Level 4 — ApplicationProbability & Statistics

Probability & Statistics

50 marksprintable — key stays hidden on paper

Level 4 — Application (novel problems, no hints) Time: 60 minutes Total Marks: 50

Answer all questions. Show full working. Use $...$ for mathematical notation. Standard normal values you may use: $\Phi(1.645)=0.95$ , $\Phi(1.96)=0.975$ , $\Phi(1)=0.8413$ .

Question 1 — Bayesian diagnostics & information (12 marks)

A spam filter classifies emails. From training data, $30\%$ of incoming emails are spam. A word "FREE" appears in $60\%$ of spam emails and in $10\%$ of legitimate emails.

(a) An email contains "FREE". Compute the posterior probability that it is spam. (4)

(b) Two independent word-features are now used: "FREE" (as above) and "WINNER" which appears in $50\%$ of spam and $5\%$ of legitimate emails. Assuming conditional independence of the two words given the class, compute the posterior probability of spam for an email containing both words. (5)

Question 2 — MLE for an exponential model (10 marks)

Server response times (in seconds) are modelled as i.i.d. $\text{Exponential}(\lambda)$ with density $f(x)=\lambda e^{-\lambda x}$ , $x\ge 0$ .

(a) Derive the maximum likelihood estimator $\hat\lambda$ for a sample $x_1,\dots,x_n$ . (4)

(b) A sample of $n=5$ gives observations $0.5, 1.0, 2.0, 1.5, 0.5$ seconds. Compute $\hat\lambda$ . (2)

(d) State (with brief justification) whether $\hat\lambda$ is the reciprocal of the sample mean, and what the MLE of the mean response time is. (2)

Question 3 — Sums, CLT and a probability bound (10 marks)

A factory produces bolts whose lengths are i.i.d. with mean $\mu = 10$ mm and standard deviation $\sigma = 0.4$ mm. A box contains $n = 100$ bolts.

(a) Let $\bar X$ be the sample mean length. State the approximate distribution of $\bar X$ and give its mean and standard deviation. (3)

(b) Compute the approximate probability that $\bar X$ exceeds $10.05$ mm. (3)

(d) A colleague claims the exact distribution of individual bolt lengths must be Normal for part (a) to hold. Is this correct? Justify in one sentence. (2)

Question 4 — Joint distribution, covariance, correlation (10 marks)

Discrete random variables $X,Y$ have joint pmf given by the table:

$X\backslash Y$	$0$	$1$
$0$	$0.1$	$0.2$
$1$	$0.3$	$0.4$

(a) Find the marginal pmfs of $X$ and $Y$ . (2)

(b) Are $X$ and $Y$ independent? Justify. (3)

(d) Compute the correlation coefficient $\rho_{XY}$ . (2)

Question 5 — Hypothesis test & confidence interval (8 marks)

A new model achieves accuracy on $n=400$ test items. It is correct on $340$ of them. The previous model had a known accuracy of $0.80$ .

(a) Estimate the new model's accuracy $\hat p$ and its standard error. (2)

(b) Test $H_0: p = 0.80$ vs $H_1: p > 0.80$ at the $5\%$ level using a one-sided z-test. State the z-statistic and conclusion. (4)

Answer keyMark scheme & solutions

Question 1

(a) Bayes. Let $S$ =spam, $F$ ="FREE". $P(S)=0.3,\ P(F|S)=0.6,\ P(F|\bar S)=0.1$ . (1) $P(F)=0.6(0.3)+0.1(0.7)=0.18+0.07=0.25$ . (1) $P(S|F)=\dfrac{0.18}{0.25}=0.72$ . (2)

(b) With conditional independence, $P(F,W|S)=0.6\times0.5=0.3$ ; $P(F,W|\bar S)=0.1\times0.05=0.005$ . (2) Numerator: $0.3\times0.3=0.09$ . (1) Evidence: $0.09+0.005\times0.7=0.09+0.0035=0.0935$ . (1) $P(S|F,W)=0.09/0.0935\approx 0.9626$ . (1)

(c) $H=-0.3\log_2 0.3 - 0.7\log_2 0.7$ . (1) $=-0.3(-1.737)-0.7(-0.5146)=0.5211+0.3602=0.8813$ bits. (2)

Question 2

(a) Likelihood $L=\prod \lambda e^{-\lambda x_i}=\lambda^n e^{-\lambda\sum x_i}$ . (1) Log-likelihood $\ell=n\ln\lambda-\lambda\sum x_i$ . (1) $\frac{d\ell}{d\lambda}=\frac{n}{\lambda}-\sum x_i=0$ . (1) $\hat\lambda=\dfrac{n}{\sum x_i}=\dfrac{1}{\bar x}$ (second deriv $-n/\lambda^2<0$ ⇒ max). (1)

(b) $\sum x_i=0.5+1.0+2.0+1.5+0.5=5.5$ , $n=5$ . $\hat\lambda=5/5.5=0.9091\ \text{s}^{-1}$ . (2)

(c) $P(X>2)=e^{-\hat\lambda\cdot 2}=e^{-1.818}\approx 0.1623$ . (2)

(d) Yes — $\hat\lambda=1/\bar x$ where $\bar x=1.1$ ; MLE of mean response time is $1/\hat\lambda=\bar x=1.1$ s (invariance of MLE). (2)

Question 3

(a) By CLT, $\bar X \approx N\!\left(\mu,\ \sigma^2/n\right)$ . Mean $=10$ , SD $=0.4/\sqrt{100}=0.04$ mm. (3)

(b) $z=\dfrac{10.05-10}{0.04}=1.25$ . (1) $P(\bar X>10.05)=1-\Phi(1.25)=1-0.8944=0.1056$ . (2) (accept ≈0.106)

(c) $10 \pm 1.96(0.04)=10\pm0.0784$ , i.e. $[9.9216,\ 10.0784]$ mm. (2)

(d) Incorrect — the CLT gives an approximate Normal distribution for $\bar X$ regardless of the individual distribution (given finite variance and large $n$ ); individual lengths need not be Normal. (2)

Question 4

(a) $P(X=0)=0.1+0.2=0.3$ , $P(X=1)=0.7$ . $P(Y=0)=0.1+0.3=0.4$ , $P(Y=1)=0.6$ . (2)

(b) Check $P(X=0,Y=0)=0.1$ vs $P(X=0)P(Y=0)=0.3\times0.4=0.12\ne0.1$ . (2) Not independent. (1)

(c) $E[X]=0.7$ , $E[Y]=0.6$ . (1) $E[XY]=\sum xy\,p = 1\cdot1\cdot0.4=0.4$ . (1) $\operatorname{Cov}=0.4-0.7\times0.6=0.4-0.42=-0.02$ . (1)

(d) $\operatorname{Var}(X)=0.7-0.49=0.21$ ; $\operatorname{Var}(Y)=0.6-0.36=0.24$ . $\rho=\dfrac{-0.02}{\sqrt{0.21\times0.24}}=\dfrac{-0.02}{0.22449}\approx -0.0891$ . (2)

Question 5

(a) $\hat p=340/400=0.85$ . $SE=\sqrt{\hat p(1-\hat p)/n}=\sqrt{0.85\cdot0.15/400}=\sqrt{0.000319}=0.01786$ . (2)

(b) Under $H_0$ , $SE_0=\sqrt{0.8\cdot0.2/400}=\sqrt{0.0004}=0.02$ . (1) $z=\dfrac{0.85-0.80}{0.02}=2.5$ . (2) Since $2.5>1.645$ (and p-value $\approx0.0062<0.05$ ), reject $H_0$ : significant improvement. (1)

(c) $0.85\pm1.96(0.01786)=0.85\pm0.0350=[0.8150,\ 0.8850]$ . (2)

[
{"claim":"Q1a posterior spam given FREE = 0.72","code":"pS=Rational(3,10); pFS=Rational(6,10); pFnS=Rational(1,10); post=pFS*pS/(pFS*pS+pFnS*(1-pS)); result = (post==Rational(72,100))"},
{"claim":"Q1b posterior both words approx 0.9626","code":"num=0.3*0.3; den=num+0.005*0.7; result = abs(num/den-0.9626)<1e-3"},
{"claim":"Q2 exp MLE lambda=5/5.5 and P(X>2)","code":"lam=Rational(5,55)*10; result = (lam==Rational(10,11)) and abs(exp(-float(lam)*2)-0.1623)<1e-3"},
{"claim":"Q3b P(Xbar>10.05) approx 0.1056","code":"from sympy import erf; z=1.25; p=1-(0.5*(1+erf(z/sqrt(2)))); result = abs(float(p)-0.1056)<2e-3"},
{"claim":"Q4 covariance -0.02 and correlation approx -0.0891","code":"cov=0.4-0.7*0.6; rho=cov/sqrt(0.21*0.24); result = abs(cov+0.02)<1e-9 and abs(rho+0.0891)<1e-3"},
{"claim":"Q5 z-statistic = 2.5","code":"z=(0.85-0.80)/sqrt(0.8*0.2/400); result = abs(z-2.5)<1e-9"}
]