How unit 5 is examined
This unit covers random-variable functions (PMF, PDF) and the Binomial, Poisson, Normal and Exponential distributions; Binomial, Poisson and Normal carry almost all the marks.
Probability Mass function
<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Not asked since 2022</span>
Definition. ==For a discrete random variable $X$, the probability mass function is $p(x)=P(X=x)$, which gives the probability of each value $x$ that $X$ can take.==
Key points.
- The PMF satisfies $p(x)\ge 0$ for every value $x$.
- The probabilities add up to one, that is $\sum_x p(x)=1$.
- The distribution function is $F(x)=P(X\le x)=\sum_{t\le x}p(t)$, and the mean is $E[X]=\sum x\,p(x)$.
- Binomial and Poisson are the two standard PMFs of this unit.
Probability Density Function
<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Not asked since 2022</span>
Definition. ==For a continuous random variable $X$, the probability density function $f(x)$ gives probabilities as areas: $P(a\le X\le b)=\int_a^b f(x)\,dx$.==
Key points.
- The PDF satisfies $f(x)\ge 0$ for all $x$, and the total area is one: $\int_{-\infty}^{\infty}f(x)\,dx=1$.
- For a continuous variable $P(X=c)=0$, so $f(x)$ itself is not a probability, only a density.
- The distribution function is $F(x)=\int_{-\infty}^{x}f(t)\,dt$, and the mean is $E[X]=\int x f(x)\,dx$.
- Normal and Exponential are the PDFs of this unit.
Discrete Distribution: Binomial
<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>
Definition. ==A random variable $X$ is Binomial with parameters $n$ and $p$ if it counts the successes in $n$ independent trials, each with the same success probability $p$, and $P(X=r)=\binom{n}{r}p^r q^{n-r}$, $r=0,1,\dots,n$, where $q=1-p$.==
Key points.
- Each trial has only two outcomes, success with probability $p$ and failure with probability $q=1-p$.
- The number of trials $n$ is fixed and the trials are independent, with $p$ constant.
- The probabilities are the terms of $(q+p)^n$, so they add up to 1.
- Mean $=np$, variance $=npq$ and standard deviation $=\sqrt{npq}$, so the variance is always less than the mean.
- If $X\sim B(n_1,p)$ and $Y\sim B(n_2,p)$ are independent, then $X+Y\sim B(n_1+n_2,p)$.
- "At least one" is best found as $1-P(0)=1-q^n$.
Derivation of the mean. $$E[X]=\sum_{r=0}^{n} r\binom{n}{r}p^r q^{n-r}$$ Use $r\binom{n}{r}=n\binom{n-1}{r-1}$ (the $r=0$ term is zero): $$E[X]=np\sum_{r=1}^{n}\binom{n-1}{r-1}p^{r-1}q^{(n-1)-(r-1)}=np\,(q+p)^{n-1}=np$$ Similarly $E[X(X-1)]=n(n-1)p^2$, so $\mathrm{Var}(X)=n(n-1)p^2+np-n^2p^2=np(1-p)=npq$.
Example (Jun 2020, Nov 2022, Jun 2026). Given $n=5$, $p=0.8$, $q=0.2$; find $P(X=3)$. $$P(X=3)=\binom{5}{3}(0.8)^3(0.2)^2=10\times0.512\times0.04$$ $P(X=3)=0.2048$. Hit at least once: $n=10$, $p=0.1$, $P=1-(0.9)^{10}=$ 0.6513.
Mean 12, SD 2 (May 2019). $np=12$, $npq=4$, so $q=\tfrac13$, $p=\tfrac23$, $n=12\div\tfrac23=$ 18.
Mean 4, variance 4/3 (Jun 2020). $q=\tfrac{4/3}{4}=\tfrac13$, $p=\tfrac23$, $n=6$. Then $P(r)=\binom6r(\tfrac23)^r(\tfrac13)^{6-r}$.
| Part | Working | Answer |
|---|---|---|
| (i) two successes | $\binom62\frac{2^2}{3^6}=\frac{60}{729}$ | 0.0823 |
| (ii) more than two | $1-[P(0)+P(1)+P(2)]=1-\frac{1+12+60}{729}$ | $\frac{656}{729}=0.8999$ |
| (iii) three or more | same event as (ii) | $\frac{656}{729}=0.8999$ |
Sum of binomials (Dec 2020). $X+Y\sim B(10,\tfrac12)$, so (i) $P(X+Y=r)=\binom{10}{r}\left(\tfrac12\right)^{10}$, $r=0,\dots,10$. (ii) $P(X+Y\ge3)=1-\frac{1+10+45}{1024}=$ $\frac{968}{1024}=0.9453$.
Balls to men (Jun 2022). Each ball goes to a man with $p=\frac{a}{a+b}$ and to a woman with $q=\frac{b}{a+b}$, so the number $k$ received by men is $B(m,p)$. $$(q+p)^m=\sum_k\binom mk p^kq^{m-k},\qquad (q-p)^m=\sum_k\binom mk(-p)^kq^{m-k}$$ Subtracting, the even terms cancel and the odd terms double: $$P(k\text{ odd})=\frac{(q+p)^m-(q-p)^m}{2}=\frac12\left[1-\frac{(b-a)^m}{(b+a)^m}\right]=\frac12\cdot\frac{(b+a)^m-(b-a)^m}{(b+a)^m}$$ since $q+p=1$ and $q-p=\frac{b-a}{a+b}$.
Answer frame. Open with the pmf and the meaning of $n,p,q$; for the mean, write $E[X]$ as a sum, apply $r\binom nr=n\binom{n-1}{r-1}$, and close with $np$; for numericals, list Given, find $n,p,q$ first (from mean and variance if needed), then substitute; close with the boxed value.
Pitfall: Dividing mean and variance in the wrong order: $q=\text{variance}/\text{mean}$, not the reverse.
Asked: [7 marks] (Jun 2020, Nov 2022, Jun 2026) A machine produces on average 80% good pieces; find the probability that out of 5 pieces 3 are good. If the probability of hitting an object is 10% and 10 shots are fired independently, find the probability that it is hit at least once. Asked: [7 marks] (May 2019) In a Binomial distribution the mean and standard deviation are 12 and 2; find $n$ and $p$. Asked: [7 marks] (Dec 2020) $X,Y$ independent binomial with $n_1=6,p=\frac12$ and $n_2=4,p=\frac12$; evaluate (i) $P(X+Y=r)$ (ii) $P(X+Y\ge3)$. Asked: [7 marks] (Jun 2020) Mean and variance of a Binomial distribution are 4 and $\frac43$; find the probability of (i) two successes (ii) more than two (iii) three or more. Asked: [7 marks] (Jun 2022) If $m$ balls are distributed among $a$ men and $b$ women, show that the probability that the number of balls received by men is odd is $\frac12\left[\frac{(b+a)^m-(b-a)^m}{(b+a)^m}\right]$. Asked: [7 marks] (Nov 2022) Find the mean of the Binomial distribution.
Poisson's distribution
<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>
Definition. ==A random variable $X$ is Poisson with parameter $\lambda$ if $P(X=r)=\dfrac{e^{-\lambda}\lambda^r}{r!}$, $r=0,1,2,\dots$; it is the limit of the Binomial when $n\to\infty$, $p\to0$ and $np=\lambda$ stays finite.==
Key points.
- It models the number of rare events in a fixed interval, such as accidents, misprints or deaths.
- The only parameter is $\lambda$, the average number of events.
- The probabilities add up to one because $\sum \lambda^r/r!=e^{\lambda}$.
- Mean $=$ variance $=\lambda$, which is the identifying property of the Poisson distribution.
- Skewness coefficient is $\gamma_1=1/\sqrt{\lambda}$ (and $\beta_1=1/\lambda$), so the distribution is always positively skewed.
- The recurrence $P(r+1)=\dfrac{\lambda}{r+1}P(r)$ gives the successive probabilities quickly.
Derivation of mean and variance. $$E[X]=\sum_{r=0}^{\infty} r\frac{e^{-\lambda}\lambda^r}{r!}=\lambda e^{-\lambda}\sum_{r=1}^{\infty}\frac{\lambda^{r-1}}{(r-1)!}=\lambda e^{-\lambda}e^{\lambda}=\lambda$$ $$E[X(X-1)]=\sum_{r=2}^{\infty} r(r-1)\frac{e^{-\lambda}\lambda^r}{r!}=\lambda^2e^{-\lambda}\sum_{r=2}^{\infty}\frac{\lambda^{r-2}}{(r-2)!}=\lambda^2$$ $$E[X^2]=\lambda^2+\lambda,\qquad \mathrm{Var}(X)=E[X^2]-\lambda^2=\lambda$$ So mean $=$ variance $=\lambda$.
Example (Dec 2020). $P(2)=9P(4)+90P(6)$: $$\frac{\lambda^2}{2}=9\frac{\lambda^4}{24}+90\frac{\lambda^6}{720}\ \Rightarrow\ 4=3\lambda^2+\lambda^4\ \Rightarrow\ (\lambda^2+4)(\lambda^2-1)=0$$ So $\lambda^2=1$, and $\lambda=1$. Coefficient of skewness $=1/\sqrt\lambda=$ 1.
Fitting (Jun 2020). Total $N=122+60+15+2+1=200$; $\sum fx=60+30+6+4=100$; mean $m=100/200=0.5$. Theoretical $f_x=200\,e^{-0.5}\dfrac{0.5^x}{x!}$ with $e^{-0.5}=0.6065$:
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Observed | 122 | 60 | 15 | 2 | 1 |
| Theoretical | 121.3 | 60.7 | 15.2 | 2.5 | 0.3 |
| Rounded | 121 | 61 | 15 | 3 | 0 |
The fit is very close, so Poisson describes the deaths well.
Answer frame. Open by writing the pmf and stating that $\lambda$ is the mean; for the derivation, do the mean, then $E[X(X-1)]$, then the variance; close with mean $=$ variance $=\lambda$. For numericals, substitute in the pmf, reduce to an equation in $\lambda$ or find $m=\sum fx/\sum f$, then compute.
Asked: [7 marks] (May 2019, Jun 2020, Jun 2023) Find the mean and variance of Poisson's distribution. Find the mean of Poisson distribution. Asked: [7 marks] (Dec 2020) $X$ is Poisson with $P(X=2)=9P(X=4)+90P(X=6)$; find (i) $\lambda$, the mean (ii) $\beta$, the coefficient of skewness. Asked: [7 marks] (Jun 2020) Find the mean and variance of the Poisson's distribution. Asked: [7 marks] (Jun 2020) Fit Poisson's distribution to deaths 0-4 with frequencies 122, 60, 15, 2, 1 and calculate the theoretical frequencies.
Continuous Distribution: Normal Distribution
<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>
Definition. ==A continuous random variable $X$ is normal with mean $\mu$ and variance $\sigma^2$ if its density is $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$, $-\infty<x<\infty$.==
Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u5-01" viewBox="0 0 596 338" width="596" height="338" role="img" aria-label="Bell-shaped normal curve, symmetric about the mean (mean = median = mode at the peak), total area 1"><style>#dsfig-u5-01 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u5-01 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u5-01 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u5-01 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u5-01 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u5-01 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u5-01 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u5-01 .t{fill:#16181D;font-weight:500}#dsfig-u5-01 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u5-01 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u5-01 .dot{fill:#16181D}#dsfig-u5-01 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u5-01 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u5-01 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u5-01 .ah{fill:#454C5A}#dsfig-u5-01 .ah.hi{fill:#2340B8}#dsfig-u5-01 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u5-01 .wl .t{font-size:12px;font-weight:700}#dsfig-u5-01 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u5-01 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u5-01 .e{stroke:#B1B7C3}html.dark #dsfig-u5-01 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u5-01 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u5-01 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u5-01 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u5-01 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u5-01 .t{fill:#E6E8ED}html.dark #dsfig-u5-01 .t.inv{fill:#0F1115}html.dark #dsfig-u5-01 .kd{stroke:#E6E8ED}html.dark #dsfig-u5-01 .dot{fill:#E6E8ED}html.dark #dsfig-u5-01 .ann{fill:#8FA3FF}html.dark #dsfig-u5-01 .lbl{fill:#858D9C}html.dark #dsfig-u5-01 .ptr{fill:#8FA3FF}html.dark #dsfig-u5-01 .ah{fill:#B1B7C3}html.dark #dsfig-u5-01 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u5-01 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u5-01 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u5-01 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah2" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh2" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M53.4,284.6 L283.2,54.8" marker-end="url(#ah2)"/><path class="e" d="M311.4,53.4 L541.2,283.2" marker-end="url(#ah2)"/><path class="e hi" d="M298,59 L298,279"/><circle class="n" cx="40" cy="298" r="18"/><text class="t" x="40" y="298" dy=".35em" text-anchor="middle">L</text><circle class="n" cx="298" cy="40" r="18"/><text class="t" x="298" y="40" dy=".35em" text-anchor="middle">M</text><circle class="n" cx="556" cy="298" r="18"/><text class="t" x="556" y="298" dy=".35em" text-anchor="middle">R</text><circle class="n" cx="298" cy="298" r="18"/><text class="t" x="298" y="298" dy=".35em" text-anchor="middle">T</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Bell-shaped normal curve, symmetric about the mean (mean = median = mode at the peak), total area 1</figcaption></figure>
Key points.
- The curve is bell-shaped and symmetric about $x=\mu$, so mean $=$ median $=$ mode $=\mu$.
- It has its maximum at $x=\mu$ and points of inflection at $x=\mu\pm\sigma$, and it approaches the $x$-axis asymptotically on both sides.
- The total area under the curve is 1, and skewness is 0.
- Empirical rule: $\mu\pm\sigma$ holds 68.27% of the area, $\mu\pm2\sigma$ holds 95.45% and $\mu\pm3\sigma$ holds 99.73%.
- The standard normal variate is $Z=\dfrac{X-\mu}{\sigma}$, with mean 0 and variance 1; areas are read from the table as $P(0\le Z\le z)$.
- A linear combination of independent normal variables is normal: if $X\sim N(\mu_1,\sigma_1^2)$, $Y\sim N(\mu_2,\sigma_2^2)$ then $aX+bY\sim N(a\mu_1+b\mu_2,\ a^2\sigma_1^2+b^2\sigma_2^2)$.
- Quartile deviation $=0.6745\sigma$, mean deviation $=\sqrt{2/\pi}\,\sigma=0.7979\sigma$ and SD $=\sigma$.
- It is used for heights, errors of measurement, marks and as the limit of Binomial and Poisson for large $n$ or $\lambda$.
Example 1 (Jun 2020). $\mu=68.22$, $\sigma=\sqrt{10.8}=3.286$, $x=6\text{ ft}=72$ in. $$z=\frac{72-68.22}{3.286}=1.15$$ $P(X>72)=0.5-0.3746=0.1254$, so expected number $=1000\times0.1254=$ 125 soldiers.
Example 2 (Jun 2022). $Z=X-Y$ has mean $1-2=-1$ and variance $9+16=25$, so $Z\sim N(-1,25)$ with pdf $$f(z)=\frac{1}{5\sqrt{2\pi}}e^{-\frac{(z+1)^2}{50}}$$ Mean $=$ median $=-1$, s.d. $=5$. $P[Z+1\le0]=P[Z\le-1]=$ 0.5, since $-1$ is the mean.
Example 3 (Jun 2026). 7% under 35 gives $z_1=-1.48$; 89% under 63 gives $z_2=+1.23$ (areas $0.43$ and $0.39$ from the mean). $$35=\mu-1.48\sigma,\qquad 63=\mu+1.23\sigma$$ Subtracting, $28=2.71\sigma$, so $\sigma\approx10.3$ and $\mu\approx50.3$.
Proof of QD : MD : SD (Jun 2022). For the normal curve, $SD=\sigma$. Quartiles are at $\mu\pm0.6745\sigma$ (area 0.25 each side of the median), so $QD=0.6745\sigma$. Mean deviation is $MD=\int|x-\mu|f(x)\,dx=2\int_{\mu}^{\infty}\frac{(x-\mu)}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx=\sigma\sqrt{2/\pi}=0.7979\sigma$. Hence $$QD:MD:SD=0.6745:0.7979:1\approx\tfrac23:\tfrac45:1=10:12:15$$ (multiplying by 15).
Answer frame. For "what is / properties", open with the pdf and parameters, draw the bell curve, then develop points 1-8 in order and close with the standard variate; for numericals, convert to $z$, use the table, and box the answer.
Pitfall: Forgetting to take the variance's square root: $\sigma=\sqrt{10.8}$, and the area beyond $z$ is $0.5-$ table value.
Asked: [7 marks] (Dec 2020) What is Normal distribution? Explain its properties. Asked: [7 marks] (Jun 2020) Mean height of soldiers 68.22 inches, variance 10.8; how many of 1000 soldiers would be over 6 ft tall? (Areas: $t=0$ to $0.35$ is 0.1368; to $1.15$ is 0.3746.) Asked: [7 marks] (Jun 2022) $X,Y$ independent normal with means 1, 2 and s.d. 3, 4; $Z=X-Y$: write the pdf of $Z$, state median, s.d., mean, and find $P[Z+1\le0]$. Asked: [7 marks] (Jun 2022) Prove that for the normal distribution QD : MD : SD :: 10 : 12 : 15. Asked: [7 marks] (Jun 2026) In an exactly normal distribution 7% of items are under 35 and 89% are under 63; find the mean and standard deviation.
Exponential Distribution
<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Low weight</span>
Definition. ==A continuous variable $X$ is exponential with parameter $\lambda>0$ if its density is $f(x)=\lambda e^{-\lambda x}$ for $x\ge0$ and $0$ otherwise.==
Key points.
- The distribution function is $F(x)=1-e^{-\lambda x}$, so $P(X>x)=e^{-\lambda x}$.
- Mean $=1/\lambda$ and variance $=1/\lambda^2$.
- It has the memoryless property: $P(X>s+t\mid X>s)=P(X>t)$.
- It models waiting times and lifetimes, such as time between calls or life of a component, and is the waiting time between Poisson events.
Asked: [7 marks] (Jun 2023) Write short note on Exponential Distribution.
Last-minute revision
- PMF: $p(x)\ge0$ and $\sum p(x)=1$; PDF: $f(x)\ge0$ and $\int f=1$, with $P(X=c)=0$.
- Binomial: $P(r)=\binom nr p^rq^{n-r}$, mean $np$, variance $npq$.
- From mean and variance: $q=\text{var}/\text{mean}$, $p=1-q$, $n=\text{mean}/p$; mean 12, SD 2 gives $n=18$, $p=\frac23$.
- Sum of independent $B(n_1,p)$ and $B(n_2,p)$ is $B(n_1+n_2,p)$; $B(10,\frac12)$: $P(\ge3)=\frac{968}{1024}$.
- Poisson: $P(r)=e^{-\lambda}\lambda^r/r!$, mean $=$ variance $=\lambda$, skewness $1/\sqrt\lambda$.
- Fitting Poisson: $m=\sum fx/\sum f$, $f_x=Ne^{-m}m^x/x!$; deaths data gives $m=0.5$.
- Normal: $z=(x-\mu)/\sigma$; 68.27%, 95.45%, 99.73% within 1, 2, 3 SD.
- Normal: QD : MD : SD $=0.6745\sigma : 0.7979\sigma : \sigma\approx10:12:15$.
- Soldiers over 6 ft: $z=1.15$, area $0.1254$, about 125; 7% and 89% question gives $\mu\approx50.3$, $\sigma\approx10.3$.
- Exponential: $f=\lambda e^{-\lambda x}$, mean $1/\lambda$, variance $1/\lambda^2$, memoryless.
Memory hooks
- Binomial "npq": variance is mean times $q$, so it is always smaller than the mean.
- Poisson: mean and variance are twins, both $\lambda$.
- Normal: "68-95-99.7" for 1, 2, 3 sigma.
- QD, MD, SD go 10, 12, 15, the same order as 0.67, 0.80, 1.
- Exponential: mean $1/\lambda$ is the inverse of the rate, and it forgets the past.
Coverage checklist
- Probability Mass function: definition, properties, mean.
- Probability Density Function: definition, properties, mean.
- Discrete Distribution: Binomial: mean derivation, machine and hit problems, $n,p$ from mean and SD, mean 4 variance 4/3, sum of binomials, balls to men proof.
- Poisson's: mean and variance derivation, $P(2)=9P(4)+90P(6)$, fitting deaths data.
- Continuous Distribution: Normal Distribution: properties, soldiers, $Z=X-Y$, QD:MD:SD proof, 7% and 89% problem.
- Exponential Distribution: short note.