Worked examples — Common continuous distributions — Uniform, Normal, Exponential, Gamma, Beta

4.9.8 · D3 · Maths › Probability Theory & Statistics › Common continuous distributions — Uniform, Normal, Exponenti

Yeh page Common continuous distributions — Uniform, Normal, Exponential, Gamma, Beta ka drill floor hai. Parent note ne tumhe stories aur formulas diye. Yahan hum har case class ko cover karte hain jo har distribution throw kar sakti hai — ek worked example per cell — taaki koi bhi exam scenario naya na lage.

Koi bhi symbol aane se pehle, woh plain-word meanings yaad karo jinpar hum rely karte hain:

• $f(x)$ = density: value $x$ ke paas kitni probability pack hai (jaise har point par paint kitni thick hai). Yeh khud ek probability nahi hai — probability paane ke liye integrate (ek strip add up) karna padta hai.
• $F(x)=P(X\le x)$ = CDF: $x$ ke left tak jama hui total probability. Picture karo ki ek broom far left se $x$ tak sweep kar rahe ho aur dekh rahe ho kitna sweep hua.
• $\mu$ = mean = density ka balance point.
• $\sigma$ = standard deviation = balance point se typical distance; ${\sigma }^2$ = variance = average squared distance.
• $\lambda$ = ek rate = events per unit time (bada $\lambda$ = events faster aate hain).

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The scenario matrix

Is topic ka har problem inhi cells mein se ek hai. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

#	Case class	Kyun tricky hai	Covering example
C1	Uniform: interval probability + CDF	flat density ka ek slice padhna	Ex 1
C2	Uniform: degenerate / zero-width limit $b\to a$	variance $\to 0$, density blows up	Ex 2
C3	Exponential: survival & memorylessness	conditional "already waited" trap	Ex 3
C4	Exponential: limiting rate $\lambda \to 0$ aur $\lambda \to \infty$	mean $\to \infty$ vs $\to 0$	Ex 4
C5	Normal: dono tails, negative $z$, symmetry	$z$ ka sign, $\Phi (-z)=1-\Phi (z)$ ka use	Ex 5
C6	Normal: inverse (probability di gayi hai, $x$ find karo)	$\Phi$ ulta padhna; quantiles	Ex 6
C7	Gamma: integer $\alpha$ = sum of exponentials	mean/var add hote hain; Exp se sanity check	Ex 7
C8	Beta: skew, aur $\alpha =\beta =1$ = Uniform edge	shape flip hoti hai; flat mein degenerate hona	Ex 8
C9	Real-world word problem do distributions mix karta hua	English → parameters translate karna	Ex 9
C10	Exam twist: wrong-looking numbers ke saath standardizing	$\sigma$ use karo ${\sigma }^2$ nahi; ranges combine karo	Ex 10

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Ex 1 — Uniform interval probability (Cell C1)

Forecast: pehle guess karo — kya $P(3\le X\le 6)$ aadhe se bada hai ya chhota?

Figure kya dikhata hai: warm-paper plot jisme flat teal density $f(x)=1/6$ ko $x=2$ se $x=8$ tak ek rectangle ke roop mein draw kiya gaya hai; $x=3$ se $x=6$ tak ki strip burnt-orange fill ki gayi hai, jis par "width = 3", "height = 1/6", aur "area = 3 × 1/6 = 1/2" label hai. Picture yeh clearly dikhati hai ki uniform probability sirf ek sub-rectangle ka area hai.

1. Flat density ki height. Density ek constant $c$ hai isliye total area (width $8-2=6$ ka ek rectangle) $1$ ke barabar hai: $c\cdot 6=1\Rightarrow c=\frac{1}{6}$. Yeh step kyun? Ek probability density ka area $1$ hona zaroori hai — yeh woh ek law hai jo saari densities follow karti hain.
2. Slice probability = sub-rectangle ka area. Figure mein $3$ se $6$ tak ki shaded strip dekho: width $6-3=3$, height $\frac{1}{6}$, to $P=3\cdot \frac{1}{6}=\frac{1}{2}$. Yeh step kyun? Flat density ke liye, probability literally width $\times$ height hai — koi calculus nahi chahiye.
3. 6 tak CDF. $F(6)=\frac{6-2}{8-2}=\frac{4}{6}=\frac{2}{3}$. Yeh step kyun? Uniform CDF linear ramp $\frac{x-a}{b-a}$ hai — $x$ ke left wale interval ka fraction.

Verify: Check karo ki poora interval $1$ sum karta hai: $F(8)-F(2)=\frac{8-2}{6}-0=1$. ✓ Units: probabilities dimensionless hain aur $[0,1]$ mein hain. ✓

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Ex 2 — Uniform degenerate limit (Cell C2)

Forecast: agar saari probability ek vanishing interval mein squeeze kar do, to density ki height upar jaati hai ya neeche?

1. Height. $f=\frac{1}{\epsilon }\to \infty$ jab $\epsilon \to 0$. Yeh step kyun? Area $1$ rehna chahiye; agar width shrink hoti hai, to compensate karne ke liye height blow up karni padti hai — ek "spike."
2. Mean. $\mu =\frac{a+b}{2}=\frac{5+(5+\epsilon )}{2}=5+\frac{\epsilon }{2}\to 5$. Yeh step kyun? Ek shrinking interval ka balance point single point $5$ par collapse ho jaata hai.
3. Variance. $\mathrm{Var}=\frac{(b-a{)}^2}{12}=\frac{{\epsilon }^2}{12}\to 0$. Yeh step kyun? Zero spread ka matlab hai "random" variable ab certain hai — ek constant.

Verify: $\epsilon =0.1$ ke saath: $\mu =5.05$, $\mathrm{Var}=0.1^2/12\approx 8.33\times 10^{-4}$ — tiny, jaisa predict kiya tha. ✓

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Ex 3 — Exponential survival + memorylessness (Cell C3)

Forecast: guess karo — 4 minutes wait karne ke baad, kya tumhara remaining wait ek fresh person se shorter hai? (Intuition haan kehti hai; maths kuch surprising kehta hai.)

1. Survival function. $P(X>t)=e^{-\lambda t}$. To $P(X>3)=e^{-0.5\cdot 3}=e^{-1.5}\approx 0.2231$. Yeh step kyun? "Wait $t$ se zyada hai" = "$[0,t]$ mein koi bus nahi" = Poisson at $k=0$ = $e^{-\lambda t}$.
2. Conditional wait memorylessness use karta hai. $P(X>7\mid X>4)=\frac{P(X>7)}{P(X>4)}=\frac{e^{-0.5\cdot 7}}{e^{-0.5\cdot 4}}=e^{-0.5\cdot 3}=e^{-1.5}\approx \mathrm{0.2231.}$ Yeh step kyun? Exponential elapsed time bhool jaata hai — $3$ aur minutes ka extra wait ki wahi probability hai chahe tum fresh ho ya 4 minutes wait kar chuke ho.

Verify: Dono answers $e^{-1.5}=0.22313$ ke barabar hain. Units: $\lambda t$ ke andar minutes cancel ho jaate hain (per-min $\times$ min), ek dimensionless probability bacha rehta hai. ✓

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Ex 4 — Exponential limiting rates (Cell C4)

Forecast: agar events almost kabhi nahi hote ($\lambda \to 0$), to mean wait $0$ jaata hai ya $\infty$?

1. Mean aur variance. $\mu =\frac{1}{\lambda }$, $\mathrm{Var}=\frac{1}{{\lambda }^2}$. Yeh step kyun? Yeh standard exponential moments hain; limits dekhne ke liye inhe symbolic rakhte hain.
2. Slow limit $\lambda \to 0^{+}$. $\mu =\frac{1}{\lambda }\to \infty$; $\mathrm{Var}=\frac{1}{{\lambda }^2}\to \infty$ (aur yeh mean se bhi faster blow up karta hai, kyunki yeh mean ka square hai); aur $P(X>1)=e^{-\lambda }\to e^0=1$. Yeh step kyun? Events almost kabhi nahi hote ⇒ essentially forever wait karo, kitna lamba hoga is par enormous uncertainty ⇒ time $1$ par almost certainly ab bhi wait kar rahe ho.
3. Fast limit $\lambda \to \infty$. $\mu =\frac{1}{\lambda }\to 0$; $\mathrm{Var}=\frac{1}{{\lambda }^2}\to 0$; aur $P(X>1)=e^{-\lambda }\to 0$. Yeh step kyun? Events instantly fire hote hain ⇒ waits $0$ par collapse ho jaate hain aur spread bhi collapse ho jaata hai ⇒ time $1$ par essentially kabhi wait nahi kar rahe.

Verify: $\lambda =0.01$ par: $\mu =100$, $\mathrm{Var}=10000$, $P(X>1)=e^{-0.01}\approx 0.990$. $\lambda =100$ par: $\mu =0.01$, $\mathrm{Var}=10^{-4}$, $P(X>1)=e^{-100}\approx 0$. ✓

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Ex 5 — Normal, dono tails, negative $z$ (Cell C5)

Forecast: kaun sa tail bada hai — $160$ ke neeche ya $185$ ke upar? (Dono $15$ aur $10$ door hain... distances dekho.)

Figure kya dikhata hai: warm paper par $N(170,10^2)$ ka teal bell curve, mean $170$ ko mark karta ek plum dashed line. $x=160$ (labelled $z=-1$) aur $x=185$ (labelled $z=1.5$) par dotted verticals ek burnt-orange shaded band bracket karte hain; caption likhta hai "P(160≤X≤185)=0.7745". Kyunki $160$ mean se $1\sigma$ neeche hai jabki $185$ mean se $1.5\sigma$ upar, shaded band peak ke baare mein symmetric nahi hai — yeh thoda zyada daayein tak pahuncha hai.

1. Upper point standardize karo. $z_1=\frac{185-170}{10}=1.5$, to $P(X\le 185)=\Phi (1.5)\approx 0.9332$. Yeh step kyun? $\Phi$ sirf standard bell $N(0,1)$ jaanta hai; $z$-score kehta hai "mean se kitne $\sigma$ upar."
2. Lower point standardize karo — negative $z$. $z_2=\frac{160-170}{10}=-1.0$. Symmetry rule $\Phi (-z)=1-\Phi (z)$ use karo: $P(X\le 160)=\Phi (-1)=1-\Phi (1)=1-0.8413=\mathrm{0.1587.}$ Yeh step kyun? Bell $0$ ke baare mein symmetric hai; $-1$ ke left ka area $+1$ ke right ke area ke barabar hai.
3. Middle band = CDFs ka difference. $P(160\le X\le 185)=\Phi (1.5)-\Phi (-1)=0.9332-0.1587=0.7745$. Yeh step kyun? Band mein probability = (right end tak pile-up) minus (left end tak pile-up) — figure mein shaded region dekho.

Verify: $\Phi (1.5)=0.93319$, $\Phi (-1)=0.15866$, band $=0.77453$. Sanity: symmetric $\pm 1\sigma$ band $[160,180]$ sirf $\approx 68\%$ hold karta hai; hamara band right edge $180$ ko zyada door wale $185$ se swap karta hai, $180$ se $185$ tak ka extra slice add karta hai, to answer $68\%$ se zyada hona chahiye — aur $77\%$ hai. ✓

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Ex 6 — Normal inverse / quantile (Cell C6)

Forecast: kya $x^{\star }$ $185$ se upar hai ya neeche?

1. "Top 10%" ko CDF condition mein translate karo. Top $10\%$ ka matlab hai $P(X>x^{\star })=0.10$, to $P(X\le x^{\star })=0.90$, yaani $\Phi (z^{\star })=0.90$. Yeh step kyun? Hum CDF ki language mein kaam karna chahte hain, jo left-tail area read karta hai.
2. $\Phi$ ulta padho. $\Phi (z)=0.90$ wala $z$ hai $z^{\star }\approx 1.2816$ ($N(0,1)$ ka $90\%$ quantile). Yeh step kyun? Yeh $\Phi$ ko "undo" karta hai — inverse question "kaun sa $z$ yeh area deta hai?"
3. Wapas cm mein un-standardize karo. $x^{\star }=\mu +z^{\star }\sigma =170+1.2816\cdot 10=182.816$ cm. Yeh step kyun? $z=\frac{x-\mu }{\sigma }\Rightarrow x=\mu +z\sigma$ standard scale ko real height mein convert karta hai.

Verify: Forward check: $z=\frac{182.816-170}{10}=1.2816$, $\Phi (1.2816)\approx 0.900$. ✓ Aur $182.8<185$, Ex 5 se consistent hai jahan $\Phi (1.5)=0.933$ already $0.90$ exceed karta hai. ✓

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Ex 7 — Gamma with integer shape (Cell C7)

Forecast: kya mean exactly ek $\text{Exp}(2)$ wait ke mean ka $3\times$ hona chahiye?

1. Direct Gamma formulas. $\mu =\frac{\alpha }{\lambda }=\frac{3}{2}=1.5$ h, $\mathrm{Var}=\frac{\alpha }{{\lambda }^2}=\frac{3}{4}=0.75$. Yeh step kyun? Yeh Gamma density ko moment integrals mein plug karne se aate hain (parent §4).
2. Sum-of-exponentials check. $X$ teeno independent $\text{Exp}(2)$ waits ka sum hai, jisme har ek ka mean $\frac{1}{2}$ aur variance $\frac{1}{4}$ hai. Means add hote hain: $3\cdot \frac{1}{2}=1.5$. Variances add hote hain (independence): $3\cdot \frac{1}{4}=0.75$. Yeh step kyun? Integer-$\alpha$ Gamma hai hi exponentials ka sum, to dono routes agree karne chahiye — ek built-in sanity net.
3. Normalizer. Density hai $f(x)=\frac{{\lambda }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\lambda x}=\frac{2^3}{\Gamma (3)}{x}^2{e}^{-2x}=\frac{8}{2}{x}^2{e}^{-2x}=4x^2{e}^{-2x}$, $\Gamma (3)=2!=2$ use karke (dekho Gamma Function). Yeh step kyun? Constant ${\lambda }^{\alpha }/\Gamma (\alpha )$ exactly wahi hai jo ${\int }_0^{\infty }f=1$ force karta hai.

Verify: Dono routes $\mu =1.5,\mathrm{Var}=0.75$ dete hain. Aur ${\int }_0^{\infty }4x^2{e}^{-2x}dx=4\cdot \frac{2!}{2^3}=4\cdot \frac{2}{8}=1$. ✓

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Ex 8 — Beta shape aur Uniform edge (Cell C8)

Forecast: zyada "failure weight" ($\beta =5>\alpha =2$) ke saath, kya mean $0.5$ se neeche hai ya upar?

Figure kya dikhata hai: warm paper par do curves. Plum curve $\text{Beta}(2,5)$ hai — ek hump left ki taraf pushed, burnt-orange dashed vertical uske mean $2/7\approx 0.286$ par. Teal curve $\text{Beta}(1,1)$ hai — $[0,1]$ mein height $1$ par ek perfectly flat horizontal line, yaani $U(0,1)$ ki density. Side by side dono dikhate hain ki do shape parameters curve ko kaise bend karte hain, aur special case $\alpha =\beta =1$ ise Uniform mein flat kar deta hai.

1. Beta mean. $\mu =\frac{\alpha }{\alpha +\beta }=\frac{2}{2+5}=\frac{2}{7}\approx 0.2857$. Yeh step kyun? Mean total weight mein "success weight" ka fraction hai; zyada failure weight ise left pull karta hai, jaisa figure mein plum curve dikhata hai.
2. Beta variance. $\mathrm{Var}=\frac{\alpha \beta }{(\alpha +\beta {)}^2(\alpha +\beta +1)}=\frac{2\cdot 5}{7^2\cdot 8}=\frac{10}{392}\approx 0.02551$. Yeh step kyun? Parent ke variance formula mein direct substitution.
3. Uniform edge. $\alpha =\beta =1$ rakho: $f(x)=\frac{x^0(1-x{)}^0}{B(1,1)}$. Upar ki definition use karke, $B(1,1)=\frac{\Gamma (1)\Gamma (1)}{\Gamma (2)}=\frac{1\cdot 1}{1}=1$, to $f(x)=1$ on $(0,1)$ — flat density of $U(0,1)$. Figure mein teal line dekho. Yeh step kyun? $\text{Beta}(1,1)$ woh boundary case hai jahan hump completely flat hokar total ignorance mein aa jaata hai. Dekho Beta Function.

Verify: $\mu =2/7\approx 0.2857$, $\mathrm{Var}=10/392\approx 0.02551$, aur $B(1,1)=1$ to Beta$(1,1)$ density constant $1$ hai — the Uniform. ✓

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Ex 9 — Real-world word problem, do distributions (Cell C9)

Forecast: kaun sa part Exponential use karta hai aur kaun sa Normal? Computing se pehle decide karo.

1. Part (a) ek waiting time hai ⇒ Exponential. $P(\text{gap}>0.5)=e^{-\lambda t}=e^{-4\cdot 0.5}=e^{-2}\approx 0.1353$. Yeh step kyun? Poisson stream mein "time until next arrival" $\text{Exp}(\lambda )$ hai; survival $e^{-\lambda t}$ hai.
2. Part (b) ek measured magnitude hai ⇒ Normal. Standardize: $z=\frac{9-6}{2}=1.5$, to $P(\text{duration}<9)=\Phi (1.5)\approx 0.9332$. Yeh step kyun? Durations jo bahut saare small sub-tasks ke sums hain Normal ki taraf tend karte hain (Central Limit Theorem); $\Phi$ use karo $z$-score ke saath, aur yaad rakho $\sigma =2$ (nahi ki ${\sigma }^2$).

Verify: $e^{-2}=0.13534$; $\Phi (1.5)=0.93319$. Units: (a) mein, $\lambda t=4{\text{h}}^{-1}\times 0.5\text{h}=2$ (dimensionless). ✓

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Ex 10 — Exam twist: sahi $\sigma$, combined range (Cell C10)

Forecast: $\sigma$ kya hai? (Yahi poora trap hai.)

1. $\sigma$ sahi se extract karo. ${\sigma }^2=16\Rightarrow \sigma =\sqrt{16}=4$ ml. $z$-score mein $4$ use karo, kabhi $16$ nahi. Yeh step kyun? Standardizing formula standard deviation $\sigma$ use karta hai, variance ${\sigma }^2$ nahi — parent note ki classic galti.
2. Band ke dono endpoints standardize karo. $z_{\text{lo}}=\frac{496-500}{4}=-1$, $z_{\text{hi}}=\frac{507-500}{4}=1.75$. $P(496\le X\le 507)=\Phi (1.75)-\Phi (-1)=0.9599-0.1587=\mathrm{0.8013.}$ Yeh step kyun? Band probability = right-end CDF minus left-end CDF, $\Phi (-1)=1-\Phi (1)$ use karke.
3. Rejection = dono tails. $[494,506]$ ke liye: $z_{\text{lo}}=\frac{494-500}{4}=-1.5$, $z_{\text{hi}}=\frac{506-500}{4}=1.5$. Accept $=\Phi (1.5)-\Phi (-1.5)=0.9332-0.0668=0.8664$, to reject $=1-0.8664=0.1336$. Yeh step kyun? "Interval ke bahar" = $1$ minus "andar"; symmetry deta hai $\Phi (-1.5)=1-\Phi (1.5)$.

Verify: $\Phi (1.75)-\Phi (-1)=0.95994-0.15866=0.80128$; reject $=1-(\Phi (1.5)-\Phi (-1.5))=1-0.86639=0.13361$. ✓ Units: sab ml mein, $z$ ke andar cancel ho jaate hain. ✓

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Recall Self-test (guess karne ke baad reveal karo)

Uniform $U(2,8)$: $P(3\le X\le 6)$? ::: $\frac{1}{2}$ Exponential $\lambda =0.5$: $P(X>7\mid X>4)$? ::: $e^{-1.5}\approx 0.2231$ (memoryless) Normal $N(170,10^2)$: top $10\%$ ke liye cutoff? ::: $\approx 182.8$ cm Gamma$(3,2)$ mean aur variance? ::: $1.5$ aur $0.75$ Beta$(1,1)$ kaun sa distribution hai? ::: $U(0,1)$, flat density Bottle $N(500,16)$: $\sigma$ kya hai? ::: $4$ (variance ka square root) $\Phi (z)$ ka matlab kya hai? ::: standard bell $N(0,1)$ ke neeche $z$ ke left ka area

Recall Yeh kahan connect hota hai

• Normal examples Central Limit Theorem par rely karte hain.
• Exponential/Gamma waits Poisson Process se aate hain; normalizer Gamma Function use karta hai.
• Beta ka constant Beta Function use karta hai; Beta as a random probability Conjugate Priors in Bayesian Inference ko power deta hai.
• Yahan ke moments Moment Generating Functions se bhi nikale ja sakte hain.