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

This page is the drill floor for Common continuous distributions — Uniform, Normal, Exponential, Gamma, Beta. The parent note gave you the stories and the formulas. Here we hit every case class each distribution can throw at you, one worked example per cell, so no exam scenario is new.

Before any symbol appears, remember the plain-word meanings we lean on:

• $f(x)$ = the density: how much probability is packed near the value $x$ (like how thick the paint is at each point). It is not a probability itself — you must integrate (add up a strip) to get a probability.
• $F(x)=P(X\le x)$ = the CDF: the total probability piled up to the left of $x$. Picture sweeping a broom from far left to $x$ and reading how much you've swept.
• $\mu$ = the mean = balance point of the density.
• $\sigma$ = the standard deviation = typical distance from the balance point; ${\sigma }^2$ = variance = average squared distance.
• $\lambda$ = a rate = events per unit time (bigger $\lambda$ = events come faster).

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The scenario matrix

Every problem in this topic is one of these cells. Each worked example below is tagged with the cell it fills.

#	Case class	What makes it tricky	Example that covers it
C1	Uniform: interval probability + CDF	reading a slice of a flat density	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 $\lambda \to 0$ and $\lambda \to \infty$	mean $\to \infty$ vs $\to 0$	Ex 4
C5	Normal: both tails, negative $z$, symmetry	sign of $z$, using $\Phi (-z)=1-\Phi (z)$	Ex 5
C6	Normal: inverse (given probability, find $x$)	reading $\Phi$ backwards; quantiles	Ex 6
C7	Gamma: integer $\alpha$ = sum of exponentials	mean/var add; sanity vs Exp	Ex 7
C8	Beta: skew, and the $\alpha =\beta =1$ = Uniform edge	shape flips; degenerate to flat	Ex 8
C9	Real-world word problem mixing two distributions	translating English → parameters	Ex 9
C10	Exam twist: standardizing with wrong-looking numbers	plug $\sigma$ not ${\sigma }^2$; combine ranges	Ex 10

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Ex 1 — Uniform interval probability (Cell C1)

Forecast: guess first — is $P(3\le X\le 6)$ bigger or smaller than one half?

What the figure shows: a warm-paper plot with the flat teal density $f(x)=1/6$ drawn as a rectangle from $x=2$ to $x=8$; the strip from $x=3$ to $x=6$ is filled burnt-orange, labelled "width = 3", "height = 1/6", and "area = 3 × 1/6 = 1/2". The picture makes clear that a uniform probability is just the area of a sub-rectangle.

1. Height of the flat density. The density is a constant $c$ so the total area (a rectangle of width $8-2=6$) equals $1$: $c\cdot 6=1\Rightarrow c=\frac{1}{6}$. Why this step? A probability density must enclose area $1$ — that's the one law all densities obey.
2. Slice probability = area of a sub-rectangle. Look at the shaded strip in the figure from $3$ to $6$: width $6-3=3$, height $\frac{1}{6}$, so $P=3\cdot \frac{1}{6}=\frac{1}{2}$. Why this step? For a flat density, probability is literally width $\times$ height — no calculus needed.
3. CDF up to 6. $F(6)=\frac{6-2}{8-2}=\frac{4}{6}=\frac{2}{3}$. Why this step? The uniform CDF is the linear ramp $\frac{x-a}{b-a}$ — the fraction of the interval to the left of $x$.

Verify: Check the whole interval sums to 1: $F(8)-F(2)=\frac{8-2}{6}-0=1$. ✓ Units: probabilities are dimensionless and lie in $[0,1]$. ✓

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Ex 2 — Uniform degenerate limit (Cell C2)

Forecast: if you squeeze all the probability into a vanishing interval, does the height of the density go up or down?

1. Height. $f=\frac{1}{\epsilon }\to \infty$ as $\epsilon \to 0$. Why this step? Area must stay $1$; if width shrinks, height must blow up to compensate — a "spike."
2. Mean. $\mu =\frac{a+b}{2}=\frac{5+(5+\epsilon )}{2}=5+\frac{\epsilon }{2}\to 5$. Why this step? The balance point of a shrinking interval collapses onto the single point $5$.
3. Variance. $\mathrm{Var}=\frac{(b-a{)}^2}{12}=\frac{{\epsilon }^2}{12}\to 0$. Why this step? Zero spread means the "random" variable is now certain — a constant.

Verify: With $\epsilon =0.1$: $\mu =5.05$, $\mathrm{Var}=0.1^2/12\approx 8.33\times 10^{-4}$ — tiny, as predicted. ✓

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Ex 3 — Exponential survival + memorylessness (Cell C3)

Forecast: guess — after waiting 4 minutes, is your remaining wait shorter than a fresh person's? (Intuition says yes; the maths says something surprising.)

1. Survival function. $P(X>t)=e^{-\lambda t}$. So $P(X>3)=e^{-0.5\cdot 3}=e^{-1.5}\approx 0.2231$. Why this step? "Wait exceeds $t$" = "no bus in $[0,t]$" = Poisson at $k=0$ = $e^{-\lambda t}$.
2. Conditional wait uses memorylessness. $P(X>7\mid X>4)=\frac{P(X>7)}{P(X>4)}=\frac{e^{-0.5\cdot 7}}{e^{-0.5\cdot 4}}=e^{-0.5\cdot 3}=e^{-1.5}\approx \mathrm{0.2231.}$ Why this step? The exponential forgets elapsed time — the extra wait of $3$ more minutes has the same probability whether you're fresh or have waited 4 minutes.

Verify: Both answers equal $e^{-1.5}=0.22313$. Units: minutes cancel inside $\lambda t$ (per-min $\times$ min), leaving a dimensionless probability. ✓

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Ex 4 — Exponential limiting rates (Cell C4)

Forecast: if events almost never happen ($\lambda \to 0$), does the mean wait go to $0$ or $\infty$?

1. Mean and variance. $\mu =\frac{1}{\lambda }$, $\mathrm{Var}=\frac{1}{{\lambda }^2}$. Why this step? These are the standard exponential moments; keep them symbolic to see the limits.
2. Slow limit $\lambda \to 0^{+}$. $\mu =\frac{1}{\lambda }\to \infty$; $\mathrm{Var}=\frac{1}{{\lambda }^2}\to \infty$ (and it blows up even faster than the mean, since it is the mean squared); and $P(X>1)=e^{-\lambda }\to e^0=1$. Why this step? Events almost never happen ⇒ you wait essentially forever, with enormous uncertainty about how long ⇒ you're almost surely still waiting at time $1$.
3. Fast limit $\lambda \to \infty$. $\mu =\frac{1}{\lambda }\to 0$; $\mathrm{Var}=\frac{1}{{\lambda }^2}\to 0$; and $P(X>1)=e^{-\lambda }\to 0$. Why this step? Events fire instantly ⇒ waits collapse to $0$ and the spread collapses too ⇒ you're essentially never still waiting at time $1$.

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

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

Forecast: which tail is bigger — below $160$ or above $185$? (Both are $15$ and $10$ away... watch the distances.)

What the figure shows: the teal bell curve of $N(170,10^2)$ on warm paper, with a plum dashed line marking the mean $170$. Dotted verticals at $x=160$ (labelled $z=-1$) and $x=185$ (labelled $z=1.5$) bracket a burnt-orange shaded band; the caption reads "P(160≤X≤185)=0.7745". Because $160$ is $1\sigma$ below the mean while $185$ is $1.5\sigma$ above, the shaded band is not symmetric about the peak — it reaches slightly further to the right.

1. Standardize the upper point. $z_1=\frac{185-170}{10}=1.5$, so $P(X\le 185)=\Phi (1.5)\approx 0.9332$. Why this step? $\Phi$ knows only the standard bell $N(0,1)$; the $z$-score says "how many $\sigma$ above the mean."
2. Standardize the lower point — negative $z$. $z_2=\frac{160-170}{10}=-1.0$. Use the symmetry rule $\Phi (-z)=1-\Phi (z)$: $P(X\le 160)=\Phi (-1)=1-\Phi (1)=1-0.8413=\mathrm{0.1587.}$ Why this step? The bell is symmetric about $0$; the area left of $-1$ equals the area right of $+1$.
3. Middle band = difference of CDFs. $P(160\le X\le 185)=\Phi (1.5)-\Phi (-1)=0.9332-0.1587=0.7745$. Why this step? Probability in a band = (pile-up to the right end) minus (pile-up to the left end) — see the shaded region in the figure.

Verify: $\Phi (1.5)=0.93319$, $\Phi (-1)=0.15866$, band $=0.77453$. Sanity: the symmetric $\pm 1\sigma$ band $[160,180]$ holds only $\approx 68\%$; our band swaps the right edge $180$ for the more distant $185$, adding the extra slice from $180$ to $185$, so the answer must exceed $68\%$ — and $77\%$ does. ✓

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Ex 6 — Normal inverse / quantile (Cell C6)

Forecast: is $x^{\star }$ above or below $185$?

1. Translate "top 10%" into a CDF condition. Top $10\%$ means $P(X>x^{\star })=0.10$, so $P(X\le x^{\star })=0.90$, i.e. $\Phi (z^{\star })=0.90$. Why this step? We must work in the language of the CDF, which reads left-tail area.
2. Read $\Phi$ backwards. The $z$ with $\Phi (z)=0.90$ is $z^{\star }\approx 1.2816$ (the $90\%$ quantile of $N(0,1)$). Why this step? This "un-does" $\Phi$ — the inverse question "which $z$ gives this area?"
3. Un-standardize back to cm. $x^{\star }=\mu +z^{\star }\sigma =170+1.2816\cdot 10=182.816$ cm. Why this step? $z=\frac{x-\mu }{\sigma }\Rightarrow x=\mu +z\sigma$ converts the standard scale back to real height.

Verify: Check forward: $z=\frac{182.816-170}{10}=1.2816$, $\Phi (1.2816)\approx 0.900$. ✓ And $182.8<185$, consistent with Ex 5 where $\Phi (1.5)=0.933$ already exceeds $0.90$. ✓

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Ex 7 — Gamma with integer shape (Cell C7)

Forecast: should the mean be exactly $3\times$ the mean of one $\text{Exp}(2)$ wait?

1. Direct Gamma formulas. $\mu =\frac{\alpha }{\lambda }=\frac{3}{2}=1.5$ h, $\mathrm{Var}=\frac{\alpha }{{\lambda }^2}=\frac{3}{4}=0.75$. Why this step? These come from plugging the Gamma density into the moment integrals (parent §4).
2. Sum-of-exponentials check. $X$ is the sum of $3$ independent $\text{Exp}(2)$ waits, each with mean $\frac{1}{2}$ and variance $\frac{1}{4}$. Means add: $3\cdot \frac{1}{2}=1.5$. Variances add (independence): $3\cdot \frac{1}{4}=0.75$. Why this step? Integer-$\alpha$ Gamma is a sum of exponentials, so both routes must agree — a built-in sanity net.
3. Normalizer. The density is $f(x)=\frac{{\lambda }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\lambda x}=\frac{2^3}{\Gamma (3)}{x}^2{e}^{-2x}=\frac{8}{2}{x}^2{e}^{-2x}=4x^2{e}^{-2x}$, using $\Gamma (3)=2!=2$ (see Gamma Function). Why this step? The constant ${\lambda }^{\alpha }/\Gamma (\alpha )$ is exactly what forces ${\int }_0^{\infty }f=1$.

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

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Ex 8 — Beta shape and the Uniform edge (Cell C8)

Forecast: with more "failure weight" ($\beta =5>\alpha =2$), is the mean below or above $0.5$?

What the figure shows: two curves on warm paper. The plum curve is $\text{Beta}(2,5)$ — a hump pushed toward the left, with a burnt-orange dashed vertical at its mean $2/7\approx 0.286$. The teal curve is $\text{Beta}(1,1)$ — a perfectly flat horizontal line at height $1$ across $[0,1]$, i.e. the density of $U(0,1)$. Side by side they show how the two shape parameters bend the curve, and how the special case $\alpha =\beta =1$ flattens it into the Uniform.

1. Beta mean. $\mu =\frac{\alpha }{\alpha +\beta }=\frac{2}{2+5}=\frac{2}{7}\approx 0.2857$. Why this step? The mean is the fraction of "success weight" among total weight; more failure weight pulls it left, as the plum curve in the figure shows.
2. Beta variance. $\mathrm{Var}=\frac{\alpha \beta }{(\alpha +\beta {)}^2(\alpha +\beta +1)}=\frac{2\cdot 5}{7^2\cdot 8}=\frac{10}{392}\approx 0.02551$. Why this step? Direct substitution into the parent's variance formula.
3. The Uniform edge. Put $\alpha =\beta =1$: $f(x)=\frac{x^0(1-x{)}^0}{B(1,1)}$. Using the definition above, $B(1,1)=\frac{\Gamma (1)\Gamma (1)}{\Gamma (2)}=\frac{1\cdot 1}{1}=1$, so $f(x)=1$ on $(0,1)$ — the flat density of $U(0,1)$. See the teal line in the figure. Why this step? $\text{Beta}(1,1)$ is the boundary case where the hump flattens completely into total ignorance. See Beta Function.

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

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

Forecast: which part uses Exponential and which uses Normal? Decide before computing.

1. Part (a) is a waiting time ⇒ Exponential. $P(\text{gap}>0.5)=e^{-\lambda t}=e^{-4\cdot 0.5}=e^{-2}\approx 0.1353$. Why this step? "Time until next arrival" in a Poisson stream is $\text{Exp}(\lambda )$; survival is $e^{-\lambda t}$.
2. Part (b) is a measured magnitude ⇒ Normal. Standardize: $z=\frac{9-6}{2}=1.5$, so $P(\text{duration}<9)=\Phi (1.5)\approx 0.9332$. Why this step? Durations that are sums of many small sub-tasks tend to Normal (Central Limit Theorem); use $\Phi$ with the $z$-score, and remember $\sigma =2$ (not ${\sigma }^2$).

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

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

Forecast: what is $\sigma$? (This is the whole trap.)

1. Extract $\sigma$ correctly. ${\sigma }^2=16\Rightarrow \sigma =\sqrt{16}=4$ ml. Use $4$, never $16$, in the $z$-score. Why this step? The standardizing formula uses the standard deviation $\sigma$, not the variance ${\sigma }^2$ — the classic parent-note mistake.
2. Standardize both endpoints of the band. $z_{\text{lo}}=\frac{496-500}{4}=-1$, $z_{\text{hi}}=\frac{507-500}{4}=1.75$. $P(496\le X\le 507)=\Phi (1.75)-\Phi (-1)=0.9599-0.1587=\mathrm{0.8013.}$ Why this step? Band probability = right-end CDF minus left-end CDF, using $\Phi (-1)=1-\Phi (1)$.
3. Rejection = both tails. For $[494,506]$: $z_{\text{lo}}=\frac{494-500}{4}=-1.5$, $z_{\text{hi}}=\frac{506-500}{4}=1.5$. Accept $=\Phi (1.5)-\Phi (-1.5)=0.9332-0.0668=0.8664$, so reject $=1-0.8664=0.1336$. Why this step? "Outside an interval" = $1$ minus "inside"; symmetry gives $\Phi (-1.5)=1-\Phi (1.5)$.

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: all in ml, cancelling inside $z$. ✓

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Recall Self-test (reveal after guessing)

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 $N(170,10^2)$: cutoff for top $10\%$? ::: $\approx 182.8$ cm Gamma$(3,2)$ mean and variance? ::: $1.5$ and $0.75$ Beta$(1,1)$ is which distribution? ::: $U(0,1)$, the flat density Bottle $N(500,16)$: what is $\sigma$? ::: $4$ (square root of the variance) What does $\Phi (z)$ mean? ::: area under the standard bell $N(0,1)$ to the left of $z$

Recall Where this connects

• The Normal examples lean on the Central Limit Theorem.
• Exponential/Gamma waits come from the Poisson Process; the normalizer uses the Gamma Function.
• Beta's constant uses the Beta Function; Beta as a random probability powers Conjugate Priors in Bayesian Inference.
• Moments here can also be pulled from Moment Generating Functions.