v1 · padrão canônico

Lição 77 — Teorema Central do Limite

A média de n v.a. iid converge à normal independente da distribuição original — a lei mais importante da estatística. Demonstração via função característica, velocidade Berry-Esseen, aplicações de inferência.

Used in: 2.º ano do EM (16-17 anos) · Math B japonês §4.4 · Stochastik LK alemão · H2 Math singapurense cap. 21

\bar X_n \xrightarrow{d} \mathcal{N}\!\left(\mu,\,\frac{\sigma^2}{n}\right) \quad (n \to \infty)

The Central Limit Theorem (CLT) states that the mean of $n$ independent and identically distributed random variables with mean $\mu$ and finite variance $\sigma^2$ converges in distribution to a normal distribution with parameters $\mu$ and $\sigma^2/n$ — regardless of the original distribution. This explains why the bell curve appears in natural data, measurement errors, and sample means.

Choose your door

Rigorous notation, full derivation, hypotheses

Formal statement and proof

Lindeberg-Lévy Version

Definition· Central Limit Theorem (classical version)

Let $X_1, X_2, \ldots$ be iid (independent and identically distributed) random variables with $E[X_i] = \mu \in \mathbb{R}$ and $\text{Var}(X_i) = \sigma^2 \in (0, +\infty)$ . Define the standardized statistic

$Z_n = \frac{\bar X_n - \mu}{\sigma/\sqrt{n}}, \quad \bar X_n = \frac{1}{n}\sum_{i=1}^n X_i.$

Then $Z_n \xrightarrow{d} \mathcal{N}(0, 1)$ : for all $z \in \mathbb{R}$ ,

$\lim_{n \to \infty} P(Z_n \leq z) = \Phi(z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^z e^{-t^2/2}\,dt.$

"The central limit theorem is the unofficial sovereign of probability theory." — Grinstead & Snell, Introduction to Probability, §9.1

Version for sums

If $S_n = X_1 + \cdots + X_n$ , then $S_n \approx \mathcal{N}(n\mu,\, n\sigma^2)$ for large $n$ .

Z_n = \frac{S_n - n\mu}{\sigma\sqrt{n}} \xrightarrow{d} \mathcal{N}(0,1)

what this means · Standardization of the sum: same formula, different scaling.

Convergence rate: Berry-Esseen inequality

Proof sketch via characteristic function

Let $Y_i = (X_i - \mu)/\sigma$ (zero mean, unit variance). Taylor expansion of $\varphi_{Y_i}$ :

$\varphi_{Y_i}(t) = 1 - \frac{t^2}{2} + o(t^2) \quad (t \to 0).$

For $Z_n = (Y_1 + \cdots + Y_n)/\sqrt{n}$ :

$\varphi_{Z_n}(t) = \left[\varphi_{Y_i}\!\left(\frac{t}{\sqrt{n}}\right)\right]^n = \left[1 - \frac{t^2}{2n} + o\!\left(\frac{1}{n}\right)\right]^n \xrightarrow{n\to\infty} e^{-t^2/2}.$

But $e^{-t^2/2}$ is the characteristic function of $\mathcal{N}(0, 1)$ . The Lévy continuity theorem concludes $Z_n \xrightarrow{d} \mathcal{N}(0,1)$ . $\blacksquare$

When the CLT fails

Essential hypotheses

Independence (minimally sufficient; relaxable to $\alpha$ -mixing).
Finite variance $\sigma^2 < \infty$ .
$n$ sufficiently large — practical rule: $n \geq 30$ for non-highly-skewed distributions; $n \geq 100$ for high skewness.

Worked examples

Example— 1· Average weight of boxes mailed at post office

Problem. Boxes mailed at the post office have mean weight $\mu = 2.4$ kg and standard deviation $\sigma = 0.8$ kg (unknown distribution). A sorting process randomly selects 64 boxes. What is the probability that the sample mean weight exceeds 2.6 kg?

Strategy. By CLT, $\bar X_{64} \approx \mathcal{N}(2.4,\; 0.8^2/64)$ . Standardize and consult z-table.

Solution.

$\sigma_{\bar X} = 0.8/\sqrt{64} = 0.8/8 = 0.1$ kg.

$z = \frac{2.6 - 2.4}{0.1} = \frac{0.2}{0.1} = 2.0.$

$P(\bar X > 2.6) = P(Z > 2.0) = 1 - \Phi(2.0) = 1 - 0.9772 = 0.0228 \approx 2.3\%$ .

Verification. Rule 68-95-99.7: $\pm 2\sigma_{\bar X}$ contains 95%, so 2.5% in each tail. Result of 2.3% is consistent.

Source. OpenStax Statistics, §7.2, Example 7.2 — CC-BY

Example— 2· Election: probability of sample proportion

Problem. In an election, 42% of voters support candidate A ( $p = 0.42$ ). An exit poll surveys 1,000 voters at random. What is the probability that the sample proportion $\hat p$ falls below 0.40?

Strategy. $\hat p$ is the mean of iid Bernoullis. By CLT, $\hat p \approx \mathcal{N}(p, p(1-p)/n)$ .

Solution.

$\sigma_{\hat p} = \sqrt{0.42 \times 0.58 / 1000} = \sqrt{0.000244} \approx 0.01562$ .

$z = \frac{0.40 - 0.42}{0.01562} \approx -1.28.$

$P(\hat p < 0.40) = \Phi(-1.28) \approx 0.1003 \approx 10\%$ .

Verification. Difference of 2 percentage points with $\sigma \approx 1.6\%$ gives $z \approx -1.25$ ; values are consistent.

Source. OpenIntro Statistics, §5.4, Exercise 5.39, p. 224 — CC-BY-SA

Example— 3· Sample size for margin of error

Problem. An institute wants to estimate the mean salary of engineers in SP with margin of error $\pm$ US $200 at 95% confidence. Pilot survey:$ s = 1,800 $reais. What is the minimum$ n$?

Strategy. By CLT, 95% CI has half-width $1.96\,\sigma/\sqrt{n}$ . Isolate $n$ .

Solution.

$1.96 \times \frac{1800}{\sqrt{n}} = 200 \implies \sqrt{n} = \frac{1.96 \times 1800}{200} = 17.64 \implies n = 17.64^2 \approx 311.2.$

Round up: $n = 312$ .

Verification. $1.96 \times 1800/\sqrt{312} \approx 199.6 < 200$ . Correct.

Source. OpenStax Statistics, §7.3, Sample Size Calculation — CC-BY

Example— 4· Sum of losses in insurance (actuarial modeling)

Problem. An insurance company has 500 auto policies. Each claim has mean cost US $3,000 and standard deviation US$ 1,200 (exponential distribution). What is the probability that total claims exceed US$1,560,000?

Strategy. $S_{500} = \sum X_i \approx \mathcal{N}(500 \times 3000,\; 500 \times 1200^2)$ by CLT (sum version).

Solution.

$E[S] = 500 \times 3000 = 1,500,000$ ; $\text{Var}(S) = 500 \times 1200^2 = 720,000,000$ .

$\sigma_S = \sqrt{720,000,000} \approx 26,833$ .

$z = \frac{1,560,000 - 1,500,000}{26,833} \approx \frac{60,000}{26,833} \approx 2.237.$

$P(S > 1,560,000) = P(Z > 2.237) \approx 1 - 0.9873 \approx 1.3\%$ .

Verification. $\pm 2\sigma_S$ corresponds to $\pm$ US $53,666;$ 60,000 > 2\sigma_S$, so less than 2.5% — consistent with 1.3%.

Source. OpenIntro Statistics, §4.4, p. 199–202 — CC-BY-SA

Example— 5· Berry-Esseen: quantifying approximation error

Problem. $X_i \sim \text{Bernoulli}(0.1)$ ; $\mu = 0.1$ , $\sigma^2 = 0.09$ , $E[|X-\mu|^3] = 0.9^3 \times 0.1 + 0.1^3 \times 0.9 = 0.0738$ . For $n = 100$ , what is the Berry-Esseen bound on the error $\sup_z |F_{Z_{100}}(z) - \Phi(z)|$ ?

Strategy. Apply the inequality directly with $C = 0.5$ .

Solution.

$\rho = E[|X - \mu|^3] = 0.0738$ ; $\sigma^3 = 0.3^3 = 0.027$ .

$\text{Bound} \leq \frac{0.5 \times 0.0738}{0.027 \times \sqrt{100}} = \frac{0.0369}{0.27} \approx 0.137.$

The maximum approximation error is less than 13.7 percentage points — conservative bound; actual error is much smaller.

Verification. For Bernoulli(0.1) with $n = 100$ , binomial vs. normal tables show actual error of 1–2% — Berry-Esseen is loose as expected.

Source. Grinstead & Snell, Introduction to Probability, §9.3, p. 346–349 — GNU FDL

Exercise list

45 exercises · 11 with worked solution (25%)

Application 23Understanding 5Modeling 12Challenge 3Proof 2

Ex. 77.1Application
$X$ exponential with $\mu = 1$ , $\sigma = 1$ . Write the approximate distribution of $\bar X_{100}$ and calculate $\sigma_{\bar X}$ .
Select the correct option
$\bar X_{100} \approx \mathcal N(1; 0.01)$ and $\sigma_{\bar X} = 0.1$ $\bar X_{100} \approx \mathcal N(1; 1)$ — same distribution as $X$ $\bar X_{100} \approx \mathcal N(0; 1)$ $\bar X_{100}$ has exponential distribution
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$X$ exponential with $\mu = 1$ and $\sigma = 1$ . By CLT: $\bar X_{100} \approx \mathcal N(1,\; 1/100) = \mathcal N(1,\; 0.01)$ . Standard deviation of the mean: $\sigma_{\bar X} = 1/\sqrt{100} = 0.1$ .
Ex. 77.2Application
$X \sim \mathcal U[0, 1]$ . Determine $\mu$ and $\sigma^2$ and write the approximate distribution of $\bar X_{50}$ by CLT.
Select the correct option
$\bar X_{50} \approx \mathcal N(0.5;\; 1/600)$ $\bar X_{50} \approx \mathcal N(0.5;\; 1/12)$ $\bar X_{50} \approx \mathcal N(0;\; 1)$ $\bar X_{50}$ is uniform on $[0, 1]$
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
$X \sim \mathcal U[0, 1]$ : $\mu = 0.5$ , $\sigma^2 = 1/12$ . By CLT: $\bar X_{50} \approx \mathcal N(0.5,\; (1/12)/50) = \mathcal N(0.5,\; 1/600)$ .
Show step-by-step (with the why)
1. Calculate parameters of $\mathcal U[0,1]$ : $\mu = 0.5$ ; $\sigma^2 = (b-a)^2/12 = 1/12$ .
2. Apply CLT with $n = 50$ : $\bar X_{50} \approx \mathcal N(\mu, \sigma^2/n) = \mathcal N(0.5,\; 1/600)$ .
3. Standard deviation: $\sigma_{\bar X} = \sqrt{1/600} \approx 0.041$ .
Trick: For uniform $[a,b]$ : $\sigma^2 = (b-a)^2/12$ .
Ex. 77.3Application
Roll 100 fair dice. Determine the approximate distribution of the sum $S_{100}$ , with $E[S]$ and $\text{Var}(S)$ .
Select the correct option
$S_{100} \approx \mathcal N(350;\; 291.67)$ $S_{100} \approx \mathcal N(3.5;\; 35/12)$ $S_{100} \approx \mathcal N(350;\; 35/12)$ $S_{100}$ is uniform — data are uniform
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Fair die: $\mu = 3.5$ , $\sigma^2 = 35/12$ . Sum of 100: $S_{100} \approx \mathcal N(100 \times 3.5,\; 100 \times 35/12) = \mathcal N(350,\; 291.67)$ .
Ex. 77.4ApplicationAnswer key
$X \sim \text{Bernoulli}(0.3)$ . Write the approximate distribution of $\bar X_{200}$ by CLT and calculate the standard deviation of the sample proportion.
Select the correct option
$\bar X_{200} \approx \mathcal N(0.3;\; 0.00105)$ , $\sigma_{\hat p} \approx 0.0324$ $\bar X_{200} \approx \mathcal N(0.3;\; 0.21)$ $\bar X_{200} \approx \mathcal N(0;\; 1)$ $\bar X_{200}$ is Bernoulli
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
$X \sim \text{Bernoulli}(0.3)$ : $\mu = 0.3$ , $\sigma^2 = 0.21$ . CLT: $\bar X_{200} \approx \mathcal N(0.3,\; 0.21/200) = \mathcal N(0.3,\; 0.00105)$ . $\sigma_{\hat p} = \sqrt{0.00105} \approx 0.0324$ .
Ex. 77.5Application
$\mu = 50$ , $\sigma = 10$ , $n = 25$ . Calculate the standard deviation of $\bar X$ .
Select the correct option
$\sigma_{\bar X} = 2$ $\sigma_{\bar X} = 10$ $\sigma_{\bar X} = 0.2$ $\sigma_{\bar X} = 50$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$\sigma_{\bar X} = \sigma/\sqrt{n} = 10/\sqrt{25} = 2$ .
Ex. 77.6Application
With $\mu = 50$ , $\sigma = 10$ , $n = 25$ , calculate $P(\bar X > 53)$ .
Select the correct option
$P(\bar X > 53) \approx 6.7\%$ $P(\bar X > 53) \approx 50\%$ $P(\bar X > 53) \approx 30\%$ $P(\bar X > 53) \approx 16\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$\sigma_{\bar X} = 10/\sqrt{25} = 2$ . $z = (53-50)/2 = 1.5$ . $P(\bar X > 53) = P(Z > 1.5) \approx 0.0668$ .
Show step-by-step (with the why)
1. Calculate $\sigma_{\bar X} = \sigma/\sqrt{n} = 10/5 = 2$ .
2. Standardize: $z = (53-50)/2 = 1.5$ .
3. Table: $P(Z > 1.5) = 1 - \Phi(1.5) \approx 0.0668$ .
Note: Always use $\sigma/\sqrt{n}$ (standard error of the mean), not $\sigma$ alone.
Ex. 77.7Application
With $\mu = 50$ , $\sigma = 10$ , $n = 25$ , calculate $P(\bar X < 47)$ .
Select the correct option
$P(\bar X < 47) \approx 6.7\%$ $P(\bar X < 47) \approx 50\%$ $P(\bar X < 47) \approx 93.3\%$ $P(\bar X < 47) \approx 16\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$z = (47-50)/2 = -1.5$ . $P(\bar X < 47) = \Phi(-1.5) \approx 0.0668$ . Same value as 77.6 by symmetry of the normal.
Ex. 77.8Application
$X$ with $\mu = 100$ , $\sigma = 20$ , $n = 100$ . Calculate $P(98 < \bar X < 102)$ .
Select the correct option
$P(98 < \bar X < 102) \approx 68.3\%$ $P(98 < \bar X < 102) \approx 95\%$ $P(98 < \bar X < 102) \approx 50\%$ $P(98 < \bar X < 102) \approx 99.7\%$
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
$\sigma_{\bar X} = 20/\sqrt{100} = 2$ . $z_1 = (98-100)/2 = -1$ , $z_2 = (102-100)/2 = 1$ . $P(-1 < Z < 1) \approx 0.6827$ — 68% rule.
Ex. 77.9Application
Sum of 50 iid r.v. with $\mu = 5$ , $\sigma = 2$ . Calculate $P(S_{50} > 270)$ .
Select the correct option
$P(S_{50} > 270) \approx 7.9\%$ $P(S_{50} > 270) \approx 50\%$ $P(S_{50} > 270) \approx 30\%$ $P(S_{50} > 270) \approx 2\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$S_{50} \approx \mathcal N(250,\; 50 \times 4)$ . $\sigma_S = \sqrt{200} \approx 14.14$ . $z = (270-250)/14.14 \approx 1.414$ . $P(S_{50} > 270) \approx 0.079$ .
Show step-by-step (with the why)
1. Distribution of the sum: $S_{50} \approx \mathcal N(n\mu,\; n\sigma^2) = \mathcal N(250,\; 200)$ .
2. $\sigma_S = \sqrt{200} \approx 14.14$ .
3. Standardize: $z = (270-250)/14.14 \approx 1.41$ .
4. $P(S > 270) = 1 - \Phi(1.41) \approx 0.079$ .
Difference between $\bar X$ and $S$ : $E[S] = n\mu$ , $\text{Var}(S) = n\sigma^2$ ; for $\bar X$ : $E[\bar X] = \mu$ , $\text{Var}(\bar X) = \sigma^2/n$ .
Ex. 77.10Application
$X$ with $\sigma = 3$ . How many observations for 95% CI with margin of error $\pm 0.5$ ?
Select the correct option
$n = 139$ observations $n = 25$ observations $n = 400$ observations $n = 1000$ observations
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
95% CI half-width = $1.96 \times \sigma/\sqrt n = 0.5$ . $\sqrt n = 1.96 \times 3 / 0.5 = 11.76$ . $n = 11.76^2 \approx 138.3$ . Round up: $n = 139$ .
Show step-by-step (with the why)
1. Margin of error at 95%: $ME = 1.96 \sigma/\sqrt{n} \leq 0.5$ .
2. Isolate $\sqrt{n} \geq 1.96 \times 3/0.5 = 11.76$ .
3. Square: $n \geq 138.3$ .
4. Round up: $n = 139$ .
Trick: To halve the margin of error, quadruple $n$ .
Ex. 77.11Understanding
When $n$ is multiplied by 4, the standard deviation of $\bar X$ ( $= \sigma/\sqrt{n}$ ):
Select the correct option
It is cut in half — $\sigma_{\bar X} = \sigma/\sqrt{n}$ , quadrupling $n$ doubles $\sqrt{n}$ It is cut to 1/4It remains the sameIt quadruples
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
$\sigma_{\bar X} = \sigma/\sqrt{n}$ . If $n \to 4n$ : $\sigma_{\bar X} \to \sigma/\sqrt{4n} = \sigma_{\bar X}/2$ . The standard deviation of the mean is cut in half.
Ex. 77.12UnderstandingAnswer key
$X$ has a very skewed distribution (skewness = 3). For what sample size is the CLT reasonable?
Select the correct option
$n \geq 100$ : high skewness slows convergence to the bell curve $n \geq 30$ is sufficient for any distributionCLT does not apply to skewed distributions $n \geq 10$ suffices if $\mu$ is large
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Berry-Esseen: error $\propto \rho/(\sigma^3 \sqrt n)$ , where $\rho = E[|X-\mu|^3]$ . High skewness increases $\rho$ , requiring larger $n$ for convergence. Rule of thumb: $n \geq 100$ for highly skewed distributions.
Ex. 77.13Application
Test scores with $\mu = 70$ , $\sigma = 15$ . Sample $n = 36$ . Calculate $P(\bar X > 75)$ .
Select the correct option
$P(\bar X > 75) \approx 2.3\%$ $P(\bar X > 75) \approx 16\%$ $P(\bar X > 75) \approx 50\%$ $P(\bar X > 75) \approx 6.7\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$\sigma_{\bar X} = 15/\sqrt{36} = 2.5$ . $z = (75-70)/2.5 = 2.0$ . $P(\bar X > 75) = 1 - \Phi(2) \approx 0.0228$ .
Show step-by-step (with the why)
1. Standard error of the mean: $\sigma_{\bar X} = 15/\sqrt{36} = 15/6 = 2.5$ .
2. Standardize the value 75: $z = (75 - 70)/2.5 = 2$ .
3. Probability: $P(\bar X > 75) = P(Z > 2) = 1 - \Phi(2) \approx 0.0228$ .
Trick: The 95-5 rule says only ~2.3% lies above $\mu + 2\sigma_{\bar X}$ .
Ex. 77.14Application
With $\mu = 70$ , $\sigma = 15$ , $n = 36$ , calculate $P(\bar X < 65)$ .
Select the correct option
$P(\bar X < 65) \approx 2.3\%$ $P(\bar X < 65) \approx 50\%$ $P(\bar X < 65) \approx 16\%$ $P(\bar X < 65) \approx 6.7\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$z = (65-70)/2.5 = -2.0$ . $P(\bar X < 65) = \Phi(-2) \approx 0.0228$ . Same value as 77.13 by symmetry of the normal.
Ex. 77.15Application
With $\mu = 70$ , $\sigma = 15$ , $n = 36$ and $\bar X = 72$ , construct a 95% CI for $\mu$ .
Select the correct option
95% CI: $(67.1;\; 76.9)$ 95% CI: $(67.5;\; 72.5)$ 95% CI: $(60;\; 80)$ 95% CI: $(70;\; 75)$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$\sigma_{\bar X} = 2.5$ . 95% CI: $72 \pm 1.96 \times 2.5 = 72 \pm 4.9$ . Interval: $(67.1;\; 76.9)$ .
Ex. 77.16Application
Package weight: $\mu = 500$ g, $\sigma = 50$ g, $n = 25$ . Calculate $P(\bar X > 520)$ .
Select the correct option
$P(\bar X > 520) \approx 2.3\%$ $P(\bar X > 520) \approx 16\%$ $P(\bar X > 520) \approx 50\%$ $P(\bar X > 520) \approx 6.7\%$
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
$\sigma_{\bar X} = 50/\sqrt{25} = 10$ . $z = (520-500)/10 = 2$ . $P(\bar X > 520) = 1 - \Phi(2) \approx 0.0228$ .
Ex. 77.17Application
With $\mu = 500$ , $\sigma = 50$ , $n = 25$ , calculate $P(485 < \bar X < 515)$ .
Select the correct option
$P(485 < \bar X < 515) \approx 86.6\%$ $P(485 < \bar X < 515) \approx 95\%$ $P(485 < \bar X < 515) \approx 68\%$ $P(485 < \bar X < 515) \approx 50\%$
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
$z_1 = (485-500)/10 = -1.5$ ; $z_2 = (515-500)/10 = 1.5$ . $P(-1.5 < Z < 1.5) \approx 0.8664$ .
Ex. 77.18ApplicationAnswer key
Response time: $\mu = 50$ ms, $\sigma = 10$ ms, $n = 100$ . What is the 95% SLA limit?
Select the correct option
95% SLA: $\approx 51.6$ ms95% SLA: $= 50$ ms (the mean)95% SLA: $\approx 66.4$ ms95% SLA: $\approx 70$ ms
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$\sigma_{\bar X} = 10/\sqrt{100} = 1$ . 95th percentile: $x = 50 + 1.645 \times 1 = 51.645$ ms.
Ex. 77.19Application
Roll a die 1,000 times. Calculate $P(\bar X > 3.6)$ .
Select the correct option
$P(\bar X > 3.6) \approx 3.2\%$ $P(\bar X > 3.6) \approx 16\%$ $P(\bar X > 3.6) \approx 50\%$ $P(\bar X > 3.6) \approx 0.1\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Die: $\mu = 3.5$ , $\sigma^2 = 35/12$ . $\sigma_{\bar X} = \sqrt{35/12000} \approx 0.054$ . $z = (3.6-3.5)/0.054 \approx 1.85$ . $P(\bar X > 3.6) \approx 0.032$ .
Ex. 77.20Application
Using the distribution of the sum $S_{1000}$ of 1,000 die rolls, calculate $P(S_{1000} > 3600)$ .
Select the correct option
$P(S_{1000} > 3600) \approx 3.2\%$ — same result as 77.19 $P(S_{1000} > 3600) \approx 50\%$ $P(S_{1000} > 3600) = 0$ $P(S_{1000} > 3600) \approx 16\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$S_{1000} \approx \mathcal N(3500,\; 1000 \times 35/12)$ . $\sigma_S = \sqrt{2916.7} \approx 54$ . $z = (3600-3500)/54 \approx 1.85$ . $P(S > 3600) \approx 0.032$ . Same result as 77.19 — confirms equivalence.
Ex. 77.21Application
$X \sim \text{Exp}(1)$ ( $\mu = 1$ , $\sigma = 1$ ). Calculate $P(\bar X_{100} > 1.1)$ .
Select the correct option
$P(\bar X_{100} > 1.1) \approx 15.9\%$ $P(\bar X_{100} > 1.1) \approx 2.3\%$ $P(\bar X_{100} > 1.1) \approx 50\%$ $P(\bar X_{100} > 1.1) \approx 6.7\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$X \sim \text{Exp}(1)$ : $\mu = 1$ , $\sigma = 1$ . $\sigma_{\bar X} = 1/\sqrt{100} = 0.1$ . $z = (1.1-1)/0.1 = 1$ . $P(\bar X > 1.1) = 1 - \Phi(1) \approx 0.1587$ .
Ex. 77.22ApplicationAnswer key
Election survey: $p = 0.40$ , $n = 1000$ . Calculate $P(\hat p > 0.43)$ .
Select the correct option
$P(\hat p > 0.43) \approx 2.6\%$ $P(\hat p > 0.43) \approx 16\%$ $P(\hat p > 0.43) \approx 50\%$ $P(\hat p > 0.43) \approx 6.7\%$
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
$\sigma_{\hat p} = \sqrt{0.40 \times 0.60/1000} \approx 0.01549$ . $z = (0.43-0.40)/0.01549 \approx 1.937$ . $P(\hat p > 0.43) \approx 0.0263$ .
Show step-by-step (with the why)
1. Standard error of the proportion: $\sigma_{\hat p} = \sqrt{p(1-p)/n} = \sqrt{0.24/1000} \approx 0.01549$ .
2. Standardize: $z = (0.43-0.40)/0.01549 \approx 1.94$ .
3. $P(\hat p > 0.43) = 1 - \Phi(1.94) \approx 0.026$ .
Trick: For sample proportion, $\sigma_{\hat p} = \sqrt{p(1-p)/n}$ — analogous to the standard error of the mean.
Ex. 77.23Modeling
You hold 50 stocks with daily return $\mu = 0.1\%$ , $\sigma = 2\%$ (independent). What is the distribution of your portfolio's mean daily return?
Select the correct option
$\bar R_{50} \approx \mathcal N(0.001;\; 8 \times 10^{-6})$ , $\sigma_{\bar R} \approx 0.28\%$ $\bar R_{50} \approx \mathcal N(0;\; 1)$ $\bar R_{50} \approx \mathcal N(0.001;\; 0.0004)$ $\bar R_{50}$ is exponential
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
50 independent stocks with daily return $\mu = 0.1\%$ , $\sigma = 2\%$ . $\bar R_{50} \approx \mathcal N(0.001,\; (0.02)^2/50) = \mathcal N(0.001,\; 8 \times 10^{-6})$ . $\sigma_{\bar R} = 2\%/\sqrt{50} \approx 0.28\%$ . Diversification reduced risk from 2% to 0.28%.
Ex. 77.24Modeling
ML model: individual error $\sigma = 0.5$ . Calculate the standard deviation of mean error over 1,000 predictions and the 95% CI margin of error.
Select the correct option
$\sigma_{\bar e} = 0.5/\sqrt{1000} \approx 0.0158$ ; 95% CI margin $\approx \pm 0.031$ $\sigma_{\bar e} = 0.5$ — does not change with $n$ $\sigma_{\bar e} = 0.5/1000$ $\sigma_{\bar e} = 0.5/100$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$\sigma_{\bar e} = 0.5/\sqrt{1000} \approx 0.01581$ . 95% CI: margin $\pm 1.96 \times 0.01581 \approx \pm 0.031$ .
Ex. 77.25Modeling
Satisfaction survey: margin of error $\pm 3\%$ at 95%, $p$ unknown. What is the minimum $n$ ?
Select the correct option
$n \approx 1067$ per survey $n \approx 100$ $n \approx 10000$ $n = 30$
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
Margin of error 3% at 95%, worst case $p = 0.5$ : $n = (1.96/0.03)^2 \times 0.25 = 65.33^2 \times 0.25 \approx 1067$ . This is why "3% margin of error" Brazilian surveys have ~1000 respondents.
Ex. 77.26Modeling
Batch of 500 items: $\mu = 100$ g, $\sigma = 5$ g. Determine the distribution of $S_{500}$ .
Select the correct option
$S_{500} \approx \mathcal N(50.000;\; 12.500)$ — standard deviation $\approx 112$ g $S_{500} \approx \mathcal N(500;\; 25)$ $S_{500}$ equals exactly 50,000 g $S_{500} \approx \mathcal N(50.000;\; 25)$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$S_{500} \approx \mathcal N(500 \times 100,\; 500 \times 25) = \mathcal N(50.000,\; 12.500)$ . $\sigma_{S} = \sqrt{12.500} \approx 111.8$ g. Expected total weight: 50 kg with standard deviation ~112 g.
Ex. 77.27Modeling
Bus wait time: $\mathcal U[0, 30]$ min. Calculate $P(\bar T_{50} > 16)$ for 50 passengers.
Select the correct option
$P(\bar T_{50} > 16) \approx 20.6\%$ $P(\bar T_{50} > 16) \approx 50\%$ $P(\bar T_{50} > 16) \approx 5\%$ $P(\bar T_{50} > 16) \approx 0.1\%$
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
$\mathcal U[0, 30]$ : $\mu = 15$ , $\sigma^2 = 75$ . $\bar T_{50} \approx \mathcal N(15,\; 75/50 = 1.5)$ . $z = (16-15)/\sqrt{1.5} \approx 0.816$ . $P(\bar T > 16) \approx 0.207$ .
Ex. 77.28Modeling
On an X-bar control chart with $n = 5$ , how are the control limits (UCL and LCL) calculated?
Select the correct option
X-bar chart: limits at $\bar{\bar X} \pm 3\sigma/\sqrt{n}$ Limits at $\bar{\bar X} \pm 3\sigma$ Limits at $\bar{\bar X} \pm \sigma/n$ Limits at $\bar{\bar X} \pm 1.96\sigma$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Shewhart X-bar chart: UCL/LCL = $\bar{\bar X} \pm 3\sigma_{\bar X} = \bar{\bar X} \pm 3\sigma/\sqrt{n}$ . The 3 comes from $P(|Z| > 3) \approx 0.27\%$ — acceptable false-alarm rate.
Show step-by-step (with the why)
1. By CLT: $\bar X$ of subgroup of size $n$ is approximately $\mathcal N(\mu, \sigma^2/n)$ .
2. 3-sigma limits capture 99.73% of means when the process is in control.
3. UCL = $\bar{\bar X} + 3\sigma/\sqrt{n}$ ; LCL = $\bar{\bar X} - 3\sigma/\sqrt{n}$ .
4. For $n = 5$ : total width = $6\sigma/\sqrt{5} \approx 2.68\sigma$ .
Ex. 77.29ModelingAnswer key
Survey with margin of error $\pm 5\%$ at 95% confidence, $p$ unknown. What is the minimum necessary $n$ ?
Select the correct option
$n = 385$ respondents for 5% margin (worst case $p = 0.5$ ) $n = 1067$ — same result as 3% margin $n = 100$ is sufficient for any margin of error $n = 10000$ always
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
Margin = $1.96 \times \sqrt{p(1-p)/n}$ . For $p = 0.5$ (worst case) and margin = 5%: $n = (1.96/0.05)^2/4 = 39.2^2 \times 0.25 \approx 384.2$ . Round up: $n = 385$ . Half the margin (3%) requires almost three times more respondents (1067).
Ex. 77.30ModelingAnswer key
Call time: $\mu = 3$ min, $\sigma = 1.5$ min, 100 calls. Calculate $P(\text{total} > 330 \text{ min})$ .
Select the correct option
$P(\text{total} > 330 \text{ min}) \approx 2.3\%$ $P(\text{total} > 330 \text{ min}) \approx 16\%$ $P(\text{total} > 330 \text{ min}) \approx 50\%$ $P(\text{total} > 330 \text{ min}) \approx 6.7\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
$S_{100} \approx \mathcal N(300, 225)$ . $\sigma_S = 15$ . $z = (330-300)/15 = 2$ . $P(S > 330) = 1 - \Phi(2) \approx 0.023$ .
Ex. 77.31ModelingAnswer key
A/B test: 10,000 visitors per variant; conversion A = 5%, B = 6%. Is the 1 p.p. lift statistically significant? Calculate $z$ and the p-value.
Select the correct option
$P(\hat p_B - \hat p_A > 0.01) \approx 99.9\%$ — yes, significant $z \approx 3.25$ and $p\text{-value} \approx 0.001$ — difference significant at 5% $z \approx 0.67$ and difference not significantWith $n = 10000$ it is impossible to test a 1% difference
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
Observed difference: 1%. SE: $\sqrt{2 \times 0.05 \times 0.95/10000} \approx 0.00308$ . $z \approx 0.01/0.00308 \approx 3.25$ . p-value < 0.001. A 1 p.p. difference is statistically significant with 10,000 users per variant.
Show step-by-step (with the why)
1. Pooled estimator: $\hat p = (500+600)/20000 = 0.055$ .
2. SE: $\sqrt{0.055 \times 0.945 \times (2/10000)} \approx 0.00322$ .
3. $z = 0.01/0.00322 \approx 3.11$ . p-value $\approx 0.002$ .
4. Conclusion: reject $H_0$ — difference is significant.
Ex. 77.32Understanding
Which option correctly describes the CLT?
Select the correct option
The sample mean converges to normal regardless of the original distributionAny r.v. with large $n$ is normalThe original distribution becomes normal as $n$ growsThe mean always has exactly normal distribution
Select an option first
Solve online OpenIntro Statistics · §5.1
Show solution
The CLT talks about the distribution of the sample mean (or standardized sum) — not the original distribution. The original distribution stays the same; what converges to normal is the statistic $Z_n = (\bar X_n - \mu)/(\sigma/\sqrt{n})$ .
Ex. 77.33Understanding
Why does the classical CLT not apply to the Cauchy distribution?
Select the correct option
Infinite variance: the proof via characteristic function requires $\sigma^2 < \infty$ Skewed distribution: CLT does not applyUnknown parameter: without knowing $\sigma$ it is impossibleHeavy tail: CLT requires original distribution to be normal
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
The Cauchy characteristic function is $e^{-|t|}$ — no second derivative at $t = 0$ (infinite variance). The Taylor expansion leading to the limit $e^{-t^2/2}$ doesn't work. The mean of Cauchys is still Cauchy.
Ex. 77.34UnderstandingAnswer key
When simulating means of die rolls for $n \in \{1, 5, 30\}$ , how does the histogram shape evolve?
Select the correct option
$n = 1$ : uniform; $n = 5$ : trapezoid; $n = 30$ : nearly perfect bell curve $n = 1$ : bell curve; $n = 5$ : uniform; $n = 30$ : exponentialAll $n$ have uniform distributionOnly $n = 100$ produces perceptible bell curve
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Simulating means of fair die rolls: $n = 1$ = flat uniform; $n = 5$ = triangular-trapezoidal shape; $n = 30$ = indistinguishable from normal. Convergence is visible and rapid.
Ex. 77.35Application
Distance of fly-balls: $\mathcal N(250, 50^2)$ ft. Sample of 49 balls. Calculate $P(\bar X < 240)$ .
Select the correct option
$\bar X_{49} \sim \mathcal N(250; 2500/49)$ ; $P(\bar X < 240) \approx 8.1\%$ $P(\bar X < 240) \approx 50\%$ $P(\bar X < 240) \approx 2.3\%$ $P(\bar X < 240) \approx 16\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Baseball fly-balls: $\mu = 250$ ft, $\sigma = 50$ . $n = 49$ . $\sigma_{\bar X} = 50/7 \approx 7.14$ . $z = (240-250)/7.14 \approx -1.4$ . $P(\bar X < 240) = \Phi(-1.4) \approx 0.081$ .
Ex. 77.36ProofAnswer key
Show that the CLT implies the Weak Law of Large Numbers.
Select the correct option
CLT implies LLN: $\text{Var}(\bar X_n) = \sigma^2/n \to 0$ , so convergence in probabilityLLN implies CLT — they are equivalentCLT and LLN are independent and do not imply each otherCLT implies that $\bar X_n \to \mathcal N(0,1)$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
From CLT: $\bar X_n \approx \mathcal N(\mu, \sigma^2/n)$ . Chebyshev: $P(|\bar X_n - \mu| > \varepsilon) \leq \sigma^2/(n\varepsilon^2) \to 0$ . So $\bar X_n \xrightarrow{P} \mu$ — Law of Large Numbers. CLT is stronger: it tells how the distribution of the error behaves as it converges.
Ex. 77.37Proof
Sketch the proof of the CLT via characteristic function, indicating the necessary hypotheses.
Select the correct option
Outline: standardize → Taylor expansion of characteristic function → limit is $e^{-t^2/2}$ Proof via MGF assumes existence of all momentsThe proof requires that the original distribution be normalThere is no rigorous proof of the CLT
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Steps: (1) Define $Y_i = (X_i - \mu)/\sigma$ . (2) $\varphi_{Y_i}(t) = 1 - t^2/2 + o(t^2)$ . (3) $\varphi_{Z_n}(t) = [\varphi_{Y_i}(t/\sqrt n)]^n \to e^{-t^2/2}$ . (4) Lévy's theorem concludes convergence in distribution. Requires only finite variance (Lindeberg-Lévy hypothesis).
Ex. 77.38ApplicationAnswer key
Average time to fill IRS Form 1040: $\mu = 10.53$ h, $\sigma = 2$ h. Sample of 36 taxpayers. Calculate the distribution of $\bar X$ and $P(\bar X > 12)$ .
Select the correct option
$\bar X_{36} \sim \mathcal N(10.53; 4/36)$ ; $P(\bar X > 12) \approx 5 \times 10^{-6}$ (practically impossible) $P(\bar X > 12) \approx 50\%$ $P(\bar X > 12) \approx 16\%$ $P(\bar X > 12) \approx 2.5\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
IRS Form 1040 time: $\mu = 10.53$ h, $\sigma = 2$ h. $n = 36$ . $\sigma_{\bar X} = 2/6 = 1/3$ . $z = (12-10.53)/(1/3) = 4.41$ . $P(\bar X > 12) \approx 5 \times 10^{-6}$ — practically impossible.
Ex. 77.39Application
Marathon times: $\mu = 145$ min, $\sigma = 14$ min. Sample of 49 runs. Calculate $P(142 < \bar X < 146)$ .
Select the correct option
$\bar X_{49} \sim \mathcal N(145; 196/49)$ ; $P(142 < \bar X < 146) \approx 62.5\%$ $P(142 < \bar X < 146) \approx 95\%$ $P(142 < \bar X < 146) \approx 68\%$ $P(142 < \bar X < 146) \approx 30\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Marathon times: $\mu = 145$ min, $\sigma = 14$ . $n = 49$ . $\sigma_{\bar X} = 14/7 = 2$ . $z_1 = (142-145)/2 = -1.5$ ; $z_2 = (146-145)/2 = 0.5$ . $P = \Phi(0.5) - \Phi(-1.5) \approx 0.6915 - 0.0668 = 0.6247$ .
Ex. 77.40Modeling
Percentage of fat calories: $\mu = 36\%$ , $\sigma = 10\%$ . Sample of 16 people. Calculate $P(\bar X > 40\%)$ .
Select the correct option
$P(\bar X > 40\%) \approx 5.5\%$ — unlikely $P(\bar X > 40\%) \approx 50\%$ $P(\bar X > 40\%) \approx 16\%$ $P(\bar X > 40\%) \approx 2.3\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
% fat calories: $\mu = 36$ , $\sigma = 10$ . $n = 16$ . $\sigma_{\bar X} = 10/4 = 2.5$ . $z = (40-36)/2.5 = 1.6$ . $P(\bar X > 40\%) = 1 - \Phi(1.6) \approx 0.0548$ .
Ex. 77.41Modeling
Gasoline price (Bay Area): $\mu = 4.59$ dollars, $\sigma = 0.10$ dollars. Sample of 16 stations. What is the distribution of $\bar X$ and $P(\bar X > 4.63)$ ?
Select the correct option
$P(\bar X > 4.63) \approx 5.5\%$ $P(\bar X > 4.63) \approx 50\%$ $P(\bar X > 4.63) \approx 2.3\%$ $P(\bar X > 4.63) \approx 15.9\%$
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Gasoline Bay Area: $\mu = 4.59$ , $\sigma = 0.10$ . $n = 16$ . $\sigma_{\bar X} = 0.10/4 = 0.025$ . $z = (4.63-4.59)/0.025 = 1.6$ . $P(\bar X > 4.63) \approx 0.0548 \approx 5.5\%$ .
Ex. 77.42ModelingAnswer key
Air conditioning service: mean time 1 h, $\sigma = 1$ h. Budget of 1.1 h per technician, 70 units. Is there sufficient budget with 80% confidence?
Select the correct option
Sufficient: $P(\bar X \leq 1.1) \approx 80\%$ — budget is adequateInsufficient: $P(\bar X \leq 1.1) \approx 16\%$ The budget is always exactly right $P(\bar X \leq 1.1) \approx 50\%$ — 50% chance of overrun
Select an option first
Solve online OpenStax Statistics · §7.H
Show solution
Air conditioning service: $\mu = 1$ h, $\sigma = 1$ h. $n = 70$ units. $\sigma_{\bar X} = 1/\sqrt{70} \approx 0.1195$ . $z = (1.1-1)/0.1195 \approx 0.837$ . $P(\bar X \leq 1.1) \approx \Phi(0.84) \approx 0.80$ . 80% chance of not overrunning the budget.
Ex. 77.43Challenge
Challenge — Lindeberg condition. The Lindeberg-Feller CLT generalizes the CLT to independent but non-i.d. variables. What is the essential condition that replaces the identical-distribution hypothesis, and what does it intuitively require?
Select the correct option
The Lindeberg condition requires that no individual variable dominates the total variance in the limit.Variables must be identically distributed; otherwise CLT does not hold.It suffices that variances are finite and the mean is zero.The Lindeberg-Feller CLT requires variables to be independent and uniformly distributed.
Select an option first
Solve online Introduction to Probability · §9.3
Show solution
Let $X_1, X_2, \ldots$ be independent (not necessarily i.d.) with $E[X_i]=\mu_i$ , $\mathrm{Var}(X_i)=\sigma_i^2$ . Define $s_n^2 = \sum_{i=1}^n \sigma_i^2$ . The Lindeberg condition requires that for every $\varepsilon > 0$ : $\frac{1}{s_n^2}\sum_{i=1}^n E\bigl[(X_i-\mu_i)^2 \mathbf{1}_{|X_i-\mu_i|>\varepsilon s_n}\bigr] \to 0$ . Intuitively: the contribution of any extreme individual value to total variance must be negligible. If satisfied, $(S_n - \sum\mu_i)/s_n \xrightarrow{d} N(0,1)$ .
Show step-by-step (with the why)
1. Motivation. With i.i.d. variables, classical CLT follows easily from characteristic functions. For non-i.d. variables, we need to prevent a single variable from "dominating" the sum.
2. Total variance. $s_n^2 = \sigma_1^2 + \sigma_2^2 + \cdots + \sigma_n^2$ . Should grow with $n$ .
3. Lindeberg condition. For each $\varepsilon > 0$ , the fraction of variance explained by large deviations (above $\varepsilon s_n$ ) must go to zero.
4. Practical consequence. If $\sigma_i^2 / s_n^2 \to 0$ for each $i$ (Feller condition), the Lindeberg condition is satisfied for variables with finite fourth moment.
5. Counterexample. If $X_n$ concentrates all variance (e.g. $\sigma_n^2 = s_n^2$ ), Lindeberg fails and CLT does not hold.
The Lindeberg-Feller CLT is the most general CLT result: it generalizes i.i.d. requiring only that no individual variable "dominates" variance asymptotically.
Ex. 77.44Challenge
Challenge — Berry-Esseen bound. The Berry-Esseen theorem provides a quantitative bound for the speed of CLT convergence. For i.i.d. variables with third absolute moment $\rho = E[|X_i|^3] < \infty$ , what is the correct form of the bound on the maximum approximation error?
Select the correct option
$|P(Z_n \leq x) - \Phi(x)| \leq C \cdot E[|X_i|^3] / (\sigma^3 \sqrt{n})$ , where $C < 1$ . $|P(Z_n \leq x) - \Phi(x)| \leq C / n$ (error inversely proportional to $n$ ). $|P(Z_n \leq x) - \Phi(x)| \leq C / \sqrt{n^2}$ (error inversely proportional to $n$ ).The error is exactly zero for $n \geq 30$ .
Select an option first
Solve online Introduction to Probability · §9.4
Show solution
The Berry-Esseen theorem states: if $X_1,\ldots,X_n$ are i.i.d. with $E[X_i]=0$ , $\sigma^2 = \mathrm{Var}(X_i) < \infty$ and $\rho = E[|X_i|^3] < \infty$ , then $\sup_x |P(Z_n \leq x) - \Phi(x)| \leq C\rho/(\sigma^3\sqrt{n})$ , with universal constant $C \leq 0.4748$ . Example: for $\mathrm{Bernoulli}(1/2)$ : $\sigma^2 = 1/4$ , $\rho = 1/8$ , error $\leq 0.4748 \cdot (1/8)/((1/2)^3 \sqrt{n}) = 0.4748/\sqrt{n}$ .
Show step-by-step (with the why)
1. Statement. Berry-Esseen: $\sup_x |F_n(x) - \Phi(x)| \leq C\rho/(\sigma^3\sqrt{n})$ with $C \leq 0.4748$ (best constant known in 2024).
2. Ingredients. $\rho = E[|X_i|^3]$ (third absolute moment); $\sigma^3$ normalizes by scale.
3. Convergence rate. The error is $O(1/\sqrt{n})$ — same rate as standard deviation of the sample mean.
4. Application Bernoulli(1/2). $\sigma = 1/2$ , $\rho = E[|X|^3] = 1/2$ (since $|X|^3 = |X|$ ). Error $\leq 0.4748/(0.125/0.125\cdot\sqrt{n}) = 0.4748/\sqrt{n}$ . For $n = 100$ : error $\leq 0.047$ — reasonable.
Berry-Esseen quantifies the "n=30 rule": the maximum approximation error decreases as $1/\sqrt{n}$ , but the constant depends on distribution skewness (via $\rho/\sigma^3$ ). More skewed distributions need larger $n$ .
Ex. 77.45Challenge
Challenge — Shewhart control chart. In statistical process control (SPC), limits $\mu \pm 3\sigma/\sqrt{n}$ are used to detect out-of-control processes. Based on the CLT, calculate the probability of a false alarm (process in control but point outside limits) and the ARL₀ (average number of subgroups until the first false alarm).
Select the correct option
If the process is in control, $P(\bar X \notin [\mu \pm 3\sigma/\sqrt{n}]) \approx 0.27\%$ — false alarm is very rare. $P(\bar X \notin [\mu \pm 3\sigma/\sqrt{n}]) \approx 5\%$ by the $z = 1.96$ rule. $P(\bar X \notin [\mu \pm 3\sigma/\sqrt{n}]) \approx 32\%$ , i.e., 1 in 3 samples triggers an alarm.The false-alarm probability is exactly 0 if the process is normal.
Select an option first
Solve online OpenStax Statistics · §7 (CLT Applications)
Show solution
By Shewhart's 3-sigma rule: control limits are $\mu \pm 3\sigma/\sqrt{n}$ . By CLT, $\bar X \sim N(\mu, \sigma^2/n)$ when the process is in control. Therefore $P(|\bar X - \mu| > 3\sigma/\sqrt{n}) = P(|Z| > 3) \approx 0.0027$ , or 0.27% — a false alarm occurs about once per 370 subgroups. The average number of subgroups until the first false alarm (ARL₀) is $1/0.0027 \approx 370$ .
Show step-by-step (with the why)
1. Shewhart control chart. Each point is the mean $\bar X$ of a subgroup of size $n$ . Upper control limit (UCL) = $\mu + 3\sigma/\sqrt{n}$ ; lower control limit (LCL) = $\mu - 3\sigma/\sqrt{n}$ .
2. CLT justifies using $N(\mu, \sigma^2/n)$ . Regardless of individual distribution (when $n \geq 4$ – $5$ for non-highly-skewed processes).
3. Probability of false alarm. $P(|Z| > 3) = 2(1 - \Phi(3)) \approx 2 \cdot 0.00135 = 0.0027$ .
4. ARL₀. Average number of subgroups until first false alarm = $1/0.0027 \approx 370$ . The higher the ARL₀, the less the chart "cries wolf".
5. Why 3σ? Trade-off: false alarm rare (0.27%) without excessive ARL₀; 2-sigma limits would give 4.5% false alarm, too frequent.
Shewhart's 3-sigma rule (1924) is the industrial application of CLT. The ARL₀ ≈ 370 is the gold standard in statistical process control (SPC).

Sources

OpenIntro Statistics (4th ed) — Diez, Çetinkaya-Rundel, Barr · 2019 · CC-BY-SA. Primary source for exercises 77.2, 77.4, 77.8, 77.11, 77.14–17, 77.22–23, 77.25–26, 77.28, 77.30, 77.33–34.
OpenStax Statistics — Illowsky, Dean · 2022 · CC-BY. Source for exercises 77.1, 77.3, 77.5–7, 77.9–10, 77.12–13, 77.18–19, 77.21, 77.24, 77.27, 77.29, 77.31, 77.35 and examples 1–3.
Introduction to Probability (Grinstead-Snell) — Grinstead, Snell · 2006 · CC Free · §9 "Central Limit Theorem". Source for challenge exercises 77.43–77.44.
Introduction to Probability (Grinstead-Snell) — Grinstead, Snell · Dartmouth · GNU FDL. Source for exercises 77.19–20, 77.26, 77.36–37 and example 5.