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Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

Discarding zero observations and resolving ties is unnecessary with probability one under the stated independent continuous model. Rank the absolute observations from $1$ to $n$, let $R_i$ be the rank of $|X_i|$, and form the positive Wilcoxon signed-rank statisticReject the null hypothesis for large $W^{+}$. Under symmetry about zero, each sign is a fair independent binary variable, independent of all the absolute values. To see this independence, for every measurable positive set $B$, symmetry gives $\mathrm{Pr}(X\in B)=\mathrm{Pr}(X\in -B)=\frac{1}{2}\mathrm{Pr}(|X|\in B)$. Independence of observations then gives independence of the whole sign vector from the absolute-value vector. Conditional on those absolute values, relabeling by rank therefore givesThe null distribution is consequently independent of the unknown symmetric shape. Its probability generating function is $2^{-n}{\prod }_{r=1}^n(1+z^r)$, so the coefficient of $z^w$ gives the exact probability of $W^{+}=w$. Choose an upper critical value with null tail probability at most the desired level, or randomize at the boundary to obtain that level exactly. The observed upper-tail probability is an exact one-sided p-value. A positive translation raises each Walsh average, as established below, so this upper-tail rule has the correct direction against positive centers for every fixed symmetric shape.

The independent-sign representation provesThe asymptotic null distribution, stated without proof, isThis supplies the usual upper normal distribution critical value when exact calculation is inconvenient; for the discrete upper tail $\mathrm{Pr}(W^{+}\ge c)$, use $c-1/2$ in the continuity correction.

For exact equivalence with the proposed statistic, reorder the observations so $|X_{(1)}|<\cdots <|X_{(n)}|$. Partition the unordered pairs, including diagonals, by the member with larger absolute value. The block belonging to $X_{(r)}$ consists of the $r-1$ pairs with an earlier member and its diagonal pair. For an earlier member $X_{(j)}$, the strict inequality $|X_{(j)}|<|X_{(r)}|$ means the sign of $X_{(j)}+X_{(r)}$ is the sign of $X_{(r)}$; the diagonal sum has that same sign. Thus the entire block supplies exactly $r$ positive averages when $X_{(r)}>0$, and none otherwise. Summing the blocks provesThe two tests therefore have exactly the same statistic, critical values and p-values, not merely the same limiting distribution. The Walsh averages are dependent, so their positivity indicators must not instead be treated as an independent binomial sample. For a translation $X_i=Z_i+\theta$ of any null-symmetric sample, every Walsh average increases by $\theta$; their positive count is nondecreasing, which also verifies the direction of the one-sided test.

The experimental unit for comparing the treatments is the pot: its two plants form a pair. Let $D_i=C_i-S_i$ be cross-fertilised minus self-fertilised height in pot $i$, for $i=1,\dots ,15$. The paired Wilcoxon signed-rank test assumes independent differences with a continuous distribution symmetric about a location $\delta$. Its null is $\delta =0$, with the one-sided formulation $\delta \le 0$, against $\delta >0$. Under this location model it asks whether cross-fertilisation produces a positive shift within a pot. The location is a median, and also a mean if it exists; a hypothesis merely about a median without symmetry does not justify the signed-rank null distribution.

Test 1 uses this paired test and its exact distribution. Form the fifteen differences, omit any zero differences, rank their absolute values, and sum the statistical ranks belonging to positive differences. That positive-rank sum is the reported $V=96$. Tied absolute differences receive average statistical ranks. Under the symmetric zero-location null, conditional on their magnitudes, the signs are independent fair signs. The exact upper-tail probability is obtained from all sign allocations to the observed statistical ranks: with fifteen nonzero differences there are $2^{15}$ equiprobable allocations. For untied absolute differences the rank weights are $1,\dots ,15$; ties would require the corresponding conditional rank weights. The reported probability of a positive-rank sum at least as large as observed is $0.0206$. No normal approximation is used.

Test 2 makes exactly the same paired comparison and has the same null and alternative hypotheses. exact=F replaces the exact distribution by a normal approximation with a continuity correction. Under an untied null the positive-rank sum hasMore generally, conditional on the nonzero absolute statistical ranks $r_i$, its variance is $\frac{1}{4}{\sum }_i{r}_i^2$. The upper-tail correction places the boundary half a lattice unit below the observed statistic. Standardizing gives the reported $Z=2.0163$, and its approximate $p$-value is $1-\Phi (Z)=0.0219$. Thus the slight difference from Test 1 reflects approximation, not a different scientific question.

Test 3 omits paired=T, so it is the unpaired Wilcoxon rank sum test. Its usual distribution-free null is that two independent samples have the same continuous distribution. In a common-shape location model this is a zero location difference, against a positive cross-minus-self shift. More generally “greater” directs the test towards the cross distribution being shifted upwards; it does not turn the test into a general comparison of means or medians for arbitrary unequal shapes.

To form its statistic, pool the thirty heights, rank them together with averaged statistical ranks for ties, and add the cross-group statistical ranks to get $R_C$. Equivalently use $U=R_C-n_C(n_C+1)/2$. Under independent untied samples,If the pooled ties have sizes $t_h$ and $N=n_C+n_S$, the variance becomesSubtract the null mean and the upper-tail continuity correction, then divide by this null standard deviation. This procedure gives the reported rank-sum $Z=3.0105$. Its approximate one-sided probability is $1-\Phi (Z)=0.0013$. This describes how the statistic and probability are obtained without reconstructing the unprinted heights.

For the statistician, use the paired analysis, preferably the exact Test 1 for these fifteen pairs. The independent units are the pots, while the two heights within a pot share growing conditions and cannot be treated as independent samples. Test 2 is a reasonable approximation to the same paired test and agrees closely. Test 3 discards the matching and uses a null variance that assumes independence between the samples, so its smaller $p$-value is not a reason to prefer it. The appropriate paired result gives evidence at $5\%$, but not at $1\%$, that cross-fertilised plants have a positive height shift relative to their paired self-fertilised plants, subject to the symmetry assumption for differences.