Statistics & Probability — Intermediate

Question 1

(a) (3 marks)
By definition of conditional probability: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ and $P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}$. (1)
From the second, $P(A\cap B)=P(B\mid A)P(A)$. (1)
Substitute into the first: $P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B)}$. (1) This is Bayes' theorem.

(b) (4 marks)
Let $D$ = disease, $+$ = positive test.
$P(D)=0.02$, $P(+\mid D)=0.95$, $P(+\mid D^c)=1-0.90=0.10$. (1)
Total: $P(+)=0.95(0.02)+0.10(0.98)=0.019+0.098=0.117$. (1)
$P(D\mid +)=\dfrac{0.019}{0.117}=0.16239\ldots$ (1) $\approx 0.1624$. (1)

(c) (2 marks)
Because the disease is rare (2%), the large healthy population produces many false positives that swamp the smaller number of true positives, so a positive result is still probably a false alarm. (2)

Question 2

(a) (2 marks)
Midpoints: 5,15,25,35,45. $\sum fx = 4(5)+10(15)+14(25)+8(35)+4(45)=20+150+350+280+180=980$. (1)
Mean $=980/40=24.5$ min. (1)

(b) (4 marks)
$N/2=20$. Cumulative freq: 4,14,28,36,40 → median class 20–30. (1)
Symbols: $L=20$ (lower boundary), $F=14$ (cf before class), $f=14$, $h=10$. (1)
Median $=20+\dfrac{20-14}{14}\times10=20+\dfrac{6}{14}(10)$. (1)
$=20+4.2857=24.29$ min. (1)

(c) (3 marks)
Modal class 20–30: $f_1=14, f_0=10, f_2=8, L=20, h=10$. (1)
Mode $=L+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h=20+\dfrac{4}{28-18}\times10$. (1)
$=20+\dfrac{4}{10}(10)=24.0$ min. (1)

(d) (2 marks)
Mean $24.5$ > Median $24.29$ > Mode $24.0$: mean$>$median$>$mode indicates a slight positive (right) skew. (2)

Question 3

(a) (4 marks)
$\sigma^2=\frac1n\sum(x_i-\bar x)^2=\frac1n\sum(x_i^2-2\bar x x_i+\bar x^2)$. (1)
$=\frac1n\sum x_i^2 -\frac{2\bar x}{n}\sum x_i+\frac1n(n\bar x^2)$. (1)
Since $\frac1n\sum x_i=\bar x$, middle term $=2\bar x\cdot\bar x=2\bar x^2$. (1)
$=\frac1n\sum x_i^2-2\bar x^2+\bar x^2=\frac1n\sum x_i^2-\bar x^2.$ (1)

(b) (4 marks)
$\bar x=(2+4+4+6+9)/5=25/5=5$. (1)
$\sum x_i^2=4+16+16+36+81=153$. (1)
$\sigma^2=153/5-25=30.6-25=5.6$. (1)
$\sigma=\sqrt{5.6}\approx2.3664$. (1)

Question 4

(a) (2 marks)
$P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$, $r=0,\dots,n$. (1) Parameters: $n$ (trials), $p$ (success prob). (1)

(b) (4 marks)
$E[X]=\sum_{r=0}^n r\binom{n}{r}p^r q^{n-r}$; $r=0$ term vanishes. (1)
Use $r\binom{n}{r}=n\binom{n-1}{r-1}$: $E[X]=\sum_{r=1}^n n\binom{n-1}{r-1}p^r q^{n-r}$. (1)
Factor $np$ and let $k=r-1$: $=np\sum_{k=0}^{n-1}\binom{n-1}{k}p^k q^{n-1-k}$. (1)
The sum $=(p+q)^{n-1}=1$, so $E[X]=np$. (1)

(c) (3 marks)
$n=12,p=1/6$: $E[X]=12/6=2$. $\mathrm{Var}=np(1-p)=12\cdot\frac16\cdot\frac56=\frac{10}{6}\approx1.6667$. (1)
$P(X=2)=\binom{12}{2}(1/6)^2(5/6)^{10}=66\cdot\frac{1}{36}\cdot(5/6)^{10}$. (1)
$=0.2961$. (1)

Question 5

(a) (3 marks)
Permutation = ordered selection; combination = unordered. (1)
Each combination of $r$ objects can be ordered in $r!$ ways, and these together give all permutations: ${}^nP_r=\binom{n}{r}\cdot r!$. (1)
Hence $\binom{n}{r}=\dfrac{{}^nP_r}{r!}$. (1)

(b) (3 marks)
BALLOON: 7 letters with L×2, O×2. Total arrangements $=\dfrac{7!}{2!2!}=1260$. (1)
LL together (glue): 6 units, O×2: $\dfrac{6!}{2!}=360$. (1)
Never adjacent $=1260-360=900$. (1)

(c) (2 marks)
General term $T_{r+1}=\binom{9}{r}(2x)^{9-r}(-3/x)^r=\binom{9}{r}2^{9-r}(-3)^r x^{9-2r}$. (0.5)
$9-2r=5\Rightarrow r=2$. (0.5)
Coeff $=\binom{9}{2}2^{7}(-3)^2=36\cdot128\cdot9=41472$. (1)

Question 6

(a) (2 marks)
(1) $P(A)\ge0$ for all events; (2) $P(\Omega)=1$; (3) for mutually exclusive $A_i$, $P(\bigcup A_i)=\sum P(A_i)$. (2)

(b) (3 marks)
$P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.4-0.7=0.2$. (1)
Independent test: $P(A)P(B)=0.5\times0.4=0.2=P(A\cap B)$ ✓ → independent. (1)
Mutually exclusive would need $P(A\cap B)=0$, false. So independent, not mutually exclusive. (1)

[
  {"claim":"Bayes posterior P(D|+)=0.1624","code":"num=Rational(95,100)*Rational(2,100); den=num+Rational(10,100)*Rational(98,100); result=round(float(num/den),4)==0.1624"},
  {"claim":"Grouped mean=24.5, median=24.2857","code":"mean=Rational(980,40); med=20+Rational(20-14,14)*10; result=(mean==Rational(49,2)) and (round(float(med),4)==24.2857)"},
  {"claim":"Variance of {2,4,4,6,9} is 5.6","code":"xs=[2,4,4,6,9]; m=Rational(sum(xs),5); v=Rational(sum(x*x for x in xs),5)-m**2; result=v==Rational(28,5)"},
  {"claim":"Binomial P(X=2), n=12,p=1/6 = 0.2961","code":"p=binomial(12,2)*Rational(1,6)**2*Rational(5,6)**10; result=round(float(p),4)==0.2961"},
  {"claim":"BALLOON non-adjacent L arrangements=900","code":"total=factorial(7)/(factorial(2)*factorial(2)); glue=factorial(6)/factorial(2); result=(total-glue)==900"},
  {"claim":"Coefficient of x^5 in (2x-3/x)^9 is 41472","code":"c=binomial(9,2)*2**7*(-3)**2; result=c==41472"}
]