records a count, so its possible values are separated integers. A normal random variable, by contrast, can take any real value. When a normal distribution is used to approximate a Poisson probability, the event must be translated into an interval that accounts for the width represented by each integer count. The continuity correction makes that translation by placing a half-unit boundary on either side of an integer.
For a Poisson count with parameter \(\lambda\), the matching normal approximation has mean \(\lambda\) and variance \(\lambda\). The approximation can be useful for probabilities of one count or a finite range of counts, but it is not an exact identity. The method is most appropriate when \(\lambda\) is sufficiently large for the normal approximation to be reasonable; there is no single cutoff that guarantees good accuracy in every application.
The Corrected-Endpoint Calculation
The standard normal cumulative distribution function is denoted by \(\Phi(z)\), where \(\Phi(z)=P(Z\le z)\) for \(Z\sim N(0,1)\). Standardizing an endpoint \(c\) for \(Y\sim N(\lambda,\lambda)\) gives \(z=(c-\lambda)/\sqrt{\lambda}\). Therefore, a finite inclusive count interval is approximated by
For one exact count, the same method uses the lower and upper boundaries \(k-0.5\) and \(k+0.5\). For an interval, the lower correction is applied to the smallest included integer, and the upper correction is applied to the largest included integer. These directions follow from the event’s endpoints, not from whether the associated \(z\)-scores are positive or negative.
Write \(X\sim\operatorname{Poisson}(\lambda)\). Use the mean for the specified exposure period, not automatically a rate stated per unit time.
For one count \(k\), use \(k-0.5\) to \(k+0.5\). For \(a\le X\le b\), use \(a-0.5\) to \(b+0.5\).
Subtract \(\lambda\) from each endpoint and divide by \(\sqrt{\lambda}\).
Subtract the lower-endpoint CDF value from the upper-endpoint CDF value, retaining enough precision to round the final probability correctly.
Why the Half-Unit Correction Works
The correction is a geometric approximation: the continuous interval from \(k-0.5\) to \(k+0.5\) has width 1 and is centered at the integer \(k\). It therefore assigns the normal curve’s area around \(k\) to that discrete count. For consecutive integers \(a,a+1,\ldots,b\), the unit-width intervals fit together without gaps:
The shared boundaries have probability zero under a continuous normal distribution, so including or excluding those boundary points does not change its probability. This explains why an inclusive integer event from \(a\) through \(b\) corresponds to the corrected continuous interval with outer endpoints \(a-0.5\) and \(b+0.5\). The resulting area approximates the discrete Poisson probability; it does not make the Poisson count itself continuous.
The normal parameters also have a direct basis. The moment generating function of a Poisson random variable is \(M_X(t)=\exp(\lambda(e^t-1))\). Differentiating gives \(M_X'(0)=\lambda\), so \(E[X]=\lambda\). Its second derivative at zero is \(M_X''(0)=\lambda^2+\lambda\), and hence
Thus the approximating normal distribution uses the Poisson mean and variance, with standard deviation \(\sqrt{\lambda}\). The correction addresses the discrete event boundaries; it does not change those normal parameters.
Worked Example: Approximating One Exact Count
A hypothetical monitoring system records the number of signal interruptions during a fixed four-hour test. Suppose the count \(X\) is modeled as \(\operatorname{Poisson}(25)\). Approximate the probability of exactly 25 interruptions. Give the probability to four decimal places.
Method. Use the Poisson normal approximation \(Y\sim N(25,25)\), so the standard deviation is \(\sqrt{25}=5\). The exact-count event is represented by the corrected interval from \(24.5\) to \(25.5\).
Standardize each endpoint:
Using unrounded standard normal values, \(\Phi(0.1)=0.5398278\) and \(\Phi(-0.1)=0.4601722\). Therefore,
The approximation is about \(0.0797\), or 7.97%. Check: the corrected interval is centered at the mean, so its two \(z\)-scores are opposites and the probability is less than 1, as expected for a single count. The common mistake avoided is using only one boundary, or treating the event \(X=25\) as a zero-width interval.
Worked Example: An Inclusive Interval Around the Mean
In a hypothetical one-day study, the number \(X\) of equipment alerts is modeled as \(\operatorname{Poisson}(36)\). Approximate the probability of observing between 29 and 43 alerts, inclusive. Give the probability to four decimal places.
Method. The normal approximation is \(Y\sim N(36,36)\), with standard deviation \(\sqrt{36}=6\). Since the count interval includes both integer endpoints, correct the lower endpoint to \(28.5\) and the upper endpoint to \(43.5\).
The standardized bounds are
Using \(\Phi(1.25)=0.8943502\) and \(\Phi(-1.25)=0.1056498\),
The approximation is \(0.7887\). Check: the corrected interval is centered at 36, and the normal distribution is symmetric around its mean, so the two tail areas outside these bounds are equal. The common mistake avoided is omitting the correction: using 29 and 43 directly would instead produce \(z\)-scores of \(-7/6\) and \(7/6\), and an approximate area of \(0.7566\), for a different continuous interval.
Worked Example: An Asymmetric Finite Count Interval
A hypothetical service desk models the number \(X\) of incoming requests in a specified shift as \(\operatorname{Poisson}(25)\). Approximate the probability that the shift has at least 18 and at most 31 requests. Give the probability to four decimal places.
Method. Use \(Y\sim N(25,25)\) with standard deviation \(5\). The inclusive integer event \(18\le X\le31\) becomes the corrected interval \(17.5<Y<31.5\).
Standardizing gives
Using \(\Phi(1.3)=0.9031995\) and \(\Phi(-1.5)=0.0668072\),
The approximation is \(0.8364\). Check: the standardized bounds are not opposites because the integer interval is not centered at 25; the probability must therefore be calculated by subtracting the two distinct CDF values. The common mistake avoided is assuming symmetry merely because the approximating distribution is normal.
Common Traps and Practical Checks
The continuity correction is simple, but errors often arise before the CDF values are evaluated. In an exam calculation, write the corrected endpoints explicitly before standardizing. This makes it easier to confirm that the event and the interval still correspond.
- Using the wrong side of an endpoint. For \(a\le X\le b\), the corrections are \(a-0.5\) and \(b+0.5\). Reversing either one changes which counts are included.
- Leaving the endpoints uncorrected. For an inclusive count interval, using \(a\) and \(b\) as normal boundaries omits half of the boundary count at each end.
- Using the rate instead of the expected count. If a Poisson rate is given per hour but the event covers several hours, first determine \(\lambda\) for the full period. The normal mean and variance both use that period-specific parameter.
- Using \(\lambda\) as the standard deviation. The Poisson variance is \(\lambda\), so the normal standard deviation is \(\sqrt{\lambda}\), not \(\lambda\).
- Rounding CDF values too early. Keep enough digits through the subtraction, then round the final probability to the requested precision.
A corrected normal probability remains an approximation. When the Poisson parameter is not large enough for a normal model to be suitable, a direct Poisson calculation may be preferable. The continuity correction improves the translation of count events into normal intervals, but it cannot guarantee accuracy when the underlying normal approximation is poor.
Check Your Understanding
For each problem, use the continuity-corrected normal approximation and round the final probability to four decimal places.
- For \(X\sim\operatorname{Poisson}(16)\), approximate \(P(X=16)\). (Answer: 0.0995)
- For \(X\sim\operatorname{Poisson}(25)\), approximate \(P(20\le X\le30)\). (Answer: 0.7287)
- For \(X\sim\operatorname{Poisson}(36)\), approximate \(P(29\le X\le43)\). (Answer: 0.7887)
- For \(X\sim\operatorname{Poisson}(100)\), approximate \(P(86\le X\le114)\). (Answer: 0.8529)