Why the Choice of Count Model Matters
A normal approximation can simplify probability calculations for a discrete count, but it is not a generic substitute for knowing how the count was generated. A binomial count records successes in a fixed number of independent trials with a common success probability. A Poisson count models events over a specified interval when the event count has a constant rate and independent increments. Both can be approximated by normal distributions when their parameters make the count sufficiently concentrated and nearly symmetric, but the appropriate mean and variance depend on the model.
The practical decision is twofold: first identify the count model from the assumptions, then judge whether its normal approximation is adequate for the particular event being asked about. A large mean alone is not a complete check. The variance, asymmetry, lower boundary at zero, and location of the target event all matter.
Compare the Parameters Before Calculating
For a binomial count, the usual normal model matches both its mean and variance: \(X\approx N(np,npq)\). For a Poisson count, it is \(X\approx N(\lambda,\lambda)\). In this notation the second argument of \(N(\mu,\sigma^2)\) is the variance, so the standard deviation is the square root of that argument. These are approximations, not equalities: the original counts are discrete and cannot be exactly normally distributed.
A continuity correction translates the integer event into an interval for a continuous normal variable. For example, \(P(X\ge k)\) is approximated by \(P(N\ge k-0.5)\), where \(N\) is the chosen normal model. The correction is applied after identifying the integer event and before standardizing. It is used for either count model.
The two distributions also differ in their skewness. For a binomial variable, the standardized skewness is \((q-p)/\sqrt{npq}\); for a Poisson variable, it is \(1/\sqrt{\lambda}\). A useful normal approximation generally has a small absolute skewness, not merely a large parameter written somewhere in the model. In particular, a binomial distribution with \(p\) close to 0 or 1 can remain asymmetric unless the expected number of outcomes on the less common side is also substantial.
A common screening rule for a binomial approximation is that both \(np\) and \(nq\) should be at least 5, with 10 often used as a more conservative threshold. For a Poisson approximation, \(\lambda\) around 10 or larger is a frequently used screening rule. These are guidelines, not guarantees of a particular error size. A target event very far into a tail can still be estimated poorly even when a general rule is met.
Why the Variance and Skewness Formulas Fit the Models
For a binomial count, write \(X=I_1+\cdots+I_n\), where each \(I_j\) equals 1 on a success and 0 otherwise. Independence gives \(E[X]=np\) and \(\operatorname{Var}(X)=npq\). A Bernoulli variable has third central moment \(pq(q-p)\); independence makes the third central moment of the sum \(npq(q-p)\). Dividing by the variance to the power \(3/2\) gives the binomial skewness:
For a Poisson variable, the mean, variance, and third central moment are each \(\lambda\). Its standardized skewness is therefore \(\lambda/\lambda^{3/2}=1/\sqrt{\lambda}\). These formulas explain why increasing the relevant expected counts tends to reduce the absolute skewness. They also show why the sign of binomial skewness is not a measure of approximation quality by itself: the magnitude is what indicates the extent of asymmetry.
Worked Examples: Apply the Model-Specific Approximation
Worked Example: A Nearly Symmetric Binomial Interval
A hypothetical inspection checks 100 components independently. Each component has probability \(0.50\) of requiring a minor adjustment. Estimate the probability that between 46 and 54 components, inclusive, require adjustment. Assume no dependence between inspections and no changes in the adjustment probability.
Method and model check. The number of adjustments is \(X\sim\operatorname{Binomial}(100,0.50)\). Thus \(n=100\), \(p=q=0.50\), the mean is \(100(0.50)=50\), and the variance is \(100(0.50)(0.50)=25\). Both expected counts are 50, and the binomial skewness is zero, so a normal approximation is reasonable for this central interval.
Apply the continuity correction. The inclusive integer interval \(46\le X\le54\) becomes \(45.5<N<54.5\), with \(N\sim N(50,25)\). The standard deviation is \(\sqrt{25}=5\), giving \(z_L=(45.5-50)/5=-0.90\) and \(z_U=(54.5-50)/5=0.90\).
Therefore, \(P(46\le X\le54)\approx\Phi(0.90)-\Phi(-0.90)=2\Phi(0.90)-1\). Using \(\Phi(0.90)=0.8159\), the estimate is \(2(0.8159)-1=0.6318\), or 0.6319 when evaluated before rounding the normal table value. The estimate is reported to four decimals.
Check and trap avoided. The corrected interval is centered on the mean, as it should be. Using boundaries 46 and 54 without the half-unit correction would estimate a narrower event. Here the equal success and failure probabilities also make the skewness zero; that symmetry should not be assumed for every binomial count.
Worked Example: Keep the Binomial Variance in a Tail Estimate
In a hypothetical batch of 200 independent sensor units, each unit has probability \(0.10\) of issuing a false alert during a test. Estimate the probability of at least 25 false alerts. Assume the same alert probability for all units and independence across units.
Method and model check. The count is \(X\sim\operatorname{Binomial}(200,0.10)\), so \(q=0.90\), \(E[X]=200(0.10)=20\), and \(\operatorname{Var}(X)=200(0.10)(0.90)=18\). The standard deviation is \(\sqrt{18}\approx4.2426\). The expected counts on both sides are \(20\) and \(180\), so the binomial screening conditions are met. Its skewness is \(0.80/\sqrt{18}\approx0.1886\), a moderate positive skew.
The event \(X\ge25\) becomes \(N\ge24.5\) under the continuity correction, where \(N\sim N(20,18)\). The standardized boundary is \(z=(24.5-20)/\sqrt{18}=4.5/\sqrt{18}\approx1.06066\). Thus the normal estimate, evaluated using the unrounded \(z\), is \(1-\Phi(1.06066)\approx\mathbf{0.1444}\), to four decimals.
A tempting alternative is to replace this binomial count by a Poisson variable with \(\lambda=np=20\), then use a normal approximation with variance 20. That gives \(z=4.5/\sqrt{20}\approx1.00623\) and an upper-tail estimate of about \(0.1571\). The discrepancy illustrates that preserving the mean does not preserve the variance: the binomial variance is 18, while the Poisson variance is 20. Since the stated model is binomial, the direct binomial normal approximation is the appropriate one of these two calculations.
Check and trap avoided. The corrected boundary is 24.5 because the event includes 25. Do not use a Poisson variance merely because \(p\) is relatively small; that replacement is an additional approximation, not part of the binomial normal method.
Worked Example: A Poisson Count With a Central Interval
Suppose a hypothetical operations desk receives a Poisson number of independent requests during one fixed shift, with mean 64 requests per shift. Estimate the probability of receiving between 56 and 72 requests, inclusive. The model is \(X\sim\operatorname{Poisson}(64)\) for that shift; no other time interval is being used.
Method and model check. For a Poisson count, the mean and variance are both \(\lambda=64\), so the normal approximation is \(N(64,64)\), with standard deviation \(8\). Its skewness is \(1/\sqrt{64}=0.125\), and the rate is large enough for a normal approximation to be plausible.
The inclusive interval becomes \(55.5<N<72.5\). Standardizing both boundaries gives \(z_L=(55.5-64)/8=-1.0625\) and \(z_U=(72.5-64)/8=1.0625\). Consequently, \(P(56\le X\le72)\approx\Phi(1.0625)-\Phi(-1.0625)\approx\mathbf{0.7120}\), to four decimals, using the unrounded standardized boundaries.
Check and trap avoided. The two corrected boundaries are equally far from 64, consistent with the symmetry of the matched normal model. The Poisson count itself is not symmetric; its skewness is positive. Do not interpret a symmetric normal approximation as a claim that the original count distribution is exactly symmetric.
A Practical Selection Process
Use the model assumptions first and the numerical checks second. A binomial model has a fixed number of trials and a common success probability; a Poisson model has a rate parameter for a count over a specified interval. If the problem states one of these models, do not switch models simply to obtain a convenient formula.
For a fixed number of independent yes-or-no trials, use the binomial parameters \(n,p\). For a Poisson count, identify \(\lambda\) for the exact time, area, or exposure interval in the question.
Use \(np\) and \(npq\) for a binomial count, or \(\lambda\) and \(\lambda\) for a Poisson count. Do not confuse a Poisson approximation to a binomial model with its direct normal approximation.
For the binomial model, inspect both \(np\) and \(nq\), as well as the magnitude of its skewness. For the Poisson model, inspect \(\lambda\) and \(1/\sqrt{\lambda}\). Treat screening thresholds as guidelines rather than guarantees.
Apply a half-unit continuity correction to the integer boundary. A far-tail event, a target near zero, or a boundary close to a model's support limit deserves extra caution even if the general parameter check looks satisfactory.
Support is another reason to be cautious. A binomial count cannot exceed \(n\), whereas a Poisson count has no finite upper limit; both counts are nonnegative. A normal distribution assigns probability to negative values and has unbounded support on both sides. When the normal model places appreciable probability outside the count's possible range, or when the target is close to a boundary, its estimate may be unreliable. A continuity correction helps align integer events with a continuous model, but it cannot repair a poor distributional fit.
Common Exam Trap: Matching the Mean but Not the Model
Suppose a problem specifies \(X\sim\operatorname{Binomial}(n,p)\). The calculation \(X\approx N(np,np)\) is not the binomial normal approximation: it has replaced the binomial variance \(npq\) by the Poisson variance \(np\). For example, in the sensor example, the incorrect variance would be \(20\), not \(18\). That change produces a different standard deviation and therefore a different tail probability.
The correction is to begin with the model actually given: use \(N(np,npq)\) for the binomial count and \(N(\lambda,\lambda)\) for a Poisson count. Then translate the event with the half-unit correction and standardize. A Poisson model may sometimes approximate a binomial count when trials are numerous and success probabilities are small, but this is a separate model approximation that should be justified rather than silently substituted.
This model-first approach is aligned with the SOA Exam P (Probability) syllabus: it prevents a correct normal calculation from being applied to the wrong count distribution. It also keeps the relevant limitation visible: normal estimates are approximations whose quality depends on parameters and on the event being estimated.
Check Your Understanding
For each problem, identify the count model, use its model-specific mean and variance, and apply a continuity correction where appropriate.
- A hypothetical fair-coin process has 100 independent trials. Use a normal approximation to estimate the probability of 46 through 54 successes, inclusive. (Answer: approximately \(0.6319\).)
- A Poisson count has mean 49 during one fixed monitoring period. Use a normal approximation to estimate the probability of fewer than 43 events. (Answer: approximately \(0.1766\).)
- A batch has 250 independent items, each with probability \(0.08\) of a specified defect. Use the direct binomial normal approximation to estimate the probability of at least 25 defects. (Answer: approximately \(0.1471\).)
- A count is known to be binomial with \(n=200\) and \(p=0.10\). Estimate the probability of at least 25 successes using the direct binomial normal approximation, then compare it with the estimate from first replacing the count by a Poisson variable with \(\lambda=np\). Which estimate uses the stated model directly? (Answer: direct binomial estimate \(\approx0.1444\); Poisson-then-normal estimate \(\approx0.1571\); the binomial estimate uses the stated model directly.)