Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
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Original Cambridge 2027 concept practice for source-backed gaps. Indicative self-marking only, not an official paper or complete syllabus coverage.
Selected subtopics: Poisson models and linear combinations · Density moments and unbiased estimation · Confidence intervals and testing errors
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Question 1Calls arrive independently at a constant average rate of 2 per minute; simultaneous events in a very short interval are negligible. Model the count in 3 minutes, find its mean, variance and probability of at most one call. Explain a Poisson approximation for a separate Binomial(1000,0.002) count and find its approximate zero-count probability. Finally use a normal approximation to estimate P(Y<=24) for Y~Poisson(20).6 indicative marks
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Calls arrive independently at a constant average rate of 2 per minute; simultaneous events in a very short interval are negligible. Model the count in 3 minutes, find its mean, variance and probability of at most one call. Explain a Poisson approximation for a separate Binomial(1000,0.002) count and find its approximate zero-count probability. Finally use a normal approximation to estimate P(Y<=24) for Y~Poisson(20).
Cambridge 2027: Direct conceptual match
Recomputed exact Poisson masses and rare-binomial parameter 2; independently evaluated the continuity-corrected normal tail at 4.5/sqrt(20).
Scale the arrival-rate parameter by the observation duration. A Poisson mean and variance share the same numerical parameter, unlike a binomial variance np(1-p).
The two approximations concern different random variables. Do not reuse parameter 6 for the rare-defect count or omit the half-unit boundary in the Poisson-to-normal calculation.
Self-marking points
The three-minute call count is Poisson(6), with mean and variance both 6.
P(count<=1)=e^-6*(1+6)=7e^-6, approximately 0.0174.
The assumptions are independent events at a constant rate; clustering or changing rates could invalidate this model.
For the separate rare-event binomial, n is large, p is small and np=2, so use Poisson(2).
Its approximate zero probability is e^-2, approximately 0.135.
For Poisson(20), use N(20,20) and continuity boundary 24.5; z=4.5/sqrt(20), giving approximately 0.843.
Examiner tip
A named distribution is a modelling assumption; explain why its event mechanism is plausible.
Question 2Independent measurements X~N(10,4) and Y~N(3,1) use variance as the second parameter. Find the distribution of W=2X-Y+5 and P(W>25). Separately independent counts U~Poisson(2), V~Poisson(5) are added; state the distribution, mean and variance of U+V.5 indicative marks
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Independent measurements X~N(10,4) and Y~N(3,1) use variance as the second parameter. Find the distribution of W=2X-Y+5 and P(W>25). Separately independent counts U~Poisson(2), V~Poisson(5) are added; state the distribution, mean and variance of U+V.
Cambridge 2027: Direct conceptual match
Independently computed normal mean 22 and variance 17, upper tail near 0.233, and the Poisson sum's parameter 7.
The additive constant shifts the mean but not the variance, and subtraction does not subtract a variance.
Normality of the combination uses the stated independent normal models; the Poisson sum also needs independence. There is no claim that the difference of independent Poisson variables is Poisson.
Self-marking points
E(W)=2*10-3+5=22.
Var(W)=2^2*4+(-1)^2*1=17; independence removes the covariance term.
A linear combination of independent normal variables is normal, so W~N(22,17).
P(W>25)=1-Phi(3/sqrt(17)), approximately 0.233.
U+V~Poisson(7), with mean 7 and variance 7.
Examiner tip
Square coefficients for variance, including a negative coefficient.
Question 3A continuous random variable has density f(x)=k*x for 0<=x<=2 and zero otherwise. Determine k, P(X>1), the median, E(X) and Var(X). Give exact answers.5 indicative marks
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A continuous random variable has density f(x)=k*x for 0<=x<=2 and zero otherwise. Determine k, P(X>1), the median, E(X) and Var(X). Give exact answers.
Cambridge 2027: Direct conceptual match
Independently integrated normalisation, tail and raw moments; squared median gives half the area, variance is exactly 2/9.
The density must be nonnegative and have total area one; its value at a point is not a point probability.
The median uses a half-area condition, while the mean weights x by the density. Both lie inside [0,2], but they need not coincide.
Self-marking points
Normalisation requires integral(0 to 2) k*x dx=2k=1, so k=1/2.
P(X>1)=integral(1 to 2) x/2 dx=3/4.
The median m satisfies integral(0 to m) x/2 dx=m^2/4=1/2, so m=sqrt(2).
E(X)=integral(0 to 2) x^2/2 dx=4/3.
E(X^2)=integral(0 to 2) x^3/2 dx=2, so Var(X)=2-16/9=2/9.
Examiner tip
Use areas directly: explicit CDF terminology is not required for 9709 Paper 6.
Question 4An independent random sample of 64 observations from a normal population with unknown mean mu and known standard deviation 8 has mean 50. Find and interpret a 95% confidence interval for mu, using z=1.96. A separate independent random sample of 100 voters contains 60 supporters: find an approximate 95% interval for the population support proportion. Explain why sampling only a voluntary social-media poll could bias that estimate and how random selection helps.6 indicative marks
Show your reasoning6 indicative marksCalculator allowed
An independent random sample of 64 observations from a normal population with unknown mean mu and known standard deviation 8 has mean 50. Find and interpret a 95% confidence interval for mu, using z=1.96. A separate independent random sample of 100 voters contains 60 supporters: find an approximate 95% interval for the population support proportion. Explain why sampling only a voluntary social-media poll could bias that estimate and how random selection helps.
Cambridge 2027: Direct conceptual match
Checked SEs 1 and sqrt(0.0024), endpoints 48.04/51.96 and 0.50398/0.69602, and distinguished random-sampling coverage from selection bias.
The mean and proportion calculations use different random samples and different standard errors. The known-sigma normal model makes the first interval's sampling calculation exact apart from the supplied rounded critical value.
The proportion interval is a large-sample approximation, supported here by 60 successes and 40 failures. It is not a valid repair for a biased sampling frame.
Self-marking points
The sample mean has expectation mu and variance 8^2/64=1, and its distribution is exactly normal under the stated population model.
The mean interval is 50 +/-1.96*1, or [48.04,51.96].
For repeated random samples, approximately 95% of intervals formed by this method contain the fixed true mu; mu is not assigned a 95% random probability after observing this interval.
The estimated proportion is 0.60; estimated SE=sqrt(0.6*0.4/100), approximately 0.0489898.
The approximate proportion interval is 0.6 +/-1.96*0.0489898, or [0.504,0.696] to 3 decimal places.
Self-selection may overrepresent engaged users; selecting individuals at random from a suitable population frame reduces selection bias but does not guarantee that every realised sample is representative.
Examiner tip
Confidence concerns the repeated-sampling procedure, not the probability that a fixed parameter moves.
Question 5Ten independent Bernoulli trials have common success probability p. A test of H0:p=0.5 against H1:p>0.5 rejects when X>=9. Find the actual significance level and describe Type I error in context. If p=0.8, find the Type II error probability and state its meaning.5 indicative marks
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Ten independent Bernoulli trials have common success probability p. A test of H0:p=0.5 against H1:p>0.5 rejects when X>=9. Find the actual significance level and describe Type I error in context. If p=0.8, find the Type II error probability and state its meaning.
Cambridge 2027: Direct conceptual match
Summed both rejection outcomes under p=0.5 and recomputed the alternative complement under p=0.8, obtaining about 0.62419.
The significance level is computed under the null, whereas the Type II probability is computed under a particular alternative.
Use the full rejection tail for both calculations. The discrete rule happens to have actual significance about 1.07%, not an automatically assumed 5%.
Self-marking points
Type I error means rejecting p=0.5 when that null is true.
Its probability is P_0.5(X>=9)=(10+1)/1024=11/1024, approximately 0.0107.
When p=0.8 the Type II event is failing to reject, namely X<=8.
P_0.8(X<=8)=1-[10*(0.8)^9*0.2+(0.8)^10], approximately 0.624.
This is the probability of missing a real increase to p=0.8 under the specified rejection rule, not the probability that H0 is true.
Examiner tip
A Type II probability cannot be calculated without specifying an alternative parameter value.
Question 6A random sample of 5 independent measurements is 2,4,4,6,9. Estimate the population mean and give an unbiased estimate of its variance. Explain what 'unbiased' means. For a different non-normal population with mean 12 and variance 25, explain an approximate model for the mean of 100 independent observations and state its parameters.6 indicative marks
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A random sample of 5 independent measurements is 2,4,4,6,9. Estimate the population mean and give an unbiased estimate of its variance. Explain what 'unbiased' means. For a different non-normal population with mean 12 and variance 25, explain an approximate model for the mean of 100 independent observations and state its parameters.
Cambridge 2027: Direct conceptual match
Verified sample sum 25, squared-deviation sum 28 and unbiased variance 7; derived the separate sampling variance 0.25 without assuming an exactly normal population.
Keep the small data set and the separate theoretical population distinct. Dividing the observed squared-deviation sum by five gives a descriptive sample second moment, not the unbiased population-variance estimate.
The second population is explicitly non-normal; the sample mean is therefore approximately, not necessarily exactly, normal. The standard deviation of that mean is 0.5.
Self-marking points
The sample mean is 25/5=5.
The sum of squared deviations from 5 is 9+1+1+1+16=28.
The unbiased variance estimate is 28/(5-1)=7, using n-1 rather than n.
Unbiased means the estimator's expectation over repeated random samples equals the target parameter; it does not guarantee any one estimate is exact.
For the separate sample mean, E(Xbar)=12 and Var(Xbar)=25/100=0.25.
An informal central-limit approximation is Xbar approximately N(12,0.25), for independent identically distributed observations of finite variance and a sufficiently large sample.
Examiner tip
Do not transfer exact normality from the population-normal case to every finite sample.
