Skewness and Boundary Effects in Normal Count Approximations

Why a Good Mean and Variance May Not Be Enough

A normal approximation uses a count distribution’s mean and variance to place and scale a symmetric curve. Those two quantities do not describe the entire shape of the count distribution. When a count is skewed, or when an event lies close to the smallest or largest possible count, the normal curve can assign probability in a way that the original model cannot. The approximation may be useful, but its error should be diagnosed rather than assumed to be small.

This tutorial focuses on three clues: the distribution’s skewness, the location of the event in relation to the mean, and the event’s proximity to a support boundary. Continuity correction remains important: it places the normal interval around the integer counts being approximated. It cannot, however, make a symmetric, unbounded normal distribution have the same shape or support as a discrete count distribution.

Definition: The skewness of a random variable $X$ with positive standard deviation $\sigma$ is its standardized third central moment: $$ \gamma_1=\frac{E[(X-E[X])^3]}{\sigma^3}. $$ Positive skewness indicates a longer or heavier right tail; negative skewness indicates a longer or heavier left tail. A normal distribution has skewness zero.

Skewness in Binomial and Poisson Counts

For a binomial count $X\sim\operatorname{Bin}(n,p)$, let $q=1-p$. Its mean is $np$, its variance is $npq$, and its skewness is

\gamma_1=\frac{1-2p}{\sqrt{npq}}.

Thus, the binomial distribution is positively skewed when $p<0.5$, symmetric when $p=0.5$, and negatively skewed when $p>0.5$. For a Poisson count $X\sim\operatorname{Poisson}(\lambda)$, the mean and variance are both $\lambda$, and its skewness is

\gamma_1=\frac{1}{\sqrt{\lambda}}.

A larger expected count generally reduces the magnitude of these skewness measures. This helps explain why normal approximations often improve as counts become larger. It does not guarantee accuracy for every event: a small probability in a far tail, or an event close to a support boundary, can remain sensitive to shape differences.

Justification for the binomial skewness formula: Write $X=B_1+\cdots+B_n$, where the independent variables $B_i$ are Bernoulli with success probability $p$. For one Bernoulli variable, $E[(B_i-p)^3]=pq(1-2p)$. When the centered independent terms are added, mixed terms in the third central moment have expectation zero, so $E[(X-np)^3]=npq(1-2p)$. The standard deviation is $\sqrt{npq}$. Dividing the third central moment by $(\sqrt{npq})^3$ gives $(1-2p)/\sqrt{npq}$.

Boundaries and Continuity Correction

A binomial count can only take values $0,1,\ldots,n$; a Poisson count can only take nonnegative integer values. The approximating normal variable can take any real value. It therefore assigns some probability below zero, and, for a binomial model, above $n$. This support mismatch matters most when the event of interest is near one of those boundaries.

Continuity correction matches an integer event to the interval of real values extending halfway to the adjacent integers. For a normal variable $Y$ used to approximate a count $X$, the standard translations include

P(X\le k)\approx P(Y<k+0.5),\qquad P(X\ge k)\approx P(Y>k-0.5), $$ $$ P(X=k)\approx P(k-0.5<Y<k+0.5).

These translations handle discreteness, not skewness or support. For example, approximating $P(X=0)$ for a Poisson count puts the normal interval from $-0.5$ to $0.5$. The lower part of that interval is impossible for the count, but the continuous normal model still assigns it probability. More generally, the normal tail shape may differ from the count tail even when the continuity-corrected boundary is placed correctly.

Check the model and support.
Identify whether the count is binomial or Poisson, its possible values, and the location of the event relative to zero and, for a binomial count, $n$.

Assess skewness.
Use the model’s skewness formula. A large magnitude signals that a symmetric normal curve may not represent the count’s tails well.

Locate the event.
Compare the event with the mean. A remote tail or a boundary event deserves particular caution, even if the mean and variance are matched.

Apply continuity correction, then qualify the result.
Translate the integer event correctly, calculate the normal probability, and decide whether skewness or support makes the approximation questionable. If feasible, compare with an exact probability.

Worked Comparisons

Each example uses the normal distribution with the same mean and variance as the count and applies a continuity correction. The comparisons show why the approximation’s numerical result should be considered alongside the distribution’s shape.

Worked Example: A Binomial Tail Near Its Upper Boundary

A hypothetical quality-control check counts defective components in a batch of 20. Assume independent outcomes, with each component defective with probability $p=0.9$. Let $X\sim\operatorname{Bin}(20,0.9)$. Approximate $P(X\ge19)$ by a continuity-corrected normal probability, compare it with the exact probability, and diagnose the difference.

Model and shape. Here $n=20$, $p=0.9$, and $q=0.1$. The mean and variance are

E[X]=np=20(0.9)=18,\qquad \operatorname{Var}(X)=npq=20(0.9)(0.1)=1.8.

The standard deviation is $\sqrt{1.8}\approx1.3416$. The skewness is

\gamma_1=\frac{1-2(0.9)}{\sqrt{20(0.9)(0.1)}} =\frac{-0.8}{\sqrt{1.8}}\approx-0.5963.

The distribution is negatively skewed, and the event $X\ge19$ is close to the upper support boundary $20$.

Continuity-corrected approximation. The event $X\ge19$ becomes $Y>18.5$, where $Y\sim N(18,1.8)$. Standardizing gives

z=\frac{18.5-18}{\sqrt{1.8}} =\frac{0.5}{\sqrt{1.8}}\approx0.3727, $$ $$ P(X\ge19)\approx1-\Phi(0.3727)\approx0.3547.

Exact check. Only counts 19 and 20 qualify:

P(X\ge19)=\binom{20}{19}(0.9)^{19}(0.1)+(0.9)^{20} =20(0.9)^{19}(0.1)+(0.9)^{20} $$ $$ \approx0.2701703+0.1215767=0.3917470\approx0.3917.

The normal approximation is about $0.3917470-0.354694\approx0.0371$ below the exact probability. The method was a continuity-corrected normal tail; the diagnostic is that substantial skewness and proximity to the upper boundary accompany a noticeable error. A common mistake is to treat matching the mean and variance, or applying the half-unit correction, as a guarantee of accuracy. Neither accounts for the count’s skewed tail.

Worked Example: A Poisson Probability at the Zero Boundary

Suppose a hypothetical monitoring system records $X\sim\operatorname{Poisson}(4)$ alerts per fixed interval. Approximate the probability of no alerts with a continuity-corrected normal distribution, compare it with the exact probability, and explain the boundary issue.

Model and shape. For $\lambda=4$, the Poisson mean and variance are both 4, so the standard deviation is 2. Its skewness is

\gamma_1=\frac{1}{\sqrt{4}}=0.5.

The count is positively skewed, and zero is the lower support boundary.

Continuity-corrected approximation. The event $X=0$ is represented by $-0.5<Y<0.5$, with $Y\sim N(4,4)$. The standardized endpoints are

z_{\mathrm{low}}=\frac{-0.5-4}{2}=-2.25,\qquad z_{\mathrm{high}}=\frac{0.5-4}{2}=-1.75.

Therefore,

P(X=0)\approx\Phi(-1.75)-\Phi(-2.25) \approx0.0400592-0.0122245 =0.0278347\approx0.02783.

Exact check. The Poisson probability of zero is

P(X=0)=e^{-4}\frac{4^0}{0!}=e^{-4}\approx0.0183156.

Thus the normal approximation is about $0.0278347-0.0183156=0.0095191$, or $0.00952$, too high. The continuity-corrected interval extends below zero, where the count has no possible values, while the normal distribution has positive density there. The skewness also matters. The method is not to subtract the impossible normal probability as if that repaired the approximation: the corrected interval is an approximation to the integer event, and its limitations should be acknowledged.

Worked Example: A Central Count With Symmetric Shape

A hypothetical audit examines 100 independent records, each having probability $0.5$ of meeting a specified criterion. Let $X\sim\operatorname{Bin}(100,0.5)$. Approximate $P(X=50)$, compare with the exact probability, and explain why this comparison differs from a boundary-tail example.

Model and shape. The mean is $100(0.5)=50$, the variance is $100(0.5)(0.5)=25$, and the standard deviation is 5. The skewness is

\gamma_1=\frac{1-2(0.5)}{\sqrt{25}}=0.

The event is at the center of the support, not near either boundary.

Continuity-corrected approximation. The event becomes $49.5<Y<50.5$, with $Y\sim N(50,25)$. The standardized endpoints are $-0.1$ and $0.1$, so

P(X=50)\approx\Phi(0.1)-\Phi(-0.1) \approx0.0796557\approx0.0797.

Exact check. Since each of the $2^{100}$ success-failure sequences is equally likely,

P(X=50)=\binom{100}{50}(0.5)^{100} =\frac{\binom{100}{50}}{2^{100}} \approx0.0795892\approx0.0796.

The approximation exceeds the exact probability by about $0.0796557-0.0795892=0.0000665$. Here the distribution is symmetric, the event is central, and the support boundary is distant. The approximation is close, though not exact. A common mistake is to conclude from one central example that the approximation must also work well in a distant tail; the event location and shape must be reassessed each time.

What the Diagnostics Can and Cannot Tell You

Skewness is a warning about shape, not an exact formula for approximation error. Positive skewness does not imply that every normal approximation will be too high or every left-tail approximation will be too low. The error depends on the particular event and on how the count distribution differs from the normal curve there. Likewise, a nearby boundary flags a support mismatch but does not quantify the resulting error.

Use the diagnostics together. A strongly skewed distribution, an event in a far tail, and proximity to a support boundary are reasons to be cautious. If the count parameters permit it, an exact binomial or Poisson calculation provides a useful check. When an exact calculation is unavailable or not requested, report the normal result as an approximation and avoid implying a precision that the model’s shape does not support.

Diagnostic	Question to ask	Why it matters
Skewness	Is $\gamma_1$ far from zero?	A symmetric normal curve may not match the count’s tail shape.
Tail location	Is the event far from the mean?	Small tail probabilities can be sensitive to shape differences.
Support boundary	Is the event near zero or, for a binomial count, near $n$?	The normal distribution assigns probability outside the count’s possible values.
Continuity correction	Were integer endpoints shifted by one-half in the right direction?	A correct correction handles discreteness, but not skewness or support mismatch.

Key takeaway: A continuity-corrected normal approximation addresses the discreteness of a count; it does not remove skewness or enforce the count’s support. Before trusting a tail or boundary probability, check the shape, event location, and possible values of the original count.

Check Your Understanding

For each problem, identify the relevant shape or boundary concern as well as the requested approximation. Use a continuity correction where asked.

Let $X\sim\operatorname{Bin}(40,0.2)$. Approximate $P(X\le3)$ using a normal distribution with matching mean and variance and a continuity correction. State whether the event is below or above the mean. (Answer: mean $8$, standard deviation $\sqrt{6.4}\approx2.5298$, event below the mean, approximation $\Phi(-1.7788)\approx0.0376$.)
Let $X\sim\operatorname{Bin}(25,0.8)$. Approximate $P(X\ge23)$ with a continuity correction, and state the binomial skewness. (Answer: skewness $-0.3000$; approximation $1-\Phi(1.25)\approx0.1056$.)
Let $X\sim\operatorname{Poisson}(9)$. Approximate $P(X\le1)$ by a continuity-corrected normal probability. Also calculate the exact probability to assess the approximation. (Answer: normal approximation $\Phi(-2.5)\approx0.0062$; exact probability $e^{-9}(1+9)\approx0.0012$, so the approximation is high for this lower-tail event.)
Let $X\sim\operatorname{Bin}(80,0.5)$. Approximate $P(38\le X\le42)$ using a matching normal distribution and continuity correction. (Answer: $2\Phi(2.5/\sqrt{20})-1\approx0.4239$.)
Let $X\sim\operatorname{Bin}(10,0.1)$. Use a continuity-corrected normal approximation for $P(X\ge1)$, then compare it with the exact probability. What makes this a questionable approximation? (Answer: normal approximation $1-\Phi(-0.5270)\approx0.7009$; exact probability $1-(0.9)^{10}\approx0.6513$. The mean is only $1$, skewness is positive, and the event lies next to the lower support boundary.)

Diagnostic	Question to ask	Why it matters
Skewness	Is \(\gamma_1\) far from zero?	A symmetric normal curve may not match the count’s tail shape.
Tail location	Is the event far from the mean?	Small tail probabilities can be sensitive to shape differences.
Support boundary	Is the event near zero or, for a binomial count, near \(n\)?	The normal distribution assigns probability outside the count’s possible values.
Continuity correction	Were integer endpoints shifted by one-half in the right direction?	A correct correction handles discreteness, but not skewness or support mismatch.

What You'll Learn