Central limit theorem: Worked examples — AI-ML deep dive

Before any example, let me re-anchor the symbols we lean on the whole page, in plain words:

Definition Expectation

$E$ and Variance $Var$ — the two operators we reuse constantly

$E [X]$ (read "expectation of X", also called the mean) — the long-run average value of a random draw $X$ , weighing each possible value by its probability. For a source with values $x_{1}, \dots, x_{k}$ occurring with probabilities $p_{1}, \dots, p_{k}$ : $E [X] = \sum_{j} x_{j} p_{j}$ Picture the balance point of the source's bar chart — the spot where the bars would balance on a fingertip. We also write $μ = E [X]$ .
$Var (X)$ (read "variance of X") — the average squared distance of a draw from its own mean, i.e. how spread out the draws are: $Var (X) = E [(X - μ)^{2}] = E [X^{2}] - μ^{2}$ Picture the typical width of the bar-chart cloud. The why of the second form: expanding $(X - μ)^{2} = X^{2} - 2 μ X + μ^{2}$ and taking $E$ term-by-term (using $E [X] = μ$ ) collapses to $E [X^{2}] - μ^{2}$ — a shortcut that only needs the average of the squares and the square of the average.
$σ = Var (X)$ (sigma) — the population standard deviation: the spread put back on the original scale (undoing the squaring).

Definition The sample mean, and the three quantities you always compute

$\overset{ˉ}{X}_{n}$ (read "X-bar-n") — the sample mean: you take $n$ actual draws $X_{1}, X_{2}, \dots, X_{n}$ , add them, and divide by $n$ : $\overset{ˉ}{X}_{n} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}$ Picture: line up your $n$ dice rolls and find their balance point. This single number is the star of every example — everything we compute is a statement about how $\overset{ˉ}{X}_{n}$ behaves.
$μ$ (mu) — the population mean $E [X]$ : the long-run average of ONE draw. The CLT says $\overset{ˉ}{X}_{n}$ hovers around $μ$ .
$σ$ (sigma) — the population standard deviation $Var (X)$ : how far one draw typically lands from $μ$ .
$SE = \frac{σ}{n}$ — the standard error: how far $\overset{ˉ}{X}_{n}$ typically lands from $μ$ . It is the width of the bell curve the CLT hands you. It shrinks like $n$ : quadruple $n$ , halve the spread.

The standardize move appears in every single example, so define it once:

The scenario matrix

Every problem this topic can throw is one of these cells. Each example below is tagged with the cell(s) it fills.

Cell	What makes it distinct	Covered by
A. Symmetric source	uniform/symmetric $X$ , moderate $n$	Ex 1
B. Skewed / Bernoulli source	lop-sided $X$ (a coin), CLT still works	Ex 2
C. Solve for $n$ (inverse)	probability given, find sample size	Ex 3
D. One-sided tail	$P (\overset{ˉ}{X} > c)$ , only one boundary	Ex 4
E. Negative-mean / sign-flip	$μ < 0$ , careful with signs of Z	Ex 5
F. Degenerate: $σ = 0$	zero-variance source, limiting behaviour	Ex 6
G. Small $n$ warning	when the bell is a bad approximation	Ex 7
H. Limit $n \to \infty$	Law of Large Numbers as CLT's shadow	Ex 8
I. Sum instead of mean	total, not average — rescale carefully	Ex 9
J. Exam twist: symmetry probability	$P(	Z

I use rounded standard-normal areas throughout: $Φ (0.70) = 0.758$ , $Φ (0.16) = 0.564$ , $Φ (1) = 0.841$ , $Φ (1.96) = 0.975$ , $Φ (2) = 0.977$ , $Φ (0.5) = 0.691$ .

Example 1 — Cell A: symmetric source (dice)

Worked example Average of 36 dice between 3.3 and 3.7

Statement. Roll a fair die $n = 36$ times. Find $P (3.3 \leq \overset{ˉ}{X}_{36} \leq 3.7)$ .

Forecast: The die is centred at $3.5$ and $3.3$ – $3.7$ is a narrow band around it. Guess: a bit above half? (It's $\approx 0.52$ .)

Population mean. $μ = E [X] = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$ . Why this step? CLT centres the bell at $μ$ ; we need the centre first.
Population variance. $σ^{2} = E [X^{2}] - μ^{2} = \frac{91}{6} - 3. 5^{2} = 2.9167$ , so $σ = 1.708$ . Why this step? The bell's width comes from $σ$ ; we use the $E [X^{2}] - μ^{2}$ shortcut defined above.
Standard error. $SE = \frac{1.708}{36} = \frac{1.708}{6} = 0.2847$ . Why this step? We are averaging, so the relevant spread is SE, not $σ$ .
Standardize both edges. $Z_{lo} = \frac{3.3 - 3.5}{0.2847} = - 0.70$ , $Z_{hi} = \frac{3.7 - 3.5}{0.2847} = + 0.70$ . Why this step? Turns "between 3.3 and 3.7" into an area on the one tabulated curve.
Read areas. $Φ (0.70) - Φ (- 0.70) = 0.758 - 0.242 = 0.516$ .

Verify: By symmetry the answer must equal $2Φ (0.70) - 1 = 2 (0.758) - 1 = 0.516$ . ✓ Both edges the same distance from centre → symmetric $Z$ 's → sanity holds.

Figure — the CLT in one picture (Example 1). The chart below overlays the source and the average: the flat blue bars are one die (uniform — every face equally likely), and the orange curve is the distribution of the average of 36 dice. Notice how flat becomes bell-shaped, centred at $μ = 3.5$ (gray dashed line). The green shaded slice is exactly the $P (3.3 \leq \overset{ˉ}{X}_{36} \leq 3.7) \approx 0.516$ we computed in step 5 — look how narrow that band is compared with the full $1$ – $6$ range of a single die, which is the shrinking effect of dividing by $n$ .

Example 2 — Cell B: skewed / Bernoulli source (a coin)

Worked example Fraction of heads in 100 flips

Statement. A biased coin lands heads with $p = 0.3$ . Flip it $n = 100$ times. What is $P (\overset{ˉ}{X}_{100} \geq 0.35)$ , where $\overset{ˉ}{X}$ is the fraction of heads?

Forecast: $0.35$ is above the true rate $0.30$ , so this should be less than half. Guess $\approx 0.14$ .

Bernoulli parameters. $μ = E [X] = p = 0.30$ , and $σ^{2} = Var (X) = p (1 - p) = 0.3 \times 0.7 = 0.21$ , so $σ = 0.4583$ . Why this step? A Bernoulli source is skewed (not symmetric like a die), yet the CLT still applies to its average — this is exactly why polls work.
Standard error. $SE = \frac{0.4583}{100} = 0.04583$ . Why this step? The question is about the fraction of heads, which is a sample mean $\overset{ˉ}{X}_{100}$ ; its spread is SE, so we must divide $σ$ by $n$ before standardizing.
Standardize the single boundary. $Z = \frac{0.35 - 0.30}{0.04583} = 1.091 \approx 1.09$ . Why this step? One-sided question → one Z value.
Tail area. $P (\overset{ˉ}{X} \geq 0.35) = 1 - Φ (1.09) = 1 - 0.862 = 0.138$ .

Verify: $0.138 < 0.5$ as forecast (above the mean → right tail). Units check: fractions are dimensionless, $Z$ dimensionless. ✓

Example 3 — Cell C: solve for $n$ (inverse problem)

Worked example How many users for a 0.5% margin?

Statement. Conversion rate $p = 0.05$ . How many users $n$ so the sample rate is within $\pm 0.005$ of the truth with 95% confidence?

Forecast: Small margins need big $n$ . Guess "thousands".

Parameters. $μ = 0.05$ , $σ^{2} = Var (X) = 0.05 \times 0.95 = 0.0475$ , $σ = 0.2179$ . Why this step? Every CLT computation starts from $μ$ and $σ$ ; here $σ$ feeds directly into the margin formula we are about to invert.
Write the 95% condition. We need $P (∣ \overset{ˉ}{X}_{n} - μ ∣ \leq 0.005) = 0.95$ . On the standard normal, 95% of area sits within $\pm 1.96$ : $Φ (1.96) - Φ (- 1.96) = 0.95$ . Why this step? We invert the usual direction — the probability is fixed, we solve for $n$ .
Set margin $=$ $1.96 \times SE$ . $0.005 = 1.96 \cdot \frac{σ}{n}$ . Why this step? The half-width of a 95% interval is $1.96 SE$ ; force it to equal the desired margin.
Solve for $n$ . $n = \frac{1.96 σ}{0.005} = \frac{1.96 \times 0.2179}{0.005} = 85.42$ , so $n = 85.4 2^{2} = 7297.6$ (round up to 7298).

Verify: Plug back — $SE = 0.2179/ 7298 = 0.002551$ , and $1.96 \times 0.002551 = 0.005000 \leq$ target margin. ✓

Example 4 — Cell D: one-sided tail (gradient safety)

Worked example Will the mini-batch gradient point the right way?

Statement. Per-sample gradient has $μ_{g} = - 0.03$ , $σ_{g}^{2} = 0.25$ , batch $n = 64$ . What is $P (\overset{g}{ˉ}_{64} > 0)$ — i.e. the chance the batch points the wrong direction (positive when truth is negative)?

Forecast: True mean is negative but noise is large, so there is a real chance of a sign flip. Guess $\approx 0.3$ .

Standard error. $σ_{g} = 0.5$ , $SE = \frac{0.5}{64} = 0.0625$ . Why this step? $\overset{g}{ˉ}_{64}$ is a sample mean, so its spread is SE; we need it before we can standardize the boundary $0$ .
Standardize $0$ . $Z = \frac{0 - ( - 0.03 )}{0.0625} = \frac{0.03}{0.0625} = 0.48$ . Why this step? " $\overset{g}{ˉ} > 0$ " is a one-sided event; only the boundary $0$ needs a Z.
Tail area. $P (\overset{g}{ˉ}_{64} > 0) = 1 - Φ (0.48) = 1 - 0.684 = 0.316$ .

Verify: Positive $Z$ (the boundary sits above the mean) → tail smaller than half, $0.316 < 0.5$ . This is why Stochastic Gradient Descent tolerates a 32% chance of a wrongly-signed batch: over many steps the mean drift wins. ✓

Example 5 — Cell E: negative mean, sign discipline

Worked example Average error inside a window around a negative target

Statement. Same gradient source ( $μ_{g} = - 0.03$ , $σ_{g} = 0.5$ , $n = 64$ ). Find $P (- 0.04 \leq \overset{g}{ˉ}_{64} \leq - 0.02)$ .

Forecast: This is a symmetric window around the mean $- 0.03$ ; expect a modest area.

SE. $SE = 0.0625$ (as above). Why this step? Same sample-mean spread reused; both edges will be divided by it.
Standardize both edges — mind the signs. $Z_{lo} = \frac{- 0.04 - ( - 0.03 )}{0.0625} = \frac{- 0.01}{0.0625} = - 0.16$ , $Z_{hi} = \frac{- 0.02 - ( - 0.03 )}{0.0625} = \frac{+ 0.01}{0.0625} = + 0.16$ . Why this step? Subtracting a negative $μ$ is adding — the classic sign trap. The window is symmetric in $Z$ because it is symmetric around $μ$ .
Area. $Φ (0.16) - Φ (- 0.16) = 0.564 - 0.436 = 0.127$ .

Verify: $2Φ (0.16) - 1 = 2 (0.564) - 1 = 0.128 \approx 0.127$ (rounding). ✓ The symmetric window gave symmetric $\pm 0.16$ Z's — sign handling correct.

Example 6 — Cell F: degenerate source, $σ = 0$

Worked example A "die" that always shows 4

Statement. $X$ always equals $4$ (a constant random variable). Roll it $n = 50$ times, average. Describe the distribution of $\overset{ˉ}{X}_{50}$ .

Forecast: No randomness in → no spread out. The "bell" collapses to a single point.

Parameters. $μ = E [X] = 4$ , $σ^{2} = Var (X) = E [(X - 4)^{2}] = 0$ , so $σ = 0$ . Why this step? A constant has zero variance by definition.
Standard error. $SE = \frac{0}{50} = 0$ . Why this step? SE is the whole width of the CLT bell; a zero here tells us the bell has no width at all.
Interpret the degenerate limit. With $SE = 0$ the distribution of $\overset{ˉ}{X}_{50}$ is not a bell at all: it places all its probability on the single value $4$ . Formally this is a degenerate (point-mass) distribution: $P (\overset{ˉ}{X}_{50} = 4) = 1$ . There is no density curve to draw — just an arrow of probability $1$ standing at $x = 4$ . And $Z = \frac{X ˉ - 4}{0}$ is undefined, which is the CLT's polite way of saying "there is nothing to standardize".

Verify: $\overset{ˉ}{X}_{50} = \frac{1}{50} (4 + 4 + \dots + 4) = 4$ with probability 1. So $P (\overset{ˉ}{X} = 4) = 1$ , consistent with all probability sitting on one point. ✓ CLT requires $0 < σ^{2} < \infty$ ; this is the edge it excludes.

Example 7 — Cell G: small $n$ — the approximation is BAD

Worked example Why

$n = 3$ dice is not yet a bell Statement. Roll a die just $n = 3$ times. Compare the true $P (\overset{ˉ}{X}_{3} \leq 1)$ with the CLT estimate.

Forecast: $\overset{ˉ}{X}_{3} \leq 1$ means all three dice show $1$ — genuinely rare. CLT (a smooth bell) may misjudge this because the true distribution is still lumpy.

True probability. $\overset{ˉ}{X}_{3} \leq 1$ forces every die to be $1$ : $P = (1/6)^{3} = 1/216 = 0.00463$ . Why this step? With tiny $n$ we can compute exactly — a fair yardstick.
CLT estimate. $SE = 1.708/ 3 = 0.9863$ ; $Z = \frac{1 - 3.5}{0.9863} = - 2.535$ ; $Φ (- 2.535) \approx 0.0056$ . Why this step? Same machine, but $n = 3$ is far from "large".
Compare. True $0.00463$ vs CLT $0.0056$ — same ballpark but noticeably off, and the CLT can't see that the average is discrete (only multiples of $1/3$ occur).

Verify: $(1/6)^{3} = 1/216$ exactly. ✓ Lesson: the bell is a large- $n$ promise; near the tails with small $n$ it lies. This is why Confidence Intervals switch to the $t$ -distribution for small samples.

Example 8 — Cell H: the limit $n \to \infty$ (Law of Large Numbers)

Example 9 — Cell I: total (sum), not average

Worked example Total score of 36 dice

Statement. Let $S = \sum_{i = 1}^{36} X_{i}$ be the sum of 36 dice. Find $P (S \leq 120)$ .

Forecast: Expected total is $36 \times 3.5 = 126$ , and $120$ is below it, so under half. Guess $\approx 0.35$ .

Mean and variance of the SUM. $E [S] = n μ = 36 \times 3.5 = 126$ ; $Var (S) = n σ^{2} = 36 \times 2.9167 = 105$ , so $sd (S) = 105 = 10.247$ . Why this step? A sum scales differently from an average: its spread is $σ n$ , not $σ / n$ . Get this right or every Z is wrong.
Standardize. $Z = \frac{120 - 126}{10.247} = - 0.5855 \approx - 0.59$ . Why this step? Turns the " $S \leq 120$ " boundary into a Z on the standard normal, using the SUM's own standard deviation from step 1.
Area (one-sided lower tail). $P (S \leq 120) = Φ (- 0.59) = 1 - Φ (0.59) = 1 - 0.7224 = 0.278$ . Why this step? $120$ is below the mean, so we want the left-tail area up to $Z = - 0.59$ ; reflect through symmetry to use the tabulated $Φ (0.59)$ .

Verify: Consistency check via the mean: $S /36 = 120/36 = 3.333$ ; using the average machine, $SE = 1.708/6 = 0.2847$ and $Z = \frac{3.333 - 3.5}{0.2847} = - 0.585$ — identical Z to step 2. ✓ Sum-route and average-route must agree, and they do. The answer $0.278$ matches the forecast " $\approx 0.35$ " ballpark (a bit lower). ✓

Example 10 — Cell J: exam twist + word problem

Worked example "Within one standard error" quick-fire

Statement. A sensor's readings have unknown skew but finite $μ, σ$ . You take $n = 400$ readings. A colleague claims: "The sample mean lands within one standard error of the truth about two-thirds of the time." Verify their claim using the CLT, then state $P (∣ \overset{ˉ}{X} - μ ∣ \leq 2 SE)$ .

Forecast: "One SE" $\Rightarrow ∣ Z ∣ \leq 1$ ; the 68–95–99.7 rule says $\approx 68%$ . So the colleague is right. Two SE $\Rightarrow\approx 95%$ .

Translate to $Z$ . $∣ \overset{ˉ}{X} - μ ∣ \leq SE$ means $\frac{X ˉ - μ}{SE} \leq 1$ , i.e. $∣ Z ∣ \leq 1$ . Why this step? The word "one standard error" is literally the definition of $∣ Z ∣ \leq 1$ — no arithmetic with $σ$ or $n$ needed. That is the exam shortcut.
Area for $∣ Z ∣ \leq 1$ . $Φ (1) - Φ (- 1) = 2Φ (1) - 1 = 2 (0.841) - 1 = 0.682$ . So the sample mean lands within one SE about $68%$ of the time — the colleague is correct. Why this step? Converts the two-sided $∣ Z ∣ \leq 1$ into a single area using $Φ$ -symmetry.
Area for $∣ Z ∣ \leq 2$ . $P (∣ \overset{ˉ}{X} - μ ∣ \leq 2 SE) = 2Φ (2) - 1 = 2 (0.977) - 1 = 0.954$ , i.e. about $95%$ . Why this step? Answers the second half by the same two-sided rule with $∣ Z ∣ \leq 2$ .

Verify: $0.682 \approx$ two-thirds → colleague confirmed. $0.954 \approx 95%$ matches the standard 95% interval half-width of $1.96$ SE we used in Example 3. ✓ Notice the answer is independent of $n$ and of the skew — the twist is that you never needed the sensor's actual $σ$ .

Recall Quick self-test

The average of $n$ i.i.d. draws has spread ::: $SE = σ / n$ (shrinks like $n$ ) To get within margin $m$ at 95% confidence you need $n \geq$ ::: $(1.96 σ / m)^{2}$ For a SUM of $n$ draws the standard deviation is ::: $σ n$ (grows, not shrinks) The CLT fails when ::: $σ = 0$ (degenerate) or $σ^{2} = \infty$ (infinite variance), or $n$ is too small for the tails $P (∣ Z ∣ \leq 1)$ and $P (∣ Z ∣ \leq 2)$ are about ::: $0.68$ and $0.95$

See also: Central limit theorem · Normal Distribution · Confidence Intervals · Sample Size Calculation · Hypothesis Testing · Law of Large Numbers · Bootstrap Methods · Monte Carlo Methods