Central Limit Theorem — statement, proof sketch, significance: Worked examples

Before anything, three symbols we will lean on the whole page — each defined in plain words:

The one move every example makes: $z = \frac{( your cutoff ) - ( mean of the thing )}{( SD of the thing )}, P (thing \leq cutoff) \approx Φ (z) .$

The scenario matrix

Every CLT problem falls into one of these cells. The examples below are tagged with the cell they cover.

Cell	What makes it distinct	Covered by
A — Symmetric discrete, sum	flat/symmetric start, sum of many	Ex 1
B — Skewed continuous, mean	heavy right-skew start, average	Ex 2
C — Discrete count, needs continuity correction	Bernoulli sum, half-shifts	Ex 3
D — Two-sided ("between") probability	need $Φ (z_{2}) - Φ (z_{1})$	Ex 4
E — Inverse problem (solve for $n$ )	given error target, find sample size	Ex 5
F — Real-world word problem	translate English → $μ, σ, n$	Ex 6
G — Degenerate: $σ = 0$	zero variance, no bell at all	Ex 7
H — Limiting / failure: infinite variance	Cauchy Distribution, CLT breaks	Ex 8
I — Exam twist: difference of two means	combine two independent averages	Ex 9

We cover signs (positive and negative $z$ ), both tails, the "between" case, the zero-input case, and the limiting case where the theorem stops working. Nothing is left unshown.

Example 1 — Cell A: symmetric discrete, a big sum

Worked example Sum of 100 fair dice

A fair die shows $1$ – $6$ each with probability $\frac{1}{6}$ . Roll it $n = 100$ times, let $S = \sum X_{i}$ . Find $P (S > 380)$ .

Forecast: $380$ is above the mean $350$ — is it a little above or a lot? Guess whether the answer is closer to $0.4$ or $0.04$ .

Find $μ$ and $σ^{2}$ of one die. $μ = \frac{1 + 2 + \dots + 6}{6} = 3.5$ . Variance $= E [X^{2}] - μ^{2} = \frac{91}{6} - 3. 5^{2} = \frac{35}{12} \approx 2.9167$ , so $σ \approx 1.708$ . Why this step? The CLT only uses these two numbers; the flat shape of the die is irrelevant to the limit.
Scale up to the sum. $E [S] = 100 \cdot 3.5 = 350$ ; $Var (S) = 100 \cdot 2.9167 = 291.67$ ; $SD (S) = 291.67 = 17.078$ . Why this step? Means add and — because the rolls are independent — variances add. SDs do not add; you must add variances first, then square-root.
Standardize the cutoff. $z = \frac{380 - 350}{17.078} = 1.756$ . Why this step? This converts " $380$ on the dice scale" into " $1.756$ standard errors above the mean," which is the language the table $Φ$ speaks.
Read the tail. $P (S > 380) \approx 1 - Φ (1.756) = 1 - 0.9605 = 0.0395$ . Why this step? $Φ$ gives the left area; we want the right tail, so subtract from $1$ .

Verify: $z = 1.756 > 0$ means the cutoff is above the mean, so the probability of exceeding it should be below $0.5$ — and $0.0395$ is. Units cancel (dice minus dice over dice = pure number). Forecast: it was the small one, $\approx 0.04$ . ✓

Example 2 — Cell B: heavily skewed continuous, a sample mean

Worked example Average of 50 exponential waiting times

$X_{i} \sim Exp (λ = 2)$ (a very right-skewed distribution). For it, $μ = 1/ λ = 0.5$ and $σ = 1/ λ = 0.5$ . Take $n = 50$ measurements and average them into $\overset{ˉ}{X}$ . Find $P (\overset{ˉ}{X} > 0.6)$ .

Forecast: the raw exponential is lopsided — will its average still be roughly bell-shaped after only 50 terms? Guess yes or no.

Get the standard error. $\frac{σ}{n} = \frac{0.5}{50} = 0.0707$ . Why this step? We are averaging, not summing, so the spread shrinks like $1/ n$ . This is the SD of $\overset{ˉ}{X}$ .
Standardize. $z = \frac{0.6 - 0.5}{0.0707} = 1.414$ . Why this step? Same bridge as before: express $0.6$ in standard-error units.
Right tail. $P (\overset{ˉ}{X} > 0.6) \approx 1 - Φ (1.414) = 1 - 0.9214 = 0.0786$ .

Verify: the raw variable is skewed, yet the CLT still applies because its variance is finite. The answer $0.0786 < 0.5$ is consistent with $0.6$ being above the mean. Forecast: yes — averaging tames the skew. ✓ (This is Cell B: skewness is allowed, it just would need larger $n$ if the answer had to be super accurate deep in the tail.)

Example 3 — Cell C: discrete count, continuity correction

Figure — Central Limit Theorem — statement, proof sketch, significance

Worked example Binomial via the normal (de Moivre–Laplace)

$X \sim Bin (n = 200, p = 0.3)$ — a sum of 200 independent Bernoulli trials. Find $P (X \leq 50)$ .

Forecast: $50$ is below the expected count $n p = 60$ . Left tail or right tail? Above or below $0.5$ ?

Mean and SD of the count. $μ = n p = 200 \cdot 0.3 = 60$ ; $σ = n p (1 - p) = 200 \cdot 0.3 \cdot 0.7 = 42 = 6.4807$ . Why this step? A Bernoulli $(p)$ has variance $p (1 - p)$ ; add 200 of them.
Apply the continuity correction. For $P (X \leq 50)$ use the cutoff $50.5$ , not $50$ . Why this step? Look at the red bar in figure s01: each integer count is a bar of width 1 centred on the integer. The bar for " $50$ " stretches from $49.5$ to $50.5$ . To capture all of it under the smooth curve, the boundary sits at its right edge, $50.5$ . Missing this is the single most common CLT error.
Standardize. $z = \frac{50.5 - 60}{6.4807} = - 1.4659$ . Why this step? Note the negative sign — the cutoff is below the mean, so $z < 0$ . Signs matter: they place you on the correct side of the bell.
Read the left tail directly. $P (X \leq 50) \approx Φ (- 1.4659) = 0.0713$ . Why this step? $Φ$ already gives the left area, so no subtraction here. Use the symmetry $Φ (- a) = 1 - Φ (a)$ : $1 - Φ (1.466) = 1 - 0.9287 = 0.0713$ .

Verify: below-mean cutoff ⇒ probability below $0.5$ ⇒ $0.0713$ ✓. Without the correction we'd have used $50$ , giving $z = - 1.543$ and $Φ = 0.0614$ — a noticeable error from ignoring the bar width.

Example 4 — Cell D: a "between" (two-sided) probability

Worked example Sample mean landing in a window

Reuse the 100 dice from Example 1 but ask for the average $\overset{ˉ}{X} = S /100$ . Find $P (3.4 < \overset{ˉ}{X} < 3.6)$ .

Forecast: this window is centred on the mean $3.5$ . Should the answer be big (most of the bell) or tiny?

Standard error of the mean. $\frac{σ}{n} = \frac{1.708}{10} = 0.1708$ . Why this step? We switched from the sum to the mean, so divide the single-die $σ$ by $100 = 10$ .
Standardize both ends. $z_{1} = \frac{3.4 - 3.5}{0.1708} = - 0.5855, z_{2} = \frac{3.6 - 3.5}{0.1708} = + 0.5855.$ Why this step? A "between" question needs two boundaries; note they are symmetric ( $\pm$ ) because the window is symmetric about the mean.
Subtract the two left-areas. $P = Φ (0.5855) - Φ (- 0.5855)$ . Why this step? $Φ (z_{2})$ is everything left of the right wall; remove $Φ (z_{1})$ (everything left of the left wall); what remains is the strip between.
Evaluate. $Φ (0.5855) = 0.7209$ , $Φ (- 0.5855) = 0.2791$ , so $P = 0.7209 - 0.2791 = 0.4418$ .

Verify: the window is roughly $\pm 0.59$ SE wide, a bit more than half a standard deviation each side — capturing near $44%$ of the bell is sensible (recall $\pm 1$ SE captures $\approx 68%$ ). Symmetric window ⇒ $z_{1} = - z_{2}$ ✓. Forecast: moderately big, not tiny. ✓

Example 5 — Cell E: the inverse problem, solve for $n$

Worked example How much data to hit an error target?

You are estimating a mean with $σ = 2$ . You want a $95%$ confidence interval whose half-width (margin of error) is at most $0.25$ . How large must $n$ be?

Forecast: you want a tighter interval than one SE — will $n$ be in the tens, hundreds, or thousands?

Write the margin of error. For $95%$ we use $z_{0.025} = 1.96$ (the point with $2.5%$ in each tail). Margin $= z_{0.025} \cdot \frac{σ}{n} = 1.96 \cdot \frac{2}{n}$ . Why this step? The CLT says $\overset{ˉ}{X}$ is normal with spread $σ / n$ ; the interval reaches $1.96$ SEs each side.
Set it $\leq 0.25$ and solve. $1.96 \cdot \frac{2}{n} \leq 0.25 \Rightarrow n \geq \frac{1.96 \cdot 2}{0.25} = 15.68 \Rightarrow n \geq 15.6 8^{2} = 245.86.$ Why this step? Isolate $n$ , then square. Squaring is exactly why halving your error costs 4× the data.
Round up. $n = 246$ (you can never round down a sample-size requirement). Why this step? $n = 245$ would give a margin slightly over $0.25$ ; you must clear the bar.

Verify: plug $n = 246$ back: margin $= 1.96 \cdot 2/ 246 = 0.2499 \leq 0.25$ ✓. Plug $n = 245$ : margin $= 0.2504 > 0.25$ ✗ — confirms 246 is the smallest that works.

Example 6 — Cell F: real-world word problem

Worked example Elevator weight limit

An elevator is rated for $1200$ kg. Adult passenger weights are independent with mean $μ = 75$ kg and SD $σ = 15$ kg (shape unknown). If $n = 15$ adults board, what is the probability the total weight exceeds the rating?

Forecast: $15 \times 75 = 1125$ kg is the expected load, just under the limit. Risky or safe?

Translate to a sum. Total load $S = \sum_{i = 1}^{15} X_{i}$ . $E [S] = 15 \cdot 75 = 1125$ ; $Var (S) = 15 \cdot 1 5^{2} = 3375$ ; $SD (S) = 3375 = 58.095$ . Why this step? "Total weight exceeds" is a statement about the sum; means and variances add over the 15 independent people.
Standardize the limit. $z = \frac{1200 - 1125}{58.095} = 1.291$ . Why this step? Express the $1200$ kg rating in standard-error units of the total.
Right tail. $P (S > 1200) \approx 1 - Φ (1.291) = 1 - 0.9017 = 0.0983$ .

Verify: units check — kg minus kg over kg gives a pure $z$ . The mean load sits just $\approx 1.29$ SDs below the cap, so a roughly $10%$ overload risk is plausible and worth worrying about. Forecast: risky-ish, about $1$ in $10$ . ✓ (Note: with only $n = 15$ , this normal approximation is decent but not exact — the CLT is asymptotic.)

Example 7 — Cell G: the degenerate input $σ = 0$

Worked example What if every value is identical?

Suppose $X_{i} = 7$ always — a "random" variable with no randomness. So $μ = 7$ and $σ^{2} = 0$ . What does the CLT say about $\overset{ˉ}{X}_{n}$ ?

Forecast: can you even form the standard score? Look at the denominator.

Compute the sample mean. Every term is $7$ , so $\overset{ˉ}{X}_{n} = 7$ exactly, for every $n$ . No spread ever. Why this step? Averaging identical numbers returns that number; there is nothing to fluctuate.
Try to standardize. $Z_{n} = \frac{X ˉ _{n} - μ}{σ / n} = \frac{7 - 7}{0/ n} = \frac{0}{0}$ — undefined. Why this step? This shows why the parent definition demands $σ^{2} > 0$ . With zero variance there is no bell to converge to; you'd be dividing zero spread by zero spread.
State the correct conclusion. There is no Gaussian limit. The distribution of $\overset{ˉ}{X}_{n}$ is a single spike at $7$ (a "degenerate" distribution) — this is exactly what the Law of Large Numbers gives you, with the CLT layer collapsing to nothing.

Verify: the hypothesis " $σ^{2} > 0$ " in the theorem statement is not decoration — this case is precisely what it excludes. $\overset{ˉ}{X}_{n} \equiv 7$ has variance $0$ , matching "spike, no bell." ✓

Example 8 — Cell H: the limiting failure, infinite variance

Worked example When the CLT breaks: the Cauchy distribution

Let $X_{i}$ be i.i.d. standard Cauchy (a bell-looking curve with fat tails). It has no finite mean and infinite variance. What happens to $\overset{ˉ}{X}_{n}$ as $n \to \infty$ ?

Forecast: more data usually helps — will the average of a million Cauchy samples settle near a value?

Check the CLT hypothesis. The theorem requires finite variance $σ^{2}$ . For Cauchy, $σ^{2} = \infty$ (the fat tails make $\int x^{2} f (x) d x$ diverge). So the CLT does not apply. Why this step? Every hypothesis of a theorem is a promise you must honour; break one and the conclusion is void. We check hypotheses before computing so we don't blindly plug into a formula that isn't licensed here.
Write the characteristic function of one term. For the standard Cauchy, the characteristic function is $φ (t) = E [e^{i tX}] = e^{- ∣ t ∣}$ . Why this step? The Cauchy has no usable mean or variance, so the standardize-and-Taylor recipe of the parent proof is unavailable. The CF is the one fingerprint that still exists and behaves cleanly under sums, so we track it instead.
Build the CF of the average. By independence, CFs multiply, and scaling $X$ by $1/ n$ scales the argument of the CF: $φ_{\overset{ˉ}{X}_{n}} (t) = [φ (\frac{t}{n})]^{n} = [e^{- ∣ t / n ∣}]^{n} = e^{- n \cdot ∣ t ∣/ n} = e^{- ∣ t ∣} .$ Why this step? This is the exact same move as Step 2 of the parent proof (independence ⇒ product ⇒ $n$ -th power), but with the average $\overset{ˉ}{X}_{n} = \frac{1}{n} \sum X_{i}$ the scaling is $1/ n$ , not $1/ n$ . The two powers of $n$ cancel perfectly.
Recognize the limit — it never changed. $φ_{\overset{ˉ}{X}_{n}} (t) = e^{- ∣ t ∣}$ is identical to the CF of a single term. So $\overset{ˉ}{X}_{n}$ has exactly the same standard-Cauchy distribution as one $X_{1}$ , for every $n$ . It is not converging to $e^{- t^{2} /2}$ (the normal's CF); it converges to nothing new at all. Why this step? Because the CF is a unique fingerprint, equal CFs ⇒ equal distributions. Averaging a billion Cauchy samples is no more accurate than looking at one — exactly the wild red curve in figure s02.

Verify: the self-reproducing CF $e^{- ∣ t ∣} \to e^{- ∣ t ∣}$ (not $e^{- t^{2} /2}$ ) proves the limit is Cauchy, not normal. The theorem's finite-variance clause is exactly what fails, and the Lindeberg–Feller Condition is the general test that formally rules such cases out. ✓

Example 9 — Cell I: exam twist, difference of two means

Worked example Comparing two independent samples

Machine A produces parts with mean length $μ_{A} = 50$ mm, SD $σ_{A} = 4$ mm; you sample $n_{A} = 64$ . Machine B has $μ_{B} = 48$ mm, SD $σ_{B} = 3$ mm; you sample $n_{B} = 36$ . The two samples are independent. Find $P (\overset{ˉ}{X}_{A} - \overset{ˉ}{X}_{B} > 3)$ .

Forecast: the true difference of means is $50 - 48 = 2$ mm. Is exceeding $3$ mm common or rare?

Mean of the difference. $E [\overset{ˉ}{X}_{A} - \overset{ˉ}{X}_{B}] = μ_{A} - μ_{B} = 50 - 48 = 2$ . Why this step? Expectation is linear: the mean of a difference is the difference of means.
Variance of the difference. Each sample mean is $\approx$ normal by the CLT with variance $σ^{2} / n$ : $Var (\overset{ˉ}{X}_{A}) = \frac{16}{64} = 0.25, Var (\overset{ˉ}{X}_{B}) = \frac{9}{36} = 0.25.$ Because the samples are independent, variances of a difference add: $Var (\overset{ˉ}{X}_{A} - \overset{ˉ}{X}_{B}) = 0.25 + 0.25 = 0.5$ . SD $= 0.5 = 0.7071$ . Why this step? $Var (U - V) = Var (U) + Var (V)$ for independent $U, V$ — the minus sign does not subtract variances (variance can never shrink from adding independent noise).
Standardize. $z = \frac{3 - 2}{0.7071} = 1.4142$ . Why this step? The difference of two normals is itself normal, so one last standardization finishes it.
Right tail. $P > 1 - Φ (1.4142) = 1 - 0.9214 = 0.0786$ .

Verify: cutoff $3$ sits one true-difference-plus-a-bit above the mean $2$ , $\approx 1.41$ SDs out ⇒ answer well below $0.5$ . Notice $0.0786$ matches Example 2's tail because both had $z = 1.414$ — a nice internal consistency check. ✓

Recall Which cell was which? (quick self-test)

Symmetric discrete sum ::: Cell A (Ex 1 — dice) Skewed continuous mean ::: Cell B (Ex 2 — exponential) Discrete count needing $+ 0.5$ shift ::: Cell C (Ex 3 — binomial) "Between" two bounds ::: Cell D (Ex 4 — $Φ (z_{2}) - Φ (z_{1})$ ) Solve for sample size $n$ ::: Cell E (Ex 5 — square to get $n$ ) English word problem ::: Cell F (Ex 6 — elevator sum) Zero variance, no bell ::: Cell G (Ex 7 — $0/0$ , degenerate) Infinite variance, CLT fails ::: Cell H (Ex 8 — Cauchy) Difference of two means ::: Cell I (Ex 9 — variances still add)