From the Central Limit Theorem to Probability Calculations
The central limit theorem is useful because it turns a broad class of sampling problems into calculations with the normal distribution. Even when individual observations are not normally distributed, their average can often be approximated by a normal random variable when the sample is sufficiently large. This tutorial uses that result to calculate probabilities for sample means and sample proportions. Such calculations are approximations, and their reliability depends on the assumptions and the shape of the underlying distribution.
Throughout, suppose \(X_1,\ldots,X_n\) are independent and identically distributed (iid) observations with finite mean \(\mu\) and finite, positive variance \(\sigma^2\). The sample mean is \(\bar{X}=n^{-1}\sum_{i=1}^n X_i\). The iid central limit theorem states that the standardized sample mean converges in distribution to a standard normal random variable as \(n\) increases. In applications, this supports an approximate normal distribution for \(\bar{X}\), centered at \(\mu\), with standard error \(\sigma/\sqrt{n}\).
The standardization follows by subtracting the approximate center and dividing by the approximate standard deviation. Since independent observations with variance \(\sigma^2\) have a sample mean with variance \(\sigma^2/n\), its standard error is \(\sigma/\sqrt n\). Thus, a cutoff \(c\) for the sample mean becomes the standard-normal cutoff \((c-\mu)/(\sigma/\sqrt n)\). For example, an upper-tail probability is calculated as \(P(\bar{X}>c)\approx P(Z>(c-\mu)/(\sigma/\sqrt n))\). The same operation applies to lower-tail probabilities and intervals.
Worked Example: An Upper-Tail Probability for a Sample Mean
Suppose a hypothetical population of daily service times has mean \(\mu=50\) minutes and standard deviation \(\sigma=12\) minutes. A sample of \(n=64\) days is drawn, and the observations are assumed to be iid from this population. Approximate the probability that the sample mean exceeds 52 minutes.
Check the conditions. The stated model assumes independence and a common population distribution across sampled days, with finite variance \(12^2\). The sample size is 64. A normal approximation is reasonable as an approximation if the population is not extremely skewed or dominated by rare, very large values; the central limit theorem does not guarantee a particular accuracy at \(n=64\).
The standard error is \(SE=\sigma/\sqrt n=12/\sqrt{64}=12/8=1.5\) minutes. Standardizing the cutoff gives \(z=(52-50)/1.5=1.3333\). Therefore,
Under the normal approximation, the probability that the mean service time for 64 sampled days exceeds 52 minutes is about 0.0912, or 9.12%. This is an approximate sampling probability under the stated hypothetical population model, not a probability that the population mean exceeds 52 minutes.
Probabilities for Intervals of Sample Means
An interval probability uses two standardized cutoffs. If \(a<b\), then \(P(a<\bar{X}<b)\) is approximated by the standard-normal probability between \((a-\mu)/SE\) and \((b-\mu)/SE\), where \(SE=\sigma/\sqrt n\). With \(\Phi(z)=P(Z\le z)\), the resulting calculation is the upper-cutoff cumulative probability minus the lower-cutoff cumulative probability.
Worked Example: A Probability Between Two Mean Cutoffs
In a hypothetical production process, a measured output \(X_i\) has population mean 100 units and standard deviation 18 units. Suppose a sample of 36 production runs is iid, the variance is finite, and the population distribution is not extremely skewed. Approximate the probability that the sample mean is between 98 and 103 units.
The standard error is \(SE=18/\sqrt{36}=3\) units. The standardized bounds are \((98-100)/3=-0.6667\) and \((103-100)/3=1.0000\). Hence,
The approximate probability that the mean output over 36 runs falls between 98 and 103 units is 0.5888, or 58.88%. The result refers to repeated samples of 36 runs under the model. It does not say that 58.88% of individual production runs fall in that interval.
Sample Proportions as Sample Means
A sample proportion is a sample mean of indicator variables. For each observation, let \(X_i=1\) if the event of interest occurs and \(X_i=0\) otherwise. If the indicators are iid Bernoulli variables with event probability \(p\), then \(E[X_i]=p\), \(\operatorname{Var}(X_i)=p(1-p)\), and their sample mean is the sample proportion \(\hat p\). Applying the central limit theorem gives an approximate normal distribution centered at \(p\), with standard error \(\sqrt{p(1-p)/n}\).
Because a proportion corresponds to an integer count, a continuity correction can improve a normal approximation to a count probability. Let \(Y=\sum_{i=1}^n X_i=n\hat p\). The normal approximation to \(Y\) has mean \(np\) and standard deviation \(\sqrt{np(1-p)}\). For an event \(Y\ge k\), the integer values begin at \(k\); the corresponding continuous normal area is conventionally taken from \(k-0.5\) upward. Thus, \(P(Y\ge k)\) is approximated by \(P(N\ge k-0.5)\), where \(N\) is the approximating normal variable. The half-unit adjustment aligns the continuous boundary with the integer count values. It is a refinement of the approximation, not a change to the event itself.
Worked Example: A Proportion at or Above a Threshold
Suppose a hypothetical process produces an event with probability \(p=0.08\) on each of \(n=200\) independent trials. Let \(\hat p\) be the observed proportion of events. Approximate \(P(\hat p\ge 0.10)\), using a continuity correction.
Check the conditions. The model specifies independent trials with the same event probability, so the indicators are iid Bernoulli. The expected event count is \(np=200(0.08)=16\), and the expected non-event count is \(n(1-p)=184\). Both are sufficiently large for a normal approximation to be plausible, although the approximation is not exact.
Let \(Y=200\hat p\), the number of events. The event \(\hat p\ge0.10\) is equivalent to \(Y\ge200(0.10)=20\). The approximating normal distribution for \(Y\) has mean \(np=16\) and variance \(np(1-p)=200(0.08)(0.92)=14.72\). Its standard deviation is \(\sqrt{14.72}\approx3.836665\). Applying the continuity correction, the upper-tail boundary is \(19.5\), so the standardized cutoff is \( (19.5-16)/\sqrt{14.72}=3.5/\sqrt{14.72}\approx0.91225\). Therefore,
The normal approximation with continuity correction gives a probability of about 0.1808, or 18.08%, that at least 10% of the 200 trials produce the event. The calculation uses the unrounded standardized cutoff before rounding the final probability.
Choosing and Interpreting the Approximation
For a sample mean, the key ingredients are the population mean, the population variance, and the sample size. For a sample proportion, the variance depends on \(p\). In a probability calculation under a specified model, use the model’s \(p\) in the standard error. The observed \(\hat p\) is a random result, not a replacement for the model parameter in the sampling distribution being calculated.
The central limit theorem is an asymptotic statement: as \(n\) grows, the distribution of the standardized mean approaches the standard normal distribution under the stated conditions. It does not say that every finite-sample distribution is normal, nor does it supply a universal sample size at which the approximation is accurate. A highly skewed distribution or one with rare extreme values can require a larger sample. For proportions, if \(p\) is close to zero or one, the count distribution can be strongly asymmetric unless the expected counts in both categories are adequate. A continuity correction can help with integer-valued counts, but it cannot repair a poor normal approximation.
A useful justification for standardization is that scaling the deviation of the mean by its standard error removes its units and gives it approximately unit variance. In fact, independence implies
Consequently, \((\bar{X}-\mu)/(\sigma/\sqrt n)\) has mean zero and variance one, and the iid central limit theorem gives its limiting standard normal distribution. The theorem supplies the approximate shape; the mean and variance calculation supplies the center and scale used in the numerical work.
Common Mistakes and Practical Checks
- Using the standard deviation instead of the standard error. For a sample mean, the relevant spread is \(\sigma/\sqrt n\), not \(\sigma\). Dividing by the wrong quantity can substantially alter the standardized cutoff.
- Forgetting the direction of a tail. If the standardized cutoff is positive and the event is “greater than,” the probability is \(1-\Phi(z)\), not \(\Phi(z)\).
- Treating the approximation as exact. The CLT describes convergence as sample size grows. A finite-sample result should be reported as approximate, and the distributional conditions should be considered.
- Applying a continuity correction to a proportion without converting to counts. First write \(Y=n\hat p\) and identify the integer cutoff. For \(Y\ge k\), use \(k-0.5\) as the normal boundary.
- Confusing a sampling probability with a parameter probability. A statement such as \(P(\bar{X}>c)\) describes how often a sample mean would exceed \(c\) under repeated sampling from a specified population; it is not the probability that the fixed population mean exceeds \(c\).
Check Your Understanding
Try these questions by identifying the sampling model before calculating.
- If iid observations have standard deviation 15 and the sample size is 25, what is the standard error of their sample mean?
- A sample mean has approximate mean 40 and standard error 2. What standard-normal cutoff corresponds to a sample-mean value of 43?
- Why can a sample proportion be treated as a sample mean of Bernoulli variables?
- If \(Y\) is an event count and the goal is to approximate \(P(Y\ge 12)\) with a normal distribution, what continuity-corrected boundary is used?
- Why does a large sample size alone not guarantee an accurate normal approximation for every finite sample?