Level 4 — ApplicationStatistics & Probability — Intermediate

Statistics & Probability — Intermediate

50 marksprintable — key stays hidden on paper

Level 4 — Application (Novel Problems, No Hints)

Time: 60 minutes Total Marks: 50

Answer all questions. Calculators permitted. Show all working.

Question 1 (11 marks)

A quality inspector records the lifetimes (in hours) of 60 light bulbs, grouped as follows:

Lifetime (h)	Frequency
$800\le t<900$	6
$900\le t<1000$	14
$1000\le t<1100$	22
$1100\le t<1200$	12
$1200\le t<1300$	6

(a) Estimate the mean lifetime. (3)

(b) Estimate the median lifetime using the grouped-median formula. (3)

(d) A rival dataset has the same mean but a larger standard deviation. State, with justification, what this implies about the reliability (consistency) of the rival's bulbs. (2)

Question 2 (10 marks)

The daily numbers of customer complaints at a call centre over 11 days were:

$4,\ 7,\ 3,\ 9,\ 12,\ 5,\ 8,\ 6,\ 15,\ 4,\ 10$

(a) Find the median, lower quartile $Q_1$ and upper quartile $Q_3$ . (4)

(b) Compute the interquartile range and use the $1.5\times\text{IQR}$ rule to determine whether any value is an outlier. (3)

Question 3 (10 marks)

A diagnostic test for a disease has the following properties: the disease affects $2\%$ of a population. The test is positive in $95\%$ of people who have the disease, and gives a false positive in $6\%$ of people who do not have the disease.

(a) A randomly chosen person tests positive. Using Bayes' theorem, find the probability that they actually have the disease. Give your answer to 4 decimal places. (5)

(b) Two independent people from the same population both test positive. Find the probability that at least one of them actually has the disease. (3)

Question 4 (10 marks)

A committee is to be formed by arranging and selecting from 7 women and 5 men.

(a) In how many ways can all 12 people be seated in a row so that no two men sit next to each other? (4)

(b) A subcommittee of 5 is chosen from the 12 people. In how many ways can this be done if the subcommittee must contain at least 3 women? (4)

(c) Using Pascal's triangle relation, verify that $\binom{12}{5}=\binom{11}{4}+\binom{11}{5}$ by evaluating all three quantities. (2)

Question 5 (9 marks)

A biased coin shows heads with probability $p=0.3$ . It is tossed 8 times independently. Let $X$ be the number of heads.

(a) Write the probability mass function of $X$ and find $P(X=3)$ to 4 decimal places. (3)

(b) Find the mean and variance of $X$ . (2)

(d) In the binomial expansion of $(0.7+0.3)^8$ , identify which term corresponds to $P(X=3)$ and state its general-term form. (2)

End of paper

Answer keyMark scheme & solutions

Question 1

(a) Mean. Use midpoints $x_i = 850,950,1050,1150,1250$ .

$\sum f_i x_i = 6(850)+14(950)+22(1050)+12(1150)+6(1250)$ $=5100+13300+23100+13800+7500 = 62800$ .

Mean $=\dfrac{62800}{60}=1046.\overline{6}\approx 1046.67$ h. (3) — [1 midpoints, 1 sum, 1 divide]

(b) Median. $N=60$ , $N/2=30$ . Cumulative freq: 6, 20, 42, 54, 60. Median class $=1000\le t<1100$ (contains 30th). $L=1000$ , $CF=20$ , $f=22$ , $h=100$ .

$\text{Median}=1000+\frac{30-20}{22}\times100 = 1000+45.45 = 1045.45\text{ h}.$ (3) — [1 class, 1 formula, 1 value]

(c) Mode. Modal class $=1000\le t<1100$ ( $f=22$ ). $f_1=22, f_0=14, f_2=12, L=1000, h=100$ .

$\text{Mode}=1000+\frac{22-14}{(22-14)+(22-12)}\times100 = 1000+\frac{8}{18}\times100 = 1044.44\text{ h}.$ (3) — [1 class, 1 formula, 1 value]

(d) Larger SD ⇒ greater spread of lifetimes about the same mean ⇒ less consistent/less reliable bulbs (more variability in how long they last). (2)

Question 2

Ordered: $3,4,4,5,6,7,8,9,10,12,15$ ( $n=11$ ).

(a) Median = 6th value $=7$ . Lower half (first 5): $3,4,4,5,6$ ⇒ $Q_1=4$ . Upper half (last 5): $8,9,10,12,15$ ⇒ $Q_3=10$ . (4) — [1 order, 1 median, 1 each quartile]

(b) IQR $=10-4=6$ . Fences: lower $=4-1.5(6)=-5$ ; upper $=10+1.5(6)=19$ . All data in $[-5,19]$ , so no outliers. (3) — [1 IQR, 1 fences, 1 conclusion]

(c) Box from 4 to 10, median line at 7; whiskers to min $=3$ and max $=15$ (no separated points since no outliers). (3)

Question 3

Let $D$ =disease, $+$ =positive. $P(D)=0.02$ , $P(+|D)=0.95$ , $P(+|D')=0.06$ .

(a) $P(+)=0.02(0.95)+0.98(0.06)=0.019+0.0588=0.0778$ .

$P(D|+)=\frac{0.019}{0.0778}=0.24421\approx 0.2442.$ (5) — [1 law of total prob setup, 1 numerator, 1 denominator, 1 Bayes formula, 1 value]

(b) $P(\text{at least one has disease} \mid \text{both}+ )$ : each person independently has $P(D|+)=0.24421$ . $P(\text{neither})=(1-0.24421)^2 = (0.75579)^2 = 0.57122$ . $P(\text{at least one})=1-0.57122=0.42878\approx 0.4288$ . (3) — [1 complement idea, 1 square, 1 value]

(c) Because the disease is rare (2%), the many false positives from the large healthy 98% (0.0588) are comparable to true positives (0.019); base rate dominates. (2)

Question 4

(a) Seat 7 women first: $7!$ ways. This creates 8 gaps; place 5 men in distinct gaps so none adjacent: $P(8,5)=\frac{8!}{3!}=6720$ . Total $=7!\times 6720 = 5040\times6720 = 33{,}868{,}800$ . (4) — [1 women arrange, 1 gaps, 1 men placement, 1 multiply]

(b) At least 3 women (of 7W,5M), choosing 5:

3W,2M: $\binom{7}{3}\binom{5}{2}=35\times10=350$
4W,1M: $\binom{7}{4}\binom{5}{1}=35\times5=175$
5W,0M: $\binom{7}{5}\binom{5}{0}=21\times1=21$

Total $=350+175+21=546$ . (4) — [1 each case, 1 sum]

(c) $\binom{12}{5}=792$ , $\binom{11}{4}=330$ , $\binom{11}{5}=462$ . $330+462=792$ . ✓ (2)

Question 5

$X\sim B(8,0.3)$ .

(a) PMF: $P(X=k)=\binom{8}{k}(0.3)^k(0.7)^{8-k}$ , $k=0,\dots,8$ . $P(X=3)=\binom{8}{3}(0.3)^3(0.7)^5=56(0.027)(0.16807)=0.25412\approx 0.2541$ . (3)

(b) Mean $=np=8(0.3)=2.4$ . Variance $=np(1-p)=8(0.3)(0.7)=1.68$ . (2)

(c) $P(X\ge2)=1-P(0)-P(1)$ . $P(0)=(0.7)^8=0.05765$ ; $P(1)=8(0.3)(0.7)^7=8(0.3)(0.08235)=0.19765$ . $P(X\ge2)=1-0.05765-0.19765=0.74470\approx 0.7447$ . (2)

(d) In $(0.7+0.3)^8=\sum_{k=0}^{8}\binom{8}{k}(0.7)^{8-k}(0.3)^k$ , $P(X=3)$ is the term with $k=3$ : $\binom{8}{3}(0.7)^{5}(0.3)^{3}$ ; general term $T_{k+1}=\binom{8}{k}(0.7)^{8-k}(0.3)^k$ . (2)

[
  {"claim":"Q1b grouped median = 1045.45","code":"m=1000+Rational(30-20,22)*100; result=abs(float(m)-1045.4545454)<1e-3"},
  {"claim":"Q1c grouped mode = 1044.44","code":"mode=1000+Rational(8,18)*100; result=abs(float(mode)-1044.4444)<1e-3"},
  {"claim":"Q3a Bayes P(D|+)=0.2442","code":"p=(0.02*0.95)/(0.02*0.95+0.98*0.06); result=abs(p-0.2442)<1e-4"},
  {"claim":"Q3b at least one=0.4288","code":"p=(0.02*0.95)/(0.02*0.95+0.98*0.06); ans=1-(1-p)**2; result=abs(ans-0.4288)<1e-4"},
  {"claim":"Q4b at least 3 women = 546","code":"from sympy import binomial; t=binomial(7,3)*binomial(5,2)+binomial(7,4)*binomial(5,1)+binomial(7,5)*binomial(5,0); result=t==546"},
  {"claim":"Q5a P(X=3)=0.2541","code":"from sympy import binomial; p=binomial(8,3)*0.3**3*0.7**5; result=abs(float(p)-0.2541)<1e-4"},
  {"claim":"Q5c P(X>=2)=0.7447","code":"from sympy import binomial; p=1-sum(binomial(8,k)*0.3**k*0.7**(8-k) for k in range(2)); result=abs(float(p)-0.7447)<1e-4"}
]