Central limit theorem: Worked examples — Hinglish deep dive

1.3.15 · D3 · AI-ML › Probability & Statistics › Central limit theorem

Koi bhi example shuru karne se pehle, us poore page mein use hone wale symbols ko simple shabdon mein ek baar phir se anchor karte hain:

Definition Expectation

$E$ aur Variance $Var$ — woh do operators jo hum baar baar use karte hain

$E [X]$ (padho "expectation of X", jise mean bhi kehte hain) — ek random draw $X$ ki long-run average value, jisme har possible value ko uski probability se weight diya jata hai. Ek aisi source ke liye jisme values $x_{1}, \dots, x_{k}$ hain jo probabilities $p_{1}, \dots, p_{k}$ se aate hain: $E [X] = \sum_{j} x_{j} p_{j}$ Source ke bar chart ka balance point socho — woh jagah jahan bars ek ungli par balance karein. Hum yeh bhi likhte hain $μ = E [X]$ .
$Var (X)$ (padho "variance of X") — ek draw ka apne mean se average squared distance, yani draws kitne spread out hain: $Var (X) = E [(X - μ)^{2}] = E [X^{2}] - μ^{2}$ Bar-chart cloud ki typical width socho. Doosre form ka kyun: $(X - μ)^{2} = X^{2} - 2 μ X + μ^{2}$ expand karo aur term-by-term $E$ lo (use karte hue $E [X] = μ$ ) jo collapse hokar $E [X^{2}] - μ^{2}$ ban jata hai — ek shortcut jo sirf squares ka average aur average ka square chahta hai.
$σ = Var (X)$ (sigma) — population standard deviation: spread ko original scale par wapas laana (squaring ko undo karna).

Definition Sample mean, aur woh teen quantities jo tum hamesha compute karte ho

$\overset{ˉ}{X}_{n}$ (padho "X-bar-n") — sample mean: tum $n$ actual draws $X_{1}, X_{2}, \dots, X_{n}$ lete ho, unhe add karte ho, aur $n$ se divide karte ho: $\overset{ˉ}{X}_{n} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}$ Socho: apne $n$ dice rolls line up karo aur unka balance point dhundo. Yeh ek number har example ka hero hai — jo bhi hum compute karte hain woh $\overset{ˉ}{X}_{n}$ ke behavior ke baare mein ek statement hai.
$μ$ (mu) — population mean $E [X]$ : EK draw ka long-run average. CLT kehta hai $\overset{ˉ}{X}_{n}$ $μ$ ke aas paas rehta hai.
$σ$ (sigma) — population standard deviation $Var (X)$ : ek draw typically $μ$ se kitna door padta hai.
$SE = \frac{σ}{n}$ — standard error: $\overset{ˉ}{X}_{n}$ typically $μ$ se kitna door padta hai. Yeh woh width hai bell curve ki jo CLT tumhe deta hai. Yeh $n$ ki tarah shrink karta hai: $n$ chaar guna karo, spread aadha ho jaata hai.

Standardize karne ka move har ek example mein aata hai, isliye ise ek baar define karo:

Scenario matrix

Is topic ke har problem ka type in cells mein se ek hai. Neeche har example us cell(s) ke saath tagged hai jo woh fill karta hai.

Cell	Kya cheez ise distinct banati hai	Covered by
A. Symmetric source	uniform/symmetric $X$ , moderate $n$	Ex 1
B. Skewed / Bernoulli source	tedha $X$ (ek coin), CLT phir bhi kaam karta hai	Ex 2
C. $n$ ke liye solve karo (inverse)	probability di gayi hai, sample size dhundo	Ex 3
D. One-sided tail	$P (\overset{ˉ}{X} > c)$ , sirf ek boundary	Ex 4
E. Negative-mean / sign-flip	$μ < 0$ , Z ke signs ke saath careful rehna	Ex 5
F. Degenerate: $σ = 0$	zero-variance source, limiting behaviour	Ex 6
G. Small $n$ warning	jab bell ek buri approximation hai	Ex 7
H. Limit $n \to \infty$	Law of Large Numbers CLT ki parchhain ke roop mein	Ex 8
I. Mean ki jagah Sum	total, average nahi — dhyan se rescale karo	Ex 9
J. Exam twist: symmetry probability	$P(	Z

Main poore page mein rounded standard-normal areas use karta hoon: $Φ (0.70) = 0.758$ , $Φ (0.16) = 0.564$ , $Φ (1) = 0.841$ , $Φ (1.96) = 0.975$ , $Φ (2) = 0.977$ , $Φ (0.5) = 0.691$ .

Example 1 — Cell A: symmetric source (dice)

Worked example 36 dice ka average 3.3 aur 3.7 ke beech

Statement. Ek fair die $n = 36$ baar roll karo. $P (3.3 \leq \overset{ˉ}{X}_{36} \leq 3.7)$ nikalo.

Forecast: Die $3.5$ par centred hai aur $3.3$ – $3.7$ uske aas-paas ka ek narrow band hai. Andaaza: thoda aadhe se upar? (Yeh $\approx 0.52$ hai.)

Population mean. $μ = E [X] = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$ . Yeh step kyun? CLT bell ko $μ$ par centre karta hai; pehle centre chahiye.
Population variance. $σ^{2} = E [X^{2}] - μ^{2} = \frac{91}{6} - 3. 5^{2} = 2.9167$ , toh $σ = 1.708$ . Yeh step kyun? Bell ki width $σ$ se aati hai; hum upar define kiya hua $E [X^{2}] - μ^{2}$ shortcut use karte hain.
Standard error. $SE = \frac{1.708}{36} = \frac{1.708}{6} = 0.2847$ . Yeh step kyun? Hum average kar rahe hain, isliye relevant spread SE hai, $σ$ nahi.
Dono edges standardize karo. $Z_{lo} = \frac{3.3 - 3.5}{0.2847} = - 0.70$ , $Z_{hi} = \frac{3.7 - 3.5}{0.2847} = + 0.70$ . Yeh step kyun? "3.3 aur 3.7 ke beech" ko ek tabulated curve par area mein convert karta hai.
Areas padho. $Φ (0.70) - Φ (- 0.70) = 0.758 - 0.242 = 0.516$ .

Verify karo: Symmetry se answer $2Φ (0.70) - 1 = 2 (0.758) - 1 = 0.516$ ke barabar hona chahiye. ✓ Dono edges centre se same distance par → symmetric $Z$ 's → sanity hold karta hai.

Figure — CLT ek picture mein (Example 1). Neeche ka chart source aur average dono overlay karta hai: flat blue bars ek die hain (uniform — har face equally likely), aur orange curve 36 dice ke average ki distribution hai. Dekho kaise flat, bell-shaped ban jaata hai, $μ = 3.5$ par centred (gray dashed line). Green shaded slice exactly woh $P (3.3 \leq \overset{ˉ}{X}_{36} \leq 3.7) \approx 0.516$ hai jo hum step 5 mein compute kiya — dekho woh band kitna narrow hai ek single die ki puri $1$ – $6$ range ke comparison mein, jo $n$ se divide karne ka shrinking effect hai.

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

Worked example 100 flips mein heads ka fraction

Statement. Ek biased coin heads $p = 0.3$ probability se land karta hai. Ise $n = 100$ baar flip karo. $P (\overset{ˉ}{X}_{100} \geq 0.35)$ kya hai, jahan $\overset{ˉ}{X}$ heads ka fraction hai?

Forecast: $0.35$ true rate $0.30$ se upar hai, isliye yeh aadhe se kam hona chahiye. Andaaza $\approx 0.14$ .

Bernoulli parameters. $μ = E [X] = p = 0.30$ , aur $σ^{2} = Var (X) = p (1 - p) = 0.3 \times 0.7 = 0.21$ , toh $σ = 0.4583$ . Yeh step kyun? Ek Bernoulli source skewed hota hai (die ki tarah symmetric nahi), phir bhi CLT uske average par apply hota hai — exactly isliye polls kaam karte hain.
Standard error. $SE = \frac{0.4583}{100} = 0.04583$ . Yeh step kyun? Question heads ke fraction ke baare mein hai, jo ek sample mean $\overset{ˉ}{X}_{100}$ hai; uska spread SE hai, isliye standardize karne se pehle $σ$ ko $n$ se divide karna zaroori hai.
Single boundary standardize karo. $Z = \frac{0.35 - 0.30}{0.04583} = 1.091 \approx 1.09$ . Yeh step kyun? One-sided question → ek Z value.
Tail area. $P (\overset{ˉ}{X} \geq 0.35) = 1 - Φ (1.09) = 1 - 0.862 = 0.138$ .

Verify karo: $0.138 < 0.5$ jaise forecast kiya gaya tha (mean se upar → right tail). Units check: fractions dimensionless hain, $Z$ dimensionless hai. ✓

Example 3 — Cell C: $n$ ke liye solve karo (inverse problem)

Worked example 0.5% margin ke liye kitne users chahiye?

Statement. Conversion rate $p = 0.05$ hai. Itne users $n$ chahiye taaki sample rate 95% confidence ke saath truth ke $\pm 0.005$ ke andar ho.

Forecast: Chote margins ke liye bada $n$ chahiye. Andaaza "hazaron".

Parameters. $μ = 0.05$ , $σ^{2} = Var (X) = 0.05 \times 0.95 = 0.0475$ , $σ = 0.2179$ . Yeh step kyun? Har CLT computation $μ$ aur $σ$ se shuru hoti hai; yahan $σ$ directly us margin formula mein jaata hai jise hum invert karne wale hain.
95% condition likho. Hume chahiye $P (∣ \overset{ˉ}{X}_{n} - μ ∣ \leq 0.005) = 0.95$ . Standard normal par, 95% area $\pm 1.96$ ke andar hota hai: $Φ (1.96) - Φ (- 1.96) = 0.95$ . Yeh step kyun? Hum usual direction ulta karte hain — probability fixed hai, hum $n$ ke liye solve karte hain.
Margin $=$ $1.96 \times SE$ set karo. $0.005 = 1.96 \cdot \frac{σ}{n}$ . Yeh step kyun? Ek 95% interval ki half-width $1.96 SE$ hoti hai; ise desired margin ke barabar force karo.
$n$ ke liye solve karo. $n = \frac{1.96 σ}{0.005} = \frac{1.96 \times 0.2179}{0.005} = 85.42$ , toh $n = 85.4 2^{2} = 7297.6$ (round up karke 7298).

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

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

Worked example Kya mini-batch gradient sahi direction mein point karega?

Statement. Per-sample gradient ka $μ_{g} = - 0.03$ , $σ_{g}^{2} = 0.25$ , batch $n = 64$ hai. $P (\overset{g}{ˉ}_{64} > 0)$ kya hai — matlab batch ke galat direction (negative hone par positive) mein point karne ka chance?

Forecast: True mean negative hai lekin noise bada hai, isliye sign flip ka real chance hai. Andaaza $\approx 0.3$ .

Standard error. $σ_{g} = 0.5$ , $SE = \frac{0.5}{64} = 0.0625$ . Yeh step kyun? $\overset{g}{ˉ}_{64}$ ek sample mean hai, isliye uska spread SE hai; boundary $0$ ko standardize karne se pehle hume yeh chahiye.
$0$ standardize karo. $Z = \frac{0 - ( - 0.03 )}{0.0625} = \frac{0.03}{0.0625} = 0.48$ . Yeh step kyun? " $\overset{g}{ˉ} > 0$ " ek one-sided event hai; sirf boundary $0$ ko Z chahiye.
Tail area. $P (\overset{g}{ˉ}_{64} > 0) = 1 - Φ (0.48) = 1 - 0.684 = 0.316$ .

Verify karo: Positive $Z$ (boundary mean se upar hai) → tail aadhe se chota, $0.316 < 0.5$ . Isliye Stochastic Gradient Descent ek galat-signed batch ka 32% chance tolerate karta hai: bahut saare steps mein mean drift jeet jaata hai. ✓

Example 5 — Cell E: negative mean, sign discipline

Worked example Negative target ke aas-paas ek window mein average error

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

Forecast: Yeh mean $- 0.03$ ke aas-paas ek symmetric window hai; expect karo ek modest area.

SE. $SE = 0.0625$ (jaise upar). Yeh step kyun? Wahi sample-mean spread reuse hoti hai; dono edges isse divide honge.
Dono edges standardize karo — signs ka dhyan rakho. $Z_{lo} = \frac{- 0.04 - ( - 0.03 )}{0.0625} = \frac{- 0.01}{0.0625} = - 0.16$ , $Z_{hi} = \frac{- 0.02 - ( - 0.03 )}{0.0625} = \frac{+ 0.01}{0.0625} = + 0.16$ . Yeh step kyun? Ek negative $μ$ subtract karna yaani add karna hai — classic sign trap. Window $Z$ mein symmetric hai kyunki woh $μ$ ke aas-paas symmetric hai.
Area. $Φ (0.16) - Φ (- 0.16) = 0.564 - 0.436 = 0.127$ .

Verify karo: $2Φ (0.16) - 1 = 2 (0.564) - 1 = 0.128 \approx 0.127$ (rounding). ✓ Symmetric window ne symmetric $\pm 0.16$ Z's diye — sign handling sahi hai.

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

Worked example Ek "die" jo hamesha 4 dikhata hai

Statement. $X$ hamesha $4$ ke barabar hota hai (ek constant random variable). Ise $n = 50$ baar roll karo, average nikalo. $\overset{ˉ}{X}_{50}$ ki distribution describe karo.

Forecast: Andar koi randomness nahi → bahar koi spread nahi. "Bell" ek point par collapse ho jaati hai.

Parameters. $μ = E [X] = 4$ , $σ^{2} = Var (X) = E [(X - 4)^{2}] = 0$ , toh $σ = 0$ . Yeh step kyun? Ek constant ki variance by definition zero hoti hai.
Standard error. $SE = \frac{0}{50} = 0$ . Yeh step kyun? SE CLT bell ki puri width hai; yahan zero matlab bell ki koi bhi width nahi hai.
Degenerate limit interpret karo. $SE = 0$ hone par $\overset{ˉ}{X}_{50}$ ki distribution bilkul bell nahi hoti: woh apni saari probability sirf value $4$ par rakhti hai. Formally yeh ek degenerate (point-mass) distribution hai: $P (\overset{ˉ}{X}_{50} = 4) = 1$ . Draw karne ke liye koi density curve nahi hai — bas $x = 4$ par probability $1$ ka ek arrow khada hai. Aur $Z = \frac{X ˉ - 4}{0}$ undefined hai, jo CLT ka yeh kehne ka ek polite tarika hai ki "standardize karne ke liye kuch nahi hai".

Verify karo: $\overset{ˉ}{X}_{50} = \frac{1}{50} (4 + 4 + \dots + 4) = 4$ with probability 1. Toh $P (\overset{ˉ}{X} = 4) = 1$ , ek point par baitha hua saara probability consistent hai. ✓ CLT ko $0 < σ^{2} < \infty$ chahiye; yeh woh edge hai jo yeh exclude karta hai.

Example 7 — Cell G: small $n$ — approximation BURI hai

Worked example

$n = 3$ dice kyun abhi bhi bell nahi hai Statement. Sirf $n = 3$ baar ek die roll karo. True $P (\overset{ˉ}{X}_{3} \leq 1)$ ko CLT estimate se compare karo.

Forecast: $\overset{ˉ}{X}_{3} \leq 1$ matlab teeno dice $1$ dikhayein — genuinely rare. CLT (ek smooth bell) galat andaaza laga sakti hai kyunki true distribution abhi bhi lumpy hai.

True probability. $\overset{ˉ}{X}_{3} \leq 1$ ke liye har die $1$ hona zaroori hai: $P = (1/6)^{3} = 1/216 = 0.00463$ . Yeh step kyun? Chote $n$ ke saath hum exactly compute kar sakte hain — ek fair yardstick.
CLT estimate. $SE = 1.708/ 3 = 0.9863$ ; $Z = \frac{1 - 3.5}{0.9863} = - 2.535$ ; $Φ (- 2.535) \approx 0.0056$ . Yeh step kyun? Wahi machine, lekin $n = 3$ "large" se bahut door hai.
Compare karo. True $0.00463$ vs CLT $0.0056$ — same ballpark mein lekin noticeably off, aur CLT yeh nahi dekh sakta ki average discrete hai (sirf $1/3$ ke multiples aate hain).

Verify karo: $(1/6)^{3} = 1/216$ exactly. ✓ Lesson: bell ek large- $n$ promise hai; chote $n$ ke saath tails ke paas woh jhooth bolta hai. Isliye Confidence Intervals chote samples ke liye $t$ -distribution par switch karte hain.

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

Example 9 — Cell I: total (sum), average nahi

Worked example 36 dice ka total score

Statement. Maano $S = \sum_{i = 1}^{36} X_{i}$ 36 dice ka sum hai. $P (S \leq 120)$ nikalo.

Forecast: Expected total $36 \times 3.5 = 126$ hai, aur $120$ usse neeche hai, isliye aadhe se kam. Andaaza $\approx 0.35$ .

SUM ka mean aur variance. $E [S] = n μ = 36 \times 3.5 = 126$ ; $Var (S) = n σ^{2} = 36 \times 2.9167 = 105$ , toh $sd (S) = 105 = 10.247$ . Yeh step kyun? Sum average se alag scale karta hai: uska spread $σ n$ hota hai, $σ / n$ nahi. Yeh sahi karo nahi toh har Z galat hoga.
Standardize karo. $Z = \frac{120 - 126}{10.247} = - 0.5855 \approx - 0.59$ . Yeh step kyun? " $S \leq 120$ " boundary ko standard normal par Z mein convert karta hai, step 1 se SUM ki apni standard deviation use karke.
Area (one-sided lower tail). $P (S \leq 120) = Φ (- 0.59) = 1 - Φ (0.59) = 1 - 0.7224 = 0.278$ . Yeh step kyun? $120$ mean se neeche hai, isliye hum $Z = - 0.59$ tak left-tail area chahte hain; tabulated $Φ (0.59)$ use karne ke liye symmetry se reflect karo.

Verify karo: Mean se consistency check: $S /36 = 120/36 = 3.333$ ; average machine use karte hue, $SE = 1.708/6 = 0.2847$ aur $Z = \frac{3.333 - 3.5}{0.2847} = - 0.585$ — step 2 se identical Z. ✓ Sum-route aur average-route agree karne chahiye, aur karte hain. Answer $0.278$ forecast " $\approx 0.35$ " ballpark se match karta hai (thoda kam). ✓

Example 10 — Cell J: exam twist + word problem

Worked example "Within one standard error" quick-fire

Statement. Ek sensor ki readings ka unknown skew hai lekin finite $μ, σ$ hai. Tum $n = 400$ readings lete ho. Ek colleague claim karta hai: "Sample mean truth ke ek standard error ke andar lagbhag do-tihai time land karta hai." CLT use karke unke claim ko verify karo, phir $P (∣ \overset{ˉ}{X} - μ ∣ \leq 2 SE)$ batao.

Forecast: "One SE" $\Rightarrow ∣ Z ∣ \leq 1$ ; 68–95–99.7 rule kehta hai $\approx 68%$ . Toh colleague sahi hai. Two SE $\Rightarrow\approx 95%$ .

$Z$ mein translate karo. $∣ \overset{ˉ}{X} - μ ∣ \leq SE$ matlab $\frac{X ˉ - μ}{SE} \leq 1$ , yani $∣ Z ∣ \leq 1$ . Yeh step kyun? "One standard error" literally $∣ Z ∣ \leq 1$ ki definition hai — $σ$ ya $n$ ke saath koi arithmetic nahi chahiye. Yahi exam shortcut hai.
$∣ Z ∣ \leq 1$ ke liye area. $Φ (1) - Φ (- 1) = 2Φ (1) - 1 = 2 (0.841) - 1 = 0.682$ . Toh sample mean ek SE ke andar lagbhag $68%$ time land karta hai — colleague sahi hai. Yeh step kyun? Two-sided $∣ Z ∣ \leq 1$ ko $Φ$ -symmetry use karke ek single area mein convert karta hai.
$∣ Z ∣ \leq 2$ ke liye area. $P (∣ \overset{ˉ}{X} - μ ∣ \leq 2 SE) = 2Φ (2) - 1 = 2 (0.977) - 1 = 0.954$ , yani lagbhag $95%$ . Yeh step kyun? Doosre half ka jawab usi two-sided rule se deta hai $∣ Z ∣ \leq 2$ ke saath.

Verify karo: $0.682 \approx$ do-tihai → colleague confirmed. $0.954 \approx 95%$ standard 95% interval half-width $1.96$ SE se match karta hai jo hum Example 3 mein use kiya tha. ✓ Dhyan do ki answer $n$ aur skew se independent hai — twist yeh hai ki tumhe sensor ka actual $σ$ kabhi nahi chahiya tha.

Recall Quick self-test

$n$ i.i.d. draws ke average ka spread hota hai ::: $SE = σ / n$ ( $n$ ki tarah shrink karta hai) 95% confidence ke saath margin $m$ ke andar aane ke liye $n \geq$ chahiye ::: $(1.96 σ / m)^{2}$ $n$ draws ke SUM ka standard deviation hota hai ::: $σ n$ (grow karta hai, shrink nahi) CLT fail karta hai jab ::: $σ = 0$ (degenerate) ya $σ^{2} = \infty$ (infinite variance), ya $n$ tails ke liye bahut chota ho $P (∣ Z ∣ \leq 1)$ aur $P (∣ Z ∣ \leq 2)$ approximately hain ::: $0.68$ aur $0.95$

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