Introduction
The central limit theorem explains why standardized sums often become approximately normal as the sample size grows. It does not say that every sample size is large enough, or that every probability calculated from the approximation is equally reliable. A sample mean from a highly skewed population can remain asymmetric at moderate \(n\), and a tail probability can be sensitive to a small error in the approximating distribution.
Practical assessment therefore combines several diagnostics: sample size, population skewness, the location of the probability being estimated, and—for Bernoulli data—the expected numbers of successes and failures. These checks help decide how cautiously to interpret a CLT calculation. They are not substitutes for an exact method when one is available, and none is a universal guarantee of accuracy. This treatment is aligned with the SOA VEE Mathematical Statistics syllabus.
Why Sample Size and Skewness Work Together
The CLT standardizes the sample mean by its standard error, \(\sigma/\sqrt n\). Increasing \(n\) reduces the effect of the population’s skewness, but the starting amount matters: a strongly skewed population may need a larger sample than a nearly symmetric one before the sampling distribution looks reasonably normal. A statement such as “\(n\) is at least 30” is not a universal accuracy criterion.
A short derivation makes the scaling precise. Write \(Y_i=X_i-\mu\). Independence and \(E[Y_i]=0\) imply that, when the cube of \(\sum_iY_i\) is expanded, every mixed term has expectation zero. Thus the third central moment of the sum is \(nE[Y_1^3]\). Its variance is \(n\sigma^2\), so its standard deviation cubed is \(n^{3/2}\sigma^3\). Dividing gives
The same standardized sum is \(Z_n\), so its skewness is \(\gamma_1/\sqrt n\). This derivation assumes independent, identically distributed observations with finite variance and third central moment. It quantifies one aspect of shape, not a complete test of normality: distributions with the same skewness can still differ in other ways.
Worked Example: A Right-Skewed Measurement and an Upper Tail
Model and task. Suppose, hypothetically, that independent measurements \(X_1,\ldots,X_{64}\) follow an exponential distribution with mean \(\mu=100\) units. For this model the standard deviation is \(100\) units and the population skewness is \(2\). Assess the normal approximation to \(P(\bar X>125)\).
Step 1: Find the sampling standard error. Under the stated independent, identical distribution model, \(\operatorname{SE}(\bar X)=\sigma/\sqrt n=100/\sqrt{64}=100/8=12.5\) units.
Step 2: Check the remaining skewness. The standardized sample mean has skewness \(2/\sqrt{64}=2/8=0.25\). The averaging has reduced the asymmetry from \(2\), but it has not made the sampling distribution exactly normal.
Step 3: Calculate the normal approximation. Standardize the threshold: \(z=(125-100)/12.5=25/12.5=2.00\). Using the standard normal distribution, \(P(\bar X>125)\approx P(Z>2.00)=1-\Phi(2.00)\approx0.0228\), to four decimal places.
Interpretation and check. The result is a CLT-based estimate, not an exact probability. The threshold is two standard errors above the mean, so the calculation concerns an upper tail; the source distribution is right-skewed, a reason to be especially cautious about assuming that the normal tail is accurate. The computed \(z\)-score is positive, as it must be for a threshold above the mean. Method: combine the standard error, the skewness diagnostic, and a standardized cutoff. Common mistake avoided: treating \(n=64\) as an automatic guarantee of accuracy without considering the population’s asymmetry and the tail being estimated.
A Finite-Sample Error Bound—and Its Limits
When the observations are independent and identically distributed with \(0<\sigma^2<\infty\) and \(E|X_1-\mu|^3<\infty\), the Berry–Esseen theorem gives a bound on the maximum absolute difference between the standardized sum’s cumulative distribution function and the standard normal cumulative distribution function. Let \(\rho=E|X_1-\mu|^3/\sigma^3\). One standard form is
This result explains why a larger sample and a smaller standardized third absolute moment support a better uniform bound on cumulative probabilities. The quantity \(\rho\) is an absolute third moment, not the signed skewness \(\gamma_1\); positive and negative deviations do not cancel in \(\rho\). The bound is often conservative and can be too wide to settle whether a particular approximation is useful. It is also an absolute error bound: a small absolute discrepancy may still be large relative to a very small tail probability. The theorem does not apply when its stated moment and independence assumptions fail.
Tail Location Changes the Practical Stakes
A normal approximation can be adequate around the center while being less dependable far into a tail. There are two related reasons. First, asymmetry or heavier-than-normal tails can make the true distribution’s extreme probabilities differ from the normal curve. Second, when the probability itself is small, even a modest absolute error can be a large fraction of the quantity being estimated. A useful report should therefore identify the approximation as an estimate, especially for an extreme cutoff.
The standardized cutoff helps describe the location: values of \(z\) near zero concern the center, while large positive or negative values concern a tail. This is a diagnostic, not an accuracy certificate. A large sample does not by itself eliminate tail concerns, and a small calculated tail probability should not be presented with more precision than the approximation supports.
Worked Example: Bernoulli Data With Few Expected Successes
Model and task. A hypothetical inspection system independently flags each item with probability \(p=0.05\). For \(n=120\) inspected items, let \(X\) be the number flagged. Assess whether the normal approximation is well supported for \(P(X\le 4)\), and calculate the continuity-corrected normal estimate for reference.
Step 1: Check the success and failure counts. Here \(X\sim\operatorname{Binomial}(120,0.05)\), so \(E[X]=np=120(0.05)=6\), and the expected number of non-flags is \(n(1-p)=120(0.95)=114\). The common success-failure diagnostic asks for both counts to be at least 10. The expected success count is only 6, so this diagnostic is not met, despite \(n=120\).
Step 2: Check skewness and spread. For a binomial variable, the skewness is \((1-2p)/\sqrt{np(1-p)}\). Substitution gives \((1-0.10)/\sqrt{120(0.05)(0.95)}=0.90/\sqrt{5.70}\). Since \(\sqrt{5.70}\approx2.3875\), the skewness is approximately \(0.3770\). The mean is 6, the variance is \(np(1-p)=5.70\), and the standard deviation is approximately \(2.3875\).
Step 3: Calculate the continuity-corrected estimate. For \(P(X\le4)\), the continuity-corrected boundary is \(4.5\). Thus \(z=(4.5-6)/\sqrt{5.70}=-1.5/2.3875\approx-0.6283\), and the normal estimate is \(\Phi(-0.6283)\approx0.2649\), to four decimal places.
Interpretation and check. This calculation illustrates how the approximation would be formed; the success-failure diagnostic and positive skewness indicate that it should not be accepted without qualification. The continuity correction places the boundary halfway between the integer outcomes 4 and 5, and the negative \(z\)-score is consistent with a cutoff below the mean 6. Method: inspect the Bernoulli counts and skewness before computing a corrected normal estimate. Common mistake avoided: declaring the approximation safe because the total sample size is 120, while ignoring that only 6 successes are expected.
The Success-Failure Check for Bernoulli Data
For independent Bernoulli observations with common success probability \(p\), the sum \(X\) has mean \(np\), variance \(np(1-p)\), and skewness \[ \frac{1-2p}{\sqrt{np(1-p)}}. \] A widely used rule of thumb for a normal approximation to a binomial count or proportion is to check that both \(np\) and \(n(1-p)\) are at least 10. Some conventions use a different cutoff, so the criterion should be understood as a practical diagnostic, not a theorem. It flags cases in which one outcome is rare enough that the distribution may remain notably discrete and asymmetric.
For example, a large \(n\) does not make the check pass if \(p\) is very small: \(np\) can remain low. Conversely, merely satisfying the count rule does not guarantee small error for every cutoff, particularly in a far tail. When the check is borderline or fails, an exact binomial calculation may be preferable if it is feasible.
Worked Example: Choosing a Sample Size Using Two Diagnostics
Model and task. A hypothetical study records independent Bernoulli outcomes with success probability \(p=0.30\). Find the smallest integer \(n\) for which (i) the standardized sample mean’s skewness is no greater than \(0.20\), and (ii) both expected outcome counts meet the rule-of-thumb threshold of 10.
Step 1: Apply the skewness condition. For one Bernoulli observation, \(\gamma_1=(1-2p)/\sqrt{p(1-p)}=0.40/\sqrt{0.21}\). Since \(\sqrt{0.21}\approx0.4583\), \(\gamma_1\approx0.8729\). The sample mean’s skewness is \(\gamma_1/\sqrt n\), so the condition is \(0.8729/\sqrt n\le0.20\). Equivalently, \(\sqrt n\ge0.8729/0.20=4.3645\), and \(n\ge(4.3645)^2\approx19.05\). Thus the smallest integer satisfying the skewness condition is \(n=20\).
Step 2: Apply the success-failure counts. The requirements are \(0.30n\ge10\) and \(0.70n\ge10\). They imply \(n\ge33.333\ldots\) and \(n\ge14.286\ldots\), respectively. Both hold at the smallest integer \(n=34\).
Step 3: Combine the requirements and verify. Both diagnostics hold for the first time at \(n=34\), not \(n=20\). At \(n=34\), the expected success and failure counts are \(34(0.30)=10.2\) and \(34(0.70)=23.8\). The sample-mean skewness is approximately \(0.8729/\sqrt{34}=0.8729/5.8310\approx0.1497\), which is below \(0.20\).
Interpretation and check. The count criterion is the binding condition. Satisfying both checks supports considering a normal approximation but does not guarantee accuracy for every tail probability. Method: solve each diagnostic separately, round sample-size requirements upward to an integer, then take the larger requirement. Common mistake avoided: stopping at \(n=20\), which meets the skewness target but leaves only 6 expected successes, below the count-rule threshold.
Practical Assessment
A disciplined assessment separates what is known from what is only a heuristic. State the sampling model and check the CLT assumptions relevant to the calculation. For an approximately continuous population, consider the population’s skewness and whether the sample size plausibly reduces it. For Bernoulli data, calculate \(np\) and \(n(1-p)\), rather than checking \(n\) alone. Then locate the requested cutoff: a central probability and an extreme-tail probability need not have comparable practical accuracy.
A normal approximation should be treated with additional caution if observations are not independent, if the distribution has no finite variance, or if a required moment condition does not hold. Those situations may invalidate assumptions behind the usual CLT reasoning or the Berry–Esseen bound. A diagnostic that looks favorable cannot repair a failed model assumption.
- Sample size: larger \(n\) generally reduces the standardized effect of finite population skewness under the IID model.
- Shape: a highly skewed population can require more observations than a roughly symmetric one; skewness alone does not describe every distributional feature.
- Tail location: extreme probabilities are often more sensitive to approximation error, particularly when the source distribution is asymmetric.
- Bernoulli counts: check expected successes and failures separately; a large total sample can still have a sparse outcome.
- Conclusion: these are evidence to weigh, not guarantees. Use an exact method when the diagnostics are poor and exact computation is practical.
Check Your Understanding
For each problem, identify the relevant diagnostic before interpreting any normal approximation.
- Independent observations follow an exponential distribution with mean and standard deviation both equal to 40 units. For a sample of 25, what is the skewness of the standardized sample mean? (Answer: \(2/\sqrt{25}=0.4000\).)
- For \(n=200\) independent Bernoulli trials with \(p=0.03\), calculate the expected success and failure counts and the binomial skewness. (Answer: counts \(6\) and \(194\); skewness \(0.94/\sqrt{5.82}\approx0.3897\).)
- For a binomial count with \(n=80\) and \(p=0.25\), do both expected counts meet the threshold of 10, and what is its skewness? (Answer: yes, counts \(20\) and \(60\); skewness \(0.50/\sqrt{15}\approx0.1291\).)
- For Bernoulli data with \(p=0.40\), find the smallest \(n\) making the sample-mean skewness at most \(0.10\), and determine whether that \(n\) meets both expected-count thresholds of 10. (Answer: skewness requires \(n\ge17\); at \(n=17\), expected successes \(6.8\), so the success-failure check fails.)
- Independent exponential measurements have mean and standard deviation \(\mu\), and \(n=100\). Estimate \(P(\bar X>1.2\mu)\) by the normal approximation, and give the sample-mean skewness. (Answer: \(z=(1.2\mu-\mu)/(\mu/10)=2\), estimated tail probability \(0.0228\), and skewness \(2/\sqrt{100}=0.2000\).)