Data analysis, probability and statistics

The definitive reference. Compiled 15 September 2026 from the VCAA archive 2006–2026 (NHT). Written for students targeting 45+ and for the people building material for them.

Everything below is evidence. Every percentage is the real state figure from corpus/mm/questions.json (pct = percentage of the cohort awarded full marks for that part). Every quoted sentence is lifted from a real VCAA paper or examination report. Where the corpus is broken, that is said out loud rather than papered over.

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0. Sources and corpus caveats specific to this area

Corpus caveats that bite this area harder than any other.

1. The 2024 November papers cannot be read. Documents_exams_mathematics_2024_2024MM1-w.txt is 16 bytes and ...2024MM2-w.txt is 32 bytes; the source PDFs are image-only (pdftotext returns nothing). Everything quoted for 2024 November therefore comes from the report, and the report's equations were MathType objects that did not survive extraction. Where a 2024 November question is cited below, the mark distribution is real and the wording is not recoverable. That is stated each time.
2. 2011 is unusable as statistics. Both 2011 reports were OCR'd from degraded scans. [QJSON] carries duplicated question labels (2011 Exam 1 Q7b–7b appears twice, with pct 1% and 3%), and the 2011 paper text is mojibake ()RUWKHFRQWLQXRXVUDQGRPYDULDEOH…). The 2011 separators are listed below for completeness but no 2011 percentage should be quoted to a student.
3. NHT papers have no statistics. All 100 NHT statistics entries carry pct: null. They are content evidence, not difficulty evidence.
4. The 2026 NHT papers are in [PAPERS] but not in [QJSON]. [QJSON] covers NHT 2017–2025 only. Every 2026 … (NHT) citation below is read directly from the paper text and carries no mark data.
5. Twelve parts are mis-tagged. [QJSON] assigns topic per question-part, and a few parts inside a statistics-numbered question are actually calculus or algebra: 2007 Exam 1 Q12 (minimum distance from origin to a line), 2010 Exam 1 Q9b (integration by parts), 2011 Exam 1 Q6b–6b (simultaneous equations, no solution), 2014 Exam 2 Section B Q1a–Q1d (a wombat population n(t) = 1200 + 400cos(πt/3) — pure circular-function modelling), 2022 Exam 2 Section B Q2ai/aii/aiii/d/h (a fox/rabbit predator-prey sinusoid). They are flagged in the separator list and excluded from the type analysis. Conversely 2016 Exam 1 Q8a ("show by differentiation that … is an antiderivative") is correctly filed — it is the engine for the pdf question that follows it.
6. topic_conf. [QJSON] grades its own confidence: 407 parts clear, 36 close, 33 read by hand. Every read by hand entry is a 2011, 2024 or 2025-NHT item where the tagger worked from images.

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1. What the study design puts in this area — and what the exams actually test

1.1 The overview, verbatim

[SD]:

"In this area of study students cover discrete and continuous random variables, their representation using tables, probability functions (specified by rule and defining parameters as appropriate); the calculation and interpretation of central measures and measures of spread; and statistical inference for sample proportions. The focus is on understanding the notion of a random variable, related parameters, properties and application and interpretation in context for a given probability distribution."

1.2 The content dot points, verbatim

[SD]:

This area of study includes:
- random variables, including the concept of a random variable as a real function defined on a sample space and examples of discrete and continuous random variables
- discrete random variables:
- specification of probability distributions for discrete random variables using graphs, tables and probability mass functions
- calculation and interpretation of mean, μ, variance, σ², and standard deviation of a discrete random variable and their use
- Bernoulli trials and the binomial distribution, Bi(n, p), as an example of a probability distribution for a discrete random variable
- effect of variation in the value(s) of defining parameters on the graph of a given probability mass function for a discrete random variable
- calculation of probabilities for specific values of a random variable and intervals defined in terms of a random variable, including conditional probability
- continuous random variables:
- construction of probability density functions from non-negative functions of a real variable
- specification of probability distributions for continuous random variables using probability density functions
- calculation and interpretation of mean, μ, variance, σ², and standard deviation of a continuous random variable and their use
- standard normal distribution, N(0, 1), and transformed normal distributions, N(μ, σ²), as examples of a probability distribution for a continuous random variable
- effect of variation in the value(s) of defining parameters on the graph of a given probability density function for a continuous random variable
- calculation of probabilities for intervals defined in terms of a random variable, including conditional probability (the cumulative distribution function may be used but is not required)
- statistical inference, including definition and distribution of sample proportions, simulations and confidence intervals:
- distinction between a population parameter and a sample statistic and the use of the sample statistic to estimate the population parameter
- simulation of random sampling, for a variety of values of p and a range of sample sizes, to illustrate the distribution of P̂ and variations in confidence intervals between samples
- concept of the sample proportion P̂ = X/n as a random variable whose value varies between samples, where X is a binomial random variable which is associated with the number of items that have a particular characteristic and n is the sample size
- approximate normality of the distribution of P̂ for large samples and, for such a situation, the mean p (the population proportion) and standard deviation, √(p(1 − p)/n)
- determination and interpretation of, from a large sample, an approximate confidence interval (p̂ − z√(p̂(1 − p̂)/n), p̂ + z√(p̂(1 − p̂)/n)), for a population proportion where z is the appropriate quantile for the standard normal distribution, in particular the 95% confidence interval as an example of such an interval where z ≈ 1.96 (the term standard error may be used but is not required)."

1.3 The key knowledge and key skills that bind the examiners

From Outcome 1 [SD], the four key-knowledge points that belong here:

• the concepts of a random variable (discrete and continuous), Bernoulli trials and probability distributions, the parameters used to define a distribution and properties of probability distributions and their graphs
• the conditions under which a Bernoulli trial or a probability distribution may be selected to suitably model various situations
• the definition of sample proportion as a random variable and key features of the distribution of sample proportions
• the concept of confidence intervals for proportions, variation in confidence intervals between samples and confidence intervals for estimates

And the five key skills:

• analyse a probability mass function or probability density function and the shape of its graph in terms of the defining parameters for the probability distribution and the mean and variance of the probability distribution
• calculate and interpret the probabilities of various events associated with a given probability distribution, by hand in cases where simple arithmetic computations can be carried out
• apply probability distributions to modelling and solving related problems
• simulate repeated random sampling and interpret the results, for a variety of population proportions and a range of sample sizes, to illustrate the distribution of sample proportions and variations in confidence intervals
• calculate sample proportions and approximate confidence intervals for population proportions

The bolded clause is the single most useful sentence in the study design for Exam 1 preparation. It is VCAA telling you the shape of every technology-free probability question: the arithmetic will be simple, the fractions will be manageable, and the answer will be exact. When you meet a technology-free probability question whose numbers look ugly, you have misread it.

1.4 Boundaries that matter

• "the cumulative distribution function may be used but is not required." You will never be required to build a cdf. You may use one.
• The word "median" was deleted in 2023. The 2016–2022 design listed "mean (μ), median, variance (σ²) and standard deviation"; the 2023 dot point drops "median". Median questions have still appeared (they are reachable through "calculation of probabilities for intervals defined in terms of a random variable") but the word is gone from the design. Treat median as in scope and de-emphasised — and note that 2021 Exam 2 Section B Q4f (median spin, 37%) and 2016 Exam 2 Section B Q3hii (median battery life, 58%) are both post-2016 median questions.
• Only two named distributions. Bi(n, p) is the only named discrete family; the normal is the only named continuous family. There is no Poisson, no geometric, no named exponential. An arbitrary table or an arbitrary pdf is fair game; a named third distribution is not. (2012 Exam 2 Section A Q20 sails close: it hands you Pr(X = k) = (1 − p)ᵏp — a geometric pmf in disguise — and asks for Pr(X > 1). 19% of the state got it.)
• Sample means are not in the course. Inference in Methods is sample proportions only. Sample means belong to Specialist.
• Simulation is in the study design and has never been examined. Grepping every paper and every report in the corpus — 97 text files, 2006 to 2026 NHT, plus 33 extracted .docx reports, guides and the specification — for simulat returns zero hits. Simulation is assessed in SACs, not in the written examinations. The nearest the papers come is 2026 Exam 2 NHT Section A Q2, which prints a nested for loop that "generates the sample space" for two dice and asks for the ninth printed output — an algorithm question wearing a probability hat.

1.5 When inference entered, and the 2020 hole

Sample proportions and confidence intervals are a 2016 addition. Searching the body of every paper (formula sheet excluded) for "confidence interval" and "sample proportion":

Two things fall out of that table.

2020 was reduced. The 2020 report [RPT] opens: "In 2020 the Victorian Curriculum and Assessment Authority produced an examination based on the VCE Mathematics Adjusted Study Design for 2020 only." Neither 2020 paper contains the words "confidence interval", "sample proportion" or the symbol P̂ anywhere in the question body. If you are using 2020 as timed practice, you are practising a paper with an entire strand missing.

Nothing before 2016 has any inference in it at all. Every statistics question in a 2006–2015 paper is discrete random variables, continuous random variables, or normal distributions. The 2006–2015 formula sheet has no sample-proportion block. Conversely, those papers are full of transition matrices (2006 Exam 2 Section B Q2b, 2008 Exam 2 Section B Q1biii, 2009 Exam 2 Section B Q3g, 2010 Exam 2 Section B Q2dii, 2012 Exam 2 Section B Q3bii, 2014 Exam 2 Section B Q4g), which are not in the current course. Skip those parts.

The binomial formula was free only from 2023. [FS] gained Pr(X = x) = ⁿCₓ pˣ(1−p)ⁿ⁻ˣ, the binomial coefficient ⁿCₓ = n!/(x!(n−x)!), μ = np and σ² = np(1−p) in the 2023 revision. From 2006 to 2022 students memorised them. The sample-proportion block — P̂ = X/n, E(P̂) = p, sd(P̂) = √(p(1−p)/n), and the confidence-interval formula — has been on the sheet since 2016.

1.6 What the exams actually weigh

[QJSON], November series only, 2006–2025:

Statistics is almost exactly a quarter of the subject, year in and year out. Marks per paper:

* incomplete extraction; see §0.

Exam 2 always carries 16–25 statistics marks. In every year of the archive, Section B contains one long statistics question — Question 3 or Question 4 — worth 11 to 19 marks, and Section A carries 4 to 6 statistics multiple-choice items. Exam 1 has swung between 5 and 13 marks; the recent pattern is two or three short questions. In 2025 it was three (Q4 discrete, 4 marks; Q6 binomial, 3 marks; Q8 pdf, 5 marks).

1.7 How hard it is, relative to everything else

[QJSON], November only, parts with a published pct:

Compared to the other three areas of study:

Statistics is not the hardest area on average — Algebra is. What statistics is, is the area with the deepest tail. Nine of the twenty lowest-scoring parts in the whole archive are statistics, and the last two parts of the Exam 2 statistics question are reliably among the three hardest items on the paper: 2021 Exam 2 Section B Q4h 2%, 2020 Exam 2 Section B Q3f 3%, 2018 Exam 2 Section B Q4g 5%, 2023 Exam 2 Section B Q4j 6%, 2024 Exam 2 Section B Q4eii 8%.

That is where a 45+ is won. A student who reliably banks the opening 8–10 marks of the statistics question is at about the 60th percentile. A student who also takes the last 2–4 marks is in the top 3%.

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2. The complete catalogue of question types

Thirty-six types. For each: the name, the literal VCAA wording template quoted from a real paper, what it is really testing, the standard method, at least three archive instances with ref and pct, the typical mark value, and the traps the reports name.

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Type 1 — Law of total probability from a two-stage tree

VCAA template (2019 Exam 1 Q3a, verbatim): "Jo has three coins in her pocket; two are unbiased and one is biased. When the biased coin is tossed, the probability of tossing a head is 1/3. Jo randomly selects a coin from her pocket and tosses it. a. Find the probability that she tosses a head."

Really testing: whether you can decompose an event across a partition, and whether you can multiply two fractions.

Method: partition on the first stage. Pr(H) = Pr(U)Pr(H|U) + Pr(B)Pr(H|B). Draw the tree; label every branch; multiply along, add across.

Traps. 2007 [RPT]: "Common incorrect responses included incorrect values on the branches of the tree or just taking one branch to obtain 0.8 × 0.4 = 0.32. Multiplication of decimals was not well done by some students." 2016 [RPT]: "Many students stated probabilities greater than 1." 2019 [RPT]: "As this question was worth two marks appropriate working was required to be shown. This could include computations or a probability tree diagram with relevant branches clearly identified."

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Type 2 — Bayes-style conditional probability from a tree

VCAA template (2018 Exam 1 Q6b, verbatim): "It is not known from which box the stone has been drawn. Given that the stone that is drawn is black, what is the probability that it was drawn from Box 1?"

Really testing: that you can reverse the direction of a tree. The tree runs box → colour; the question runs colour → box.

Method: Pr(Box 1 | Black) = Pr(Box 1 ∩ Black) / Pr(Black). The denominator is the previous part. The numerator is one branch of it.

Traps. 2007 [RPT] is blunt: "This was a poorly answered question, with many students obtaining no marks. Again, students struggled with conditional probability. Despite copying the formula from the formula sheet, students did not apply it to their values." 2018 [RPT]: "Some students incorrectly worked Pr(Black|Box 1), resulting in a probability greater than 1, which is not feasible." 2014 [RPT]: "Many students were able to identify the conditional probability and use their answer to part bi. in the denominator; however, used an incorrect numerator." Distribution for 2007 Exam 1 Q11b: 62% scored zero, 19% scored one, 19% scored two.

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Type 3 — Conditional and set probability from a Venn diagram or Karnaugh map

VCAA template (2015 Exam 1 Q8, verbatim): "For events A and B, Pr(A|B) = 3/4 and Pr(B) = 1/3. a. Calculate Pr(A ∩ B). b. Calculate Pr(A′ ∩ B). c. If events A and B are independent, calculate Pr(A ∪ B)." Also 2020 Exam 1 Q2a: "State the probability that at any given six-month service model X will require an air filter change without an oil change."

Really testing: the addition rule, complements, and the difference between "and not" and "given not".

Method: build a 2×2 Karnaugh table with row and column totals. Every one-mark answer is then a single cell or a single margin.

Traps. 2007 [RPT]: "Students who used a Venn diagram or Karnaugh map usually obtained the correct answer. Many students either just added or multiplied Pr(A) and Pr(B)." 2020 [RPT]: "The most common incorrect answer was … obtained by incorrectly assuming that the events F (air filter change) and (without an oil change) were independent." 2017 [RPT]: "Many students assumed that events A and B were independent, hence incorrectly used Pr(A ∩ B) = Pr(A) × Pr(B)."

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Type 4 — Independence: test it, or use it

VCAA template (2018 Exam 2 Section B Q4bii, verbatim): "Are the events H and S independent? Justify your answer." Multiple-choice form (2009 Exam 2 Section A Q17, verbatim): "The sample space when a fair twelve-sided die is rolled is {1, 2, …, 12}. Each outcome is equally likely. For which one of the following pairs of events are the events independent?"

Really testing: that Pr(A ∩ B) = Pr(A)Pr(B) is the definition, and that independence has nothing to do with mutual exclusivity.

Method: compute Pr(A)Pr(B) and compare with Pr(A ∩ B); or compare Pr(A|B) with Pr(A). State both numbers and the conclusion. A one-word answer scores nothing.

Traps. 2018 [RPT]: "A mathematical explanation was required. Some students confused mutually exclusive events with independent events." 2007 [RPT]: "A very popular incorrect response was 4/15, obtained by confusing independent events with mutually exclusive events." 2011 [RPT] on Exam 1 Q8b: "The most common incorrect response was due to students interpreting 'mutually exclusive' as 'independent'."

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Type 5 — Conditional probability in terms of a parameter

VCAA template (2015 Exam 1 Q9, verbatim): "a. Find, in terms of p, the probability that … b.i. Find, in terms of p, Pr(B|W). b.ii. If Pr(B|W) = 0.3, find p." Also 2021 Exam 1 Q6b: "Let g be the number of glazed doughnuts in Box A. Find the probability, in terms of g, that the doughnut comes from Box B given that it is glazed."

Really testing: algebraic fractions under exam pressure. The probability is easy; (1−p)/5 ÷ (2p+1)/5 is where the marks go.

Method: build the tree with the parameter on the branches, form the quotient, cancel the common factor, then solve.

Traps. 2015 [RPT]: "While most students recognised that this question involved conditional probability, many could not apply it within the context of the specific question. Algebraic fractions were not handled well." And on the follow-up: "Many students missed the specific connection of this part with the previous part." 2021 [RPT]: "Many students did not use g as stated in the question; few could find Pr(glazed). Those who drew a probability diagram, either a tree diagram or Karnaugh table, tended to have better success."

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Type 6 — "At least one" via the complement

VCAA template (2016 Exam 1 Q4b, verbatim): "What is the probability that at least one tagged sheep is selected on a given day?" Also 2025 Exam 2 Section B Q3bii: "If k = 47, find the probability that the driver will be late on at least one day in a five-day working week. Give your answer correct to four decimal places."

Really testing: the reflex Pr(≥ 1) = 1 − Pr(0).

Method: 1 − (1−p)ⁿ. Never enumerate.

Traps. 2016 [RPT]: "Students identified that the answer to this part of the question was simply the complement of their previous answer. However, some students wasted time in finding the sum of four probabilities." 2014 [RPT]: "Many students attempted to use matrices but did not recognise the basic nature of the problem." 2020 [RPT] on Q3ei — expressing 1 − 0.85ⁿ in terms of n — records a 24% full-mark rate for one mark.

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Type 7 — Sequential trials with changing probability (not binomial)

VCAA template (2017 Exam 1 Q5, verbatim): "For Jac to log on to a computer successfully, Jac must type the correct password… The probability of success on any attempt is 2/5… A maximum of three attempts can be made. c. Calculate the probability that Jac logs on to the computer successfully on the second or on the third attempt." Also 2008 Exam 1 Q8: two of the next three Fridays, with probabilities that depend on the previous day.

Really testing: whether you notice that the trials are not identical or not independent, so the binomial does not apply.

Method: list the qualifying sequences; multiply along each; add. Three sequences is the usual count.

Traps. 2017 [RPT]: "Common errors included use of conditional probability, use of binomial theorem or not realising that once Jac logged in, there was no need to keep attempting." 2008 [RPT]: "Some students were unable to assign correct probabilities… A small number of students attempted to use binomial probability." 2012 [RPT]: "Other students did not read or interpret the statement 'Katrina answers Question 1 incorrectly'."

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Type 8 — Discrete distribution table: find the unknown parameter

VCAA template (2025 Exam 1 Q4a, verbatim): "The probability distribution for the discrete random variable X is given in the table below, where k is a positive real number. a. Show that k = 10 or k = 15." Also 2010 Exam 1 Q8 and 2013 Exam 1 Q7a.

Really testing: that Σ Pr(X = x) = 1, and that you can solve the quadratic this produces without a calculator.

Method: sum the table, set equal to 1, clear denominators, form ax² + bx + c = 0, factorise or use the quadratic formula, reject solutions outside [0, 1] or outside the stated constraints.

Traps. 2025 [RPT]: "It was not sufficient to verify the solutions of k = 10 or k = 15 by substitution… Some students incorrectly used the formula for E(X) instead of using the fact that the probabilities must sum to 1." 2010 [RPT]: "Although most students understood that the probabilities should add to 1, a number of students incorrectly worked with E(X) = 1 instead… Use of the quadratic formula, with which students should be familiar, was poor." 2022 [RPT]: "There were, however, a significant number who did not recognise that the probabilities had to sum to one."

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Type 9 — Mean and variance of a discrete random variable

VCAA template (2025 Exam 1 Q4bii / Q6a, verbatim): "Find E(X)." / "Find var(X)." Section B form (2022 Exam 2 Section B Q3aiv, verbatim): "Find the expected value and the standard deviation for X."

Really testing: E(X) = Σx·p(x) and var(X) = E(X²) − μ². On the formula sheet — so the only thing being tested is arithmetic and whether you square the mean.

Method: two-row table: x·p(x) and x²·p(x). Sum both. σ² = Σx²p(x) − (Σxp(x))². Then take the square root if the question said standard deviation.

Traps. 2009 [RPT]: "The majority of students started well and wrote down the correct rule for variance. However, many then neglected to square the mean to obtain the correct final answer." 2022 [RPT]: "Many students did not give … the standard deviation as well as the expected value in Question 3aiv." 2025 Exam 1 [RPT]: "Common mistakes included finding the standard deviation instead of the variance." 2013 [RPT]: "Many students had difficulty adding decimals and fractions to give an answer."

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Type 10 — Mode, and probabilities relative to the mean

VCAA template (2008 Exam 1 Q7a, verbatim): "State the mode." Also 2013 Exam 1 Q7bii: "Find Pr(X ≥ E(X))."

Really testing: whether you know the mode is the value of x, not the probability; and whether you will actually evaluate E(X) before comparing.

Traps. 2008 [RPT]: "Some students found the instruction regarding the mode difficult. Some attempted to find the mean or median while others stated probability values or wrote Pr(X = 3) = 0.4 without specifying which part was the mode." 2013 [RPT] on Q7bii: "Many students gave 0.5 as the answer." The 2025 Section A item asks which of four printed probability mass functions have a given mean; the report's comment is "Probability mass functions II and IV both have a mean equal to 3."

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Type 11 — Conditional probability inside a discrete distribution

VCAA template (2009 Exam 1 Q7a, verbatim): "Find Pr(X > 1 | X ≤ 3)." Also 2024 Exam 2 NHT Section A Q8: "Given that a visitor has previously visited the zoo, what is the probability that they have previously visited more than once?"

Really testing: reduced sample space. The denominator is a sum of table entries, not 1.

Method: numerator = Pr({X > 1} ∩ {X ≤ 3}) = entries for x = 2, 3. Denominator = entries for x = 0, 1, 2, 3, or 1 − Pr(X = 4).

Traps. 2009 [RPT]: "The intersection was correctly identified by most students. What to do with the 'condition' eluded many." 2012 Exam 1 Q4c is the lowest-scoring discrete-probability part in the archive: 3 marks, 3% full marks, distribution {0: 41%, 1: 13%, 2: 43%, 3: 3%}. [RPT]: "Most students were not aware that the condition of 'receives telephone calls on both Monday and Tuesday' would affect the result. A significant number of students incorrectly thought that two calls on Monday and two calls on Tuesday was different from two calls on Tuesday and two calls on Monday."

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Type 12 — Abstract / parameterised discrete distribution (multiple choice)

VCAA template (2016 Exam 2 Section A Q19, verbatim): "Consider the discrete probability distribution with random variable X shown in the table below. [x: −1, 0, b, 2b, 4 / Pr: a, b, b, 2b, 0.2] The smallest and largest possible values of E(X) are respectively…" Also 2023 Exam 2 Section A Q12: "The probability mass function for the discrete random variable X is shown below. [Pr: k², 3k, k, −k²−4k+1] The maximum possible value for the mean of X is…"

Really testing: that a probability table carries two constraints — sum to 1, and every entry in [0, 1] — and that E(X) is then a function of one parameter to be optimised over a feasible interval.

Method: impose Σp = 1 to eliminate one parameter; impose 0 ≤ p ≤ 1 on every cell to get the feasible interval; write E(X) as a function of the surviving parameter; optimise on that interval (endpoints included).

Traps. These are the hardest multiple-choice items in the subject. 2020 Exam 2 Section A Q19 gives the binomial pmf for the number of sixes in 20 rolls and asks for the pmf of the number of non-sixes — the answer is p(20 − w), and only 15% of the state found it. The failure mode is always the same: students optimise E(X) without checking that the optimising parameter keeps every probability non-negative.

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Type 13 — Straight binomial probability

VCAA template (2018 Exam 2 Section B Q4ci, verbatim): "Find the probability that a random sample of 16 Mathsland adults will contain exactly one person with a slow heart rate. Give your answer correct to three decimal places." Exam 1 form (2020 Exam 1 Q5a, verbatim): "In a randomly selected group of four people, what is the probability that three or more people have the SPGE1 gene?"

Really testing: recognising Bernoulli trials (fixed n, constant p, independent, count of successes) and stating the distribution.

Method (Exam 2): write X ~ Bi(n, p) with the numbers, then binomCdf/binomPdf. Writing the distribution with its parameters is the method mark. Method (Exam 1): ⁿCₓ pˣ(1−p)ⁿ⁻ˣ, summed over the relevant x.

Traps. The single most repeated instruction in the whole archive is this, from 2023 [RPT]: "In Question 4d. students were expected to identify and write down the n and p values for the binomial distribution, not just the answer." 2012 [RPT]: "Some students gave only the answer. For questions worth more than one mark, working must be shown. Identifying the distribution with the correct parameters is sufficient working." 2015 [RPT]: "Some students wrote 3% as 0.3. Others had the incorrect value for n." 2017 [RPT]: "Some used Pr(X > 3)." — the > versus ≥ error is chronic.

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Type 14 — Binomial "at least" and the strict/non-strict boundary

VCAA template (2017 Exam 2 Section B Q3ei, verbatim): "Find the probability that Jennifer spends more than 50 minutes on her homework on more than three of seven randomly chosen days, correct to four decimal places."

Really testing: whether you translate English into the right inequality. "More than three of seven" is Pr(X ≥ 4); "at least two" is Pr(X ≥ 2); "fewer than half of eight" is Pr(X ≤ 3).

Traps. 2018 [RPT] general comments state the rule in general terms: "Be familiar with the order properties of the real number system, for example… in Question 4f. Pr(X < 15) did not mean Pr(X ≤ 14)" — careless for a continuous variable, decisive for a discrete one. 2020 [RPT]: "A common incorrect answer was…" (symbol lost in extraction) — the reported error is the off-by-one.

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Type 15 — Binomial conditional probability

VCAA template (2016 Exam 2 Section B Q3b, verbatim): "A teacher observes that at least one of the returned laptops is not correctly plugged into the trolley. Given this, find the probability that fewer than five laptops are not correctly plugged in. Give your answer correct to four decimal places." Also 2017 Exam 2 Section B Q3eii, verbatim: "Find the probability that Jennifer spends more than 50 minutes on her homework on at least two of seven randomly chosen days, given that she spends more than 50 minutes on her homework on at least one of those days."

Really testing: that {X < 5} ∩ {X ≥ 1} = {1 ≤ X ≤ 4}, which is not {X < 5}.

Method: Pr(1 ≤ X ≤ 4) / Pr(X ≥ 1). Compute both with the cdf. Do not round the numerator before dividing.

Traps. 2016 [RPT]: "Many students recognised that this was a conditional probability question but had the incorrect numerator or denominator… Others rounded too soon and gave 0.9312 as the answer." 2017 [RPT]: "Some wrote Pr(X ≥ 2)/Pr(X ≥ 1)… Others rounded incorrectly, giving 0.7625 as the answer." 2022 [RPT]: "Many students had the correct denominator but evaluated [the wrong set] in the numerator."

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Type 16 — Find the smallest n (binomial inequality)

VCAA template (2019 Exam 2 Section B Q4fii, verbatim): "The probability that n or more butterflies, in a random sample of 36 Lorenz birdwing butterflies from Town A, are very large is less than 1%. Find the smallest value of n, where n is an integer." Also 2015 Exam 2 Section B Q3dii, 2014 Exam 2 Section B Q4e: "…find the minimum number of tomato plants…"

Really testing: solving an inequality over the integers, and then reporting an integer.

Method (calculator): define f(n) = Pr(...) and either solve f(n) = target for real n and round in the correct direction, or tabulate. Both are accepted. State the integer.

Traps. 2015 [RPT]: "Some students rounded their answer to 22. Others did not state the minimum value, leaving their answer as n ≥ 22.7566. Some students used the trial and error methods and this was acceptable." 2019 [RPT]: "A common incorrect answer was n = 6… Trial and error is an acceptable method." 2020 [RPT]: "Some students left their answer as 18.43 or rounded down to 18." 2025 [RPT]: "An integer value was required. Many students tried to solve [the equation directly]. Some students correctly used trial and error. Others just gave the answer, without showing appropriate working as required."

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Type 17 — Express a binomial probability as a polynomial in p, then optimise it

VCAA template (2017 Exam 2 Section B Q3f–g, verbatim): "Let p be the probability that on any given day Jennifer spends more than d minutes on her homework. Let q be the probability that on two or three days out of seven randomly chosen days she spends more than d minutes on her homework. f. Express q as a polynomial in terms of p. g.i. Find the maximum value of q, correct to four decimal places, and the value of p for which this maximum occurs." 2025 variant (Exam 2 Section B Q1f, verbatim): "Let X ~ Bi(4, p) be a binomial random variable. Show that Pr(X ≥ 3) = g(p) for all p ∈ [0, 1]" — where g(x) = 4x³ − 3x⁴ was defined at the top of a calculus question.

Really testing: that a binomial probability with n fixed and p free is a polynomial, and that calculus applies to it.

Method: q(p) = ⁿC₂p²(1−p)⁵ + ⁿC₃p³(1−p)⁴. Expand or leave factored. Differentiate, solve q′(p) = 0 on (0, 1), discard roots outside. Report both p and q.

Traps. 2017 [RPT]: "Some students knew to solve q′(p) = 0 if they had an equation in Question 3f. Others found only p. Some gave exact values for their answers" — where four decimal places were demanded. 2007 [RPT]: "Some students gave p = 2 + √2, which is greater than one." 2012 [RPT]: "Suitable working was required as this was a 'show that' question." 2025 [RPT]: "This was a 'show that' question and appropriate working needed to be shown. Some students worked out Pr(X = 3) instead of Pr(X ≥ 3)."

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Type 18 — Binomial parameters recovered from mean/variance

VCAA template (2017 Exam 2 Section A Q18, verbatim): "Let X be a discrete random variable with binomial distribution X ~ Bi(n, p). The mean and the standard deviation of this distribution are equal. Given that 0 < p < 1, the smallest number of trials, n, such that p ≤ 0.01 is…" Also 2020 Exam 2 Section A Q8, verbatim: "Items are packed in boxes of 25 and the mean number of defective items per box is 1.4. Assuming that the probability of an item being defective is binomially distributed, the probability that a box contains more than three defective items…"

Really testing: μ = np, σ² = np(1−p) used backwards.

Traps. 2012 [RPT]: "Some students used the standard deviation, which was not necessary. Others did not know the formula for the variance." Note the 2026 NHT item: it asks for the qualitative effect of increasing p on E(X) and var(X) — with p > 0.5, the mean increases and the variance decreases. That is a "effect of variation in the value(s) of defining parameters" question straight out of the dot point, and it is the newest style in the archive.

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Type 19 — Binomial by hand, in a specified algebraic form

VCAA template (2025 Exam 1 Q6b, verbatim): "Consider the binomial random variable X ~ Bi(6, ¼). b. Determine Pr(X ≥ 5). Give your answer in the form a/2ᵇ, where a, b ∈ Z." Also 2019 Exam 1 Q6b: "Express your answer in the form a(b)ⁿ, where a and b are positive rational numbers and n is a positive integer." Also 2016 Exam 1 Q4c: "Express your answer in the form aᶜ/b, where a, b and c are positive integers."

Really testing: index laws and prime factorisation under time pressure, with no calculator.

Method: write the two (or three) terms, factor out the common power, simplify to the demanded form. Recognise 4096 = 2¹², 1024 = 2¹⁰, 243 = 3⁵.

Traps. 2025 [RPT]: "The answer was required to be stated in a particular form, but some students were not able to reduce 4096 down to the prime factorisation of 2¹². Instead, many students gave the answer as [a decimal]. Many students calculated only one term… instead of evaluating [both]." 2019 [RPT]: "Most students recognised this as a binomial distribution; however, few managed to correctly find the two component expressions. Even fewer successfully managed to manipulate these expressions to the format specified by the question." Distribution for 2019 Exam 1 Q6b: {0: 59%, 1: 29%, 2: 12%}. 2024 [RPT]: "…some responses either included an extra zero or missed a zero."

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Type 20 — Find the constant k in a probability density function

VCAA template (2023 Exam 1 Q8a, verbatim): "Suppose that the queuing time, T (in minutes), at a customer service desk has a probability density function given by f(t) = kt(16 − t²) for 0 ≤ t ≤ 4, 0 elsewhere, for some k ∈ R. a. Show that k = 1/64." Also 2021 Exam 1 Q7a: "Show that k = 2." Also 2025 Exam 2 Section A Q14: a two-piece hybrid k sin(x) / k cos(x) pdf, "The value of k is".

Really testing: ∫f = 1 over the support, plus by-hand antidifferentiation.

Method: ∫ₐᵇ f(x)dx = 1. Antidifferentiate, substitute, solve for k. For a hybrid, two integrals, summed.

Traps. 2023 [RPT]: "This was a 'show that' question, so students were expected to be explicit and clear with their workings… Common errors involved omitting the dt in the integral statement or writing dx instead. Students are reminded to be consistent in their use of variables." 2010 [RPT]: "Some students differentiated instead of anti-differentiated and others had difficulty subtracting the two fractions with different denominators." 2006 [RPT] is the origin of a long-running VCAA grudge: "It was disappointing when students failed to include dx at the end of their integral expressions — this will be penalised in the future."

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Type 21 — Probability from a pdf by definite integration

VCAA template (2017 Exam 2 Section B Q3b, verbatim): "Find Pr(25 ≤ T ≤ 55)." Exam 1 form (2006 Exam 1 Q6a): a x/12 density on [1, 5], "Find Pr(X < 3)."

Really testing: choosing the right terminals — particularly for a hybrid pdf, where the interval straddles the join.

Method: Pr(a ≤ X ≤ b) = ∫ₐᵇ f(x)dx. For a hybrid, split at the join. On CAS, define the hybrid function once at the start of the question and use its name thereafter.

Traps. 2015 [RPT]: "Some students omitted the dx. Some had incorrect terminals such as ∫₆⁸, ∫₇.₀₀₀₁⁸ or ∫₆.₉₉₉₉⁸." 2017 [RPT]: "Some students had the incorrect terminals. 44 instead of 45 was occasionally given… Others used 20 as the lower limit instead of 25." 2017 [RPT] advice: "Define functions on the technology at the start of each question in Section B. This saves time, especially when dealing with probability questions that involve hybrid functions." 2021 [RPT]: "Students who used f(x) when writing out the definite integral were more successful with the method mark."

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Type 22 — Conditional probability with a continuous random variable

VCAA template (2023 Exam 1 Q8c, verbatim): "What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute?" Also 2019 Exam 2 Section B Q4c, verbatim: "What is the probability that a Lorenz birdwing butterfly lives for at least four weeks, given that it lives for at least two weeks, correct to four decimal places?" Also 2017 Exam 2 Section B Q3c: "Find Pr(T ≤ 25 | T ≤ 55)."

Really testing: translating an English time-condition into a set, and then intersecting correctly. "Already queued for one minute" means {T > 1}, not {T = 1}.

Method: Pr(T > 2 | T > 1) = ∫₂⁴f / ∫₁⁴f. When one event contains the other, the intersection collapses to the smaller set.

Traps. 2023 [RPT]: "Common errors included writing the conditional probability as [Pr(T > 2 | T = 1)], where students had incorrectly interpreted the mathematical meaning of 'already queued for one minute'… There were also errors where students incorrectly identified the terminals of integration." Distribution: {0: 63%, 1: 13%, 2: 13%, 3: 11%}. 2008 [RPT]: "Few students used symmetry of the probability density function to obtain the denominator… Many recognised the need for conditional probability but thought the required intersection was between ¼ and ½ rather than between 0 and ¼." 2014 [RPT]: "The most common error was assuming that the value of m (obtained in part a.) was equivalent to Pr(X ≤ m)."

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Type 23 — Median, quantile, or an unknown terminal of a continuous distribution

VCAA template (2017 Exam 2 Section B Q3d, verbatim): "Find a such that Pr(T ≥ a) = 0.7, correct to four decimal places." Also 2025 Exam 1 Q8a, verbatim: "Find k such that Pr(X > k) = 9/16." Also 2021 Exam 2 Section B Q4f: "Find the median spin, in revolutions per second, correct to one decimal place." Also Exam 1 sample Q7 [SAMP]: "Find the value of k such that 90% of telemarketing calls last less than k minutes. Express your answer in the form (a/b)logₑ(c)…"

Really testing: setting up ∫ = a number with the unknown in a terminal, and then solving — which by hand means a quadratic or an indicial equation, and rejecting the root outside the support.

Method (by hand): antidifferentiate, substitute the variable terminal, set equal to the target, solve, reject out-of-domain roots. Method (CAS): solve(∫ₐᵏ f(x)dx = c, k). Add the domain constraint or you will get spurious roots.

Traps. 2006 [RPT]: "Including −10 (outside the domain) was a common error." 2012 [RPT]: "The quadratic formula and null factor law or factorisation can be used only if the quadratic is equated to zero. A number of students established a quadratic that was not equal to zero and used guesswork." 2014 [RPT]: "Some students confused median with the mean." 2016 [RPT] records a beautiful CAS failure: "Some students wrote down the correct formula but did not delete the x on their technology from the previous computation for E(X), solving ∫₀ᵐ x·f(x)dx = ½, getting m = 75.58. Students should check their answers to see if they make sense as 75.58 is very different from 170.01." 2017 [RPT]: "∫₂₀ᵃf(t)dt = 0.7, a = 50.6351 was a common incorrect answer" — the tail was the wrong tail. 2025 [RPT]: "The quadratic expression was readily factorised by inspection, but a large proportion of students used the quadratic formula."

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Type 24 — Mean of a continuous random variable

VCAA template (2019 Exam 2 Section B Q4a, verbatim): "Find the mean life span of the Lorenz birdwing butterfly." Also 2023 Exam 2 Section B Q4i: "Find the exact mean serving speed for grade A balls, in metres per second." Also 2025 Exam 2 Section B Q3ai: "Find the mean time taken, in minutes, for the driver to travel to work each day."

Really testing: E(X) = ∫ x f(x) dx — the x inside the integral is the whole test.

Traps. 2015 [RPT]: "Some students worked out the median, solving ∫f = 0.5, instead of the mean. Others evaluated ∫f(x)dx, leaving out the x." 2013 [RPT]: "Some students did not include x in the formula." 2018 [RPT]: "Some students found the median or the mode. Others found the area under the curve." 2013 Exam 1 Q8 is the technology-free version and it is brutal (21%): the pdf is (π/4)cos(πx/4)-flavoured and the question hands you d/dx[x sin(πx/4)] so that you can do integration by parts without naming it. [RPT]: "the key was to recognise that the given relation … provided the required anti-derivative."

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Type 25 — Variance and standard deviation of a continuous random variable

VCAA template (2025 Exam 2 Section B Q3aii, verbatim): "Find the standard deviation of the time taken, in minutes, for the driver to travel to work each day." Also 2021 Exam 2 Section B Q4g: "Find the standard deviation of the spin, in revolutions per second, correct to one decimal place."

Really testing: σ² = ∫x²f(x)dx − μ² (or ∫(x−μ)²f(x)dx, both on [FS]), followed by the square root.

Traps. 2021 [RPT]: "Some students worked out the variance instead of the standard deviation." 2025 [RPT]: "…while they worked out the variance, they did not proceed to compute the standard deviation… Some students gave the approximate answer 5.34 and were not awarded full marks." 2014 Exam 2 Section A Q16 is the one-line multiple-choice version: given mean 2 and variance 5, find ∫x²p(x)dx. The answer is 5 + 2² = 9; 46% got it.

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Type 26 — Sketch a probability density function

VCAA template (2017 Exam 2 Section B Q3a, verbatim): "Sketch the graph of f on the axes provided below." (3 marks, a two-piece linear "tent" pdf on [20, 70].) Also 2026 Exam 2 NHT Section B Q3ai: "Sketch the graph of y = f(x) for 0 ≤ x ≤ 50 on the axes below."

Really testing: that f = 0 outside the support is part of the graph, and that a linear piece needs a ruler.

Traps. 2017 [RPT]: "Many students did not draw their graphs along the t-axis, ignoring f(t) = 0. Some had an open circle at (45, 0.04). Others had an open circle over a closed circle. Many students did not use rulers to draw the line segments. Some graphs looked like parabolas." 2007 [RPT]: "A common error was that (20, 1) were given as the coordinates of the maximum, not (20, 0.1)." This type has the worst full-mark rate of any "easy" statistics task in the archive.

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Type 27 — Transformation of a probability density function

VCAA template (2023 Exam 2 Section B Q4j, verbatim): "A transformation maps the graph of f to the graph of g, where g(w) = a·f(w/b). If the mean serving speed for a grade B ball is 22 + 8 metres per second, find the values of a and b." Also 2021 Exam 2 Section B Q4h, verbatim: "The teacher adjusts the spin setting so that the median spin becomes 30 revolutions per second… g(x) = a·f(x/b). Find the values of a and b for which the new median spin is 30 revolutions per second." Also 2022 Exam 2 Section B Q3biii: "g(d) = f(rd + s)… Find the values of r and s." Also 2025 Exam 1 Q8b, verbatim: "The function h(x) is a transformation of f(x) such that h(x) = m·f(x) + n where m and n are real numbers. Find ∫₀⁴ 3h(x) dx in terms of m and n."

Really testing: that a transformed pdf must still integrate to 1 — which forces a = 1/b in the a·f(x/b) family — and that dilating the domain dilates the mean and the median by the same factor.

Method: two equations. (i) Area: ∫ a·f(x/b)dx = ab∫f = 1, so ab = 1. (ii) The stated statistic: new mean = b × old mean, new median = b × old median. Solve. Alternatively set up the two definite integrals with transformed terminals and solve simultaneously — and remember to multiply the terminals by b.

Traps. This is the hardest single idea in the area of study; all five instances are separators, three of them below 7%. 2023 [RPT]: "Others were able to recognise that ab = 1 but were unable to find their values. A common incorrect answer was a = 1 and b = 1. Many of those who attempted the second method did not multiply the terminals by b." 2021 [RPT]: "Many students were unable to set up the correct equations. The terminals were often incorrect." Distribution for 2021 Q4h: {0: 86%, 1: 12%, 2: 2%}. 2025 [RPT] on the Exam 1 version: "Most students correctly rewrote [the integral]. However, many students incorrectly proceeded to factor out m from the entire integral without noticing that this was not algebraically valid. The students who recognised that the total probability is equal to 1 and applied it, were generally successful."

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Type 28 — Expected count: "how many of N are expected to…"

VCAA template (2019 Exam 2 Section B Q4b, verbatim): "In a sample of 80 Lorenz birdwing butterflies, how many butterflies are expected to live longer than two weeks, correct to the nearest integer?" Also 2014 Exam 2 Section B Q4b: 2000 basil plants; 2013 Exam 2 Section B Q2cii: 200 members.

Really testing: that expected count = N × Pr(event), rounded at the end.

Traps. 2019 [RPT]: "Some students found the probability but did not multiply by 80… Other students rounded to 74." 2014 [RPT]: "Some students had incorrect working… Some students used Pr(X > 8.9) or Pr(X > 8). Some rounded incorrectly. Some used technology syntax in their working." 2013 [RPT]: "Some students rounded their answers to 53."

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Type 29 — Straight normal probability

VCAA template (2023 Exam 2 Section B Q4a, verbatim): "The diameter of the tennis balls is a normally distributed random variable D, which has a mean of 6.7 cm and a standard deviation of 0.1 cm. a. Find Pr(D > 6.8), correct to four decimal places."

Really testing: typing the right numbers into the right calculator fields, and knowing that σ is entered, not σ².

Traps. 2016 [RPT] names the unit trap that costs a whole question: "Some students thought 3 hours and 10 minutes was 3.1 hours and 6 minutes was 0.6 hours. Others had the standard deviation as 10 minutes." 2023 [RPT]: "0.1586 was sometimes seen" — rounding. 2021 [RPT]: "A common error was 0.228" — the complement.

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Type 30 — Inverse normal: find the value given the probability

VCAA template (2023 Exam 2 Section B Q4b, verbatim): "Find the minimum diameter of a tennis ball that is larger than 90% of all tennis balls produced. Give your answer in centimetres, correct to two decimal places." Also 2019 Exam 2 Section B Q4e: "A Lorenz birdwing butterfly is considered to be very small if its wingspan is in the smallest 5%… Find the greatest possible wingspan." Also 2026 Exam 2 NHT Section B Q3bii: "In app B, the hiking trails that are classified as difficult are the longest 15%. Find the minimum distance for a hiking trail from app B to be classified as difficult."

Really testing: which tail. "Larger than 90% of all" is invNorm(0.9); "smallest 5%" is invNorm(0.05); "longest 15%" is invNorm(0.85).

Traps. 2023 [RPT]: "A common error was that students solved [the complementary equation], giving [the wrong tail] as the answer." 2014 [RPT] names the units disaster: "Many students thought 100 mm = 1 cm, giving their final answer as 1913 mm. Others had incorrect units, such as 19.1 mm." 2019 [RPT]: "A common incorrect answer was 9.9."

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Type 31 — Find σ (or μ) from a stated probability

VCAA template (2023 Exam 2 Section B Q4f, verbatim): "The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A. Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places." Also 2016 Exam 2 Section B Q3e: "…only 12% of such laptops work for more than three hours and ten minutes. Find the standard deviation…" Also 2013 Exam 2 Section A Q22: "If, from a population of 2000 newly hatched butterflies, 150 are expected to die in the first 90 days, then the value of σ is closest to…" Two-unknown form (2025 Exam 2 Section A Q12, verbatim): "For a normal random variable X, it is known that Pr(X > 200) = 0.325 and Pr(180 < X < 200) = 0.589. The mean and standard deviation of X are closest to…"

Really testing: either standardising (z = (x − μ)/σ) with an inverse-normal z, or letting CAS solve for the parameter.

Method: solve(normCdf(lower, upper, μ, s) = p, s). Or z = invNorm(p) then σ = (x − μ)/z. For two unknowns, two equations, solve({…}, {μ, σ}).

Traps. 2009 [RPT]: "Many students used 0.99 as the required probability instead of 0.995, obtaining an answer of 0.60. Some used 0.99 as their z value. Other students used 0.05 instead of 0.005." 2023 [RPT]: "Trial and error could be used but students must make sure they show some appropriate working. Drawing a diagram and showing the probabilities was acceptable." 2020 Exam 2 Section B Q3c (6%) is the two-sided version — find the values of k such that a translated normal puts 46.48% in a shifted interval; [RPT]: "k can be found by a translation of 1.5 units in the direction of the negative t-axis, use symmetry to find second value for k… Many students did not find [both values]."

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Type 32 — Standardising by hand (Exam 1 normal)

VCAA template (2018 Exam 1 Q4, verbatim): "Let X be a normally distributed random variable with a mean of 6 and a variance of 4. Let Z be a random variable with the standard normal distribution. a. Find Pr(X > 6). b. Find b such that Pr(X > 7) = Pr(Z < b)." Also 2010 Exam 1 Q5b, 2015 Exam 1 Q6a, 2026 Exam 2 NHT Section A Q3, verbatim: "Assume that the time … is a normal random variable with mean 15 and variance 4. Let Z be a standard normal random variable. The probability that it takes the student more than 20 minutes … is equal to [options in terms of Pr(Z < …)]."

Really testing: z = (x − μ)/σ with σ = √variance, plus the symmetry Pr(Z > c) = Pr(Z < −c).

Traps. The same error every year. 2018 [RPT]: "Some students did not standardise … or mistook the variance to be the standard deviation." 2010 [RPT]: "Most students realised that transformation to the standard normal distribution was needed but many substituted the variance instead of the standard deviation. The symmetry properties of the normal distribution eluded many." 2015 [RPT]: "Those students who drew a diagram of a 'normal' curve with relevant areas shaded found this helpful." 2012 [RPT] on the symmetry version: "There were many unsuccessful attempts to allocate an area to 1.5 standard deviations above or below the mean. Those who drew a diagram and realised the symmetry … were most successful."

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Type 33 — Conditional probability with the normal distribution

VCAA template (2023 Exam 2 Section B Q4e, verbatim): "Given that a tennis ball can fit through the opening at the top of the container, find the probability that it is classed as grade A. Give your answer correct to four decimal places." Also 2015 Exam 2 Section B Q3c: "Find Pr(O > 85 | O > 74)."

Really testing: intersecting two normal intervals correctly, which for a symmetric distribution often collapses to something you can write down.

Traps. 2009 [RPT]: "Many students did not recognise that the question involved conditional probability. Some students gave 0.8385 as the answer" — i.e. the numerator. 2015 [RPT]: "Some students evaluated 0.38918/0.49999 or 0.889188/0.5." 2020 [RPT]: "Others had 0.77 as the numerator and 0.5 as the denominator, creating an answer greater than 1." 2016 [RPT] on the 21% item: "Most students used the conditional probability formula but tried to use the normal distribution rather than the binomial distribution" — the condition was on a sample proportion, not on the underlying normal.

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Type 34 — Sample proportion: E(P̂), sd(P̂), and Pr(P̂ ∈ interval)

VCAA template (2019 Exam 2 Section B Q4fiii, verbatim): "For random samples of 36 Lorenz birdwing butterflies in Town A, P̂ is the random variable that represents the proportion of butterflies that are very large. Find the expected value and the standard deviation of P̂, correct to four decimal places." And Q4fiv, verbatim: "What is the probability that a sample proportion of butterflies that are very large lies within one standard deviation of 0.0527, correct to four decimal places? Do not use a normal approximation." Also 2021 Exam 2 Section B Q4d: "Use the binomial distribution to find Pr(P̂ > 0.1), correct to three decimal places." Also 2025 Exam 2 Section B Q3biii, verbatim: "For k = 47, let P̂ be the proportion of days the driver is late in any five-day working week. Find Pr(0.4 ≤ P̂ ≤ 0.6) correct to four decimal places."

Really testing: that P̂ = X/n where X ~ Bi(n, p), so every probability about P̂ is a binomial probability in disguise. E(P̂) = p; sd(P̂) = √(p(1−p)/n) — both on [FS].

Method: convert the P̂ interval to an X interval by multiplying by n, then round inwards to integers, then use the binomial cdf. Pr(0.4 ≤ P̂ ≤ 0.6) with n = 5 is Pr(2 ≤ X ≤ 3).

Traps. 2019 [RPT]: "Some students found E(X) = 36 × 0.0527 = 1.8972" — confusing E(X) with E(P̂). The same report's general comments: "in Question 4fiii., E(P̂) cannot be greater than one." On the interval version: "Many students were able to find the first interval. Some students used the normal distribution." 2018 [RPT] on Q4cii: "There was poor use of variables, for example, Pr(P̂ > 0.1) = Pr(X > 1.6) = Pr(X ≥ 2)" written without the intermediate reasoning. On Q4ciii (8%): "Many students appeared to be confused by the terminology Pr(P̂ₙ > 1/n)."

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Type 35 — Confidence intervals: build, invert, and interpret

This is really four sub-types sharing one formula. [FS]: (p̂ − z√(p̂(1−p̂)/n), p̂ + z√(p̂(1−p̂)/n)).

(a) Build one. 2016 Exam 2 Section B Q3g, verbatim: "The laptop supplier finds that, in a particular sample of 100 laptops, six of them have a battery life of less than three hours. Determine the 95% confidence interval for the supplier's estimate of the proportion of interest. Give values correct to two decimal places." Also 2022 Exam 2 Section B Q3cii: "If p̂ = 0.4, find an approximate 95% confidence interval for p, correct to three decimal places."

(b) Recover p̂. 2023 Exam 1 Q6a, verbatim: "From a sample of randomly selected households… an approximate 95% confidence interval for the proportion p of households having solar panels installed was determined to be (0.04, 0.16). a. Find the value of p̂ that was used to obtain this approximate 95% confidence interval." Method: p̂ is the midpoint.

(c) Recover n. 2025 Exam 2 Section A Q8, verbatim: "A random sample of n Victorian households is taken… The approximate 95% confidence interval calculated using this sample is (0.248, 0.552), correct to three decimal places. The number of households, n, in the sample is…" Also 2019 Exam 2 Section B Q4g, 2023 Exam 1 Q6b: "Use z = 2 to approximate the 95% confidence interval. Find the size of the sample…" Method: half-width = z√(p̂(1−p̂)/n); solve for n.

(d) Recover the confidence level. 2023 Exam 2 Section B Q4g, verbatim: "An inspector takes a random sample of 32 tennis balls… The confidence interval is (0.7382, 0.9493), correct to 4 decimal places. Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer." Method: half-width gives z; then confidence = 2Φ(z) − 1.

(e) Interpret it. 2018 Exam 2 Section B Q4dii, verbatim: "Explain why this confidence interval suggests that the proportion of adults with a slow heart rate in Statsville could be different from the proportion in Mathsland." Also 2022 Exam 2 NHT Section B: "Interpret the confidence interval you found in part g.ii. in relation to the proportion of customers who said that the bakery's doughnuts are delicious."

Traps. 2016 [RPT]: "Students were not expected to write out the formula; the relevant computation could be done directly using technology. There were some rounding errors. A common incorrect interval was (0.01, 0.12)." 2018 [RPT] on recovering p̂: "Many students tried to find the sample size rather than the proportion. n = 900 was often given." On the interpretation part (11%): "The confidence interval needed to be referred to in the answer." The model answer is: "The 95% confidence interval for Statsville, (0.102, 0.145), does not contain the Mathsland proportion, which is 0.1587." 2019 [RPT]: "Many students had the proportion as 0.0527 or 0.55 instead of 0.055. Others did not include the 1.96." 2023 [RPT]: "Many students calculated z incorrectly… Some had the correct z value but then gave the answer as 95%."

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Type 36 — Reasoning about the confidence interval itself

VCAA template (2023 Exam 1 Q6c, verbatim): "A larger sample of households is selected, with a sample size four times the original sample. The sample proportion of households having solar panels installed is found to be the same. By what factor will the increased sample size affect the width of the confidence interval?" Also 2022 Exam 2 Section B Q3ciii, verbatim: "Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips. If p̂ = 0.4, how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?" Also 2022 Exam 2 Section B Q3ci, verbatim: "Is the random variable P̂ discrete or continuous? Justify your answer." Also 2017 Exam 1 Q4, verbatim: "Find the smallest integer value of n such that the standard deviation of P̂ is less than or equal to 1/100."

Really testing: that width ∝ 1/√n. Quadruple n, halve the width. Halve the width, quadruple n.

Traps. 2023 [RPT]: "This question was not responded to well… A factor of ¼ was a common incorrect answer." Distribution: {0: 77%, 1: 23%}. 2022 [RPT] on Q3ciii: "Common incorrect answers were 0, 10, 11, 50 and 101" — the answer is 100. 2017 [RPT]: "Most students identified the correct formula; however, many were unable to correctly transpose the inequality to solve for n… Some students had poor use of notation work, in that they did not extend the square root sign to include n." 2022 [RPT] on the discrete/continuous justification: the accepted answer is "Discrete, countable" — P̂ takes only the values 0, 1/n, 2/n, …, 1.

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3. The standard wordings — VCAA's recurring sentences

These are the fixed phrases. Each has a meaning that does not change from year to year, and each is a place where marks are routinely lost.

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4. The separators — every statistics part with pct ≤ 50

204 rows, 203 distinct ref values — the 2011 OCR duplicate 2011 Exam 1 Q7b–7b is recorded twice in [QJSON], at 1% and 3%, and is listed once below. Listed by type, ref — pct% — description. Marks are in the tables of §2. [2011] marks a part from the degraded 2011 extraction; [MIS-TAG] marks a part the corpus filed under this topic that is not actually statistics; [2024 text lost] marks a part whose wording could not be recovered.

4.1 Conditional probability — discrete, tree, Venn or table (28)

• 2012 Exam 1 Q4c — 3% — conditional over two days, given calls on both days
• 2021 Exam 1 Q6b — 11% — Pr(Box B | glazed) in terms of a parameter g
• 2017 Exam 1 Q8c — 10% — largest interval for p given Pr(A ∪ B) ≤ 1/5
• 2007 Exam 1 Q11b — 19% — Pr(F | T) from a two-stage tree
• 2015 Exam 1 Q9bii — 19% — solve Pr(B|W) = 0.3 for p
• 2007 Exam 1 Q6b — 23% — Pr(A|B) for mutually exclusive events
• 2014 Exam 1 Q9bii — 27% — Pr(P|W) using the previous part as the denominator
• 2015 Exam 1 Q8c — 28% — Pr(A ∪ B) when A and B are independent
• 2015 Exam 1 Q9bi — 28% — Pr(B|W) in terms of p
• 2019 Exam 2 Section A Q11 — 30% — condition for independence given Pr(B|A) = m, Pr(B|A′) = n
• 2020 Exam 1 Q2b — 30% — find m in terms of n from a Venn relation
• 2007 Exam 1 Q6a — 31% — Pr(A′ ∩ B) from a Venn diagram
• 2022 Exam 1 Q4c — 31% — exactly two of the next three red, given the first is blue, p = 1/3
• 2016 Exam 1 Q7b — 32% — Pr(A | faulty), reduced sample space, answer in the form 1/c
• 2011 Exam 1 Q6b–6b — 33% — [MIS-TAG] [2011] values of k giving unique solutions to a linear system
• 2017 Exam 1 Q8b — 36% — Pr(A′ ∩ B′) in terms of p via a Karnaugh table
• 2022 Exam 1 Q4b — 36% — exactly two of the next three red given the first is blue, p = 1/2
• 2009 Exam 1 Q5c — 40% — conditional on a restricted two-die sample space
• 2011 Exam 1 Q8b–8b — 44% — [2011] Pr(A|B) = 0 for mutually exclusive events
• 2019 Exam 1 Q3b — 45% — Pr(unbiased | head)
• 2016 Exam 1 Q7a — 46% — total probability across two assembly lines, answer as 1/b
• 2013 Exam 2 Section A Q17 — 49% — Pr(B|A) from Pr(A|B) = p, Pr(B) = p², Pr(A) = p^(1/3)
• 2007 Exam 1 Q11a — 49% — law of total probability from a tree
• 2014 Exam 2 Section A Q14 — 45% — Pr(X < 5 | X < 8) in terms of a and b
• 2011 Exam 2 Section A Q21 — 15% — [2011] characterise independence from Pr(P ∩ Q) = Pr(P′ ∩ Q′)
• 2009 Exam 2 Section A Q17 — 31% — which pair of events on a 12-sided die is independent
• 2021 Exam 2 Section A Q20 — 39% — Pr(A′ ∪ B′) for independent events with Pr(A) = p, Pr(B) = p²
• 2018 Exam 2 Section B Q4bii — 44% — justify whether the events H and S are independent

4.2 Multi-stage and dependent-trial probability (13)

• 2020 Exam 2 Section B Q3f — 3% — minimum and maximum of y given an overall probability of 0.75
• 2018 Exam 2 Section B Q4g — 5% — (1/7)(0.05) + (6/7)x = 0.0266, solve for x
• 2008 Exam 1 Q8 — 39% — two of the next three Fridays with day-dependent probabilities
• 2012 Exam 2 Section B Q3bi — 32% — two cases, exact answer required
• 2014 Exam 1 Q9bi — 44% — sum of two products across a two-stage chain
• 2014 Exam 1 Q9a — 50% — "at least one morning walk" via the complement
• 2017 Exam 1 Q5c — 50% — success on the second or third attempt
• 2023 Exam 2 Section A Q8 — 49% — Pr(at least one green in 8 selections) in terms of n and m
• 2014 Exam 2 Section A Q22 — 37% — ratio of two "at least once" probabilities
• 2010 Exam 2 Section A Q21 — 43% — Pr(A′ ∩ B′) for mutually exclusive events
• 2020 Exam 2 Section B Q3ei — 24% — express Pr(one or more late) in terms of n
• 2019 Exam 2 Section A Q17 — 43% — two marbles the same colour, drawn without replacement
• 2020 Exam 2 Section B Q3eii — 23% — minimum n with Pr(one or more) ≥ 0.95

4.3 Legacy transition-matrix / steady-state (not in the current course) (12)

• 2010 Exam 2 Section B Q2dii — 10% — expected number of superior statues from a chain
• 2010 Exam 2 Section B Q2e — 15% — smallest n with Pr(X ≥ 2) ≥ 0.9
• 2008 Exam 2 Section B Q1biii — 17% — 8th state of a transition matrix
• 2009 Exam 2 Section B Q3g — 23% — find n from a matrix power
• 2012 Exam 2 Section B Q3bii — 30% — T²⁴ applied to an initial state
• 2010 Exam 2 Section B Q2di — 36% — "show that" p = 0.75 for a chain
• 2011 Exam 2 Section B Q2b — 37% — [2011] mean of a hybrid distribution
• 2014 Exam 2 Section B Q4g — 43% — probability the fifth pot is smooth
• 2006 Exam 2 Section B Q2b — 46% — long-run proportion of nights at the pool
• 2008 Exam 2 Section B Q1bii — 46% — three cases with dependent probabilities
• 2008 Exam 2 Section B Q1biv — 46% — steady state as a percentage
• 2010 Exam 2 Section B Q2a — 49% — probability the third statue is regular

4.4 Discrete random variables: parameter, mean, variance (12)

• 2013 Exam 1 Q7bi — 29% — E(X) mixing decimals and fractions
• 2010 Exam 1 Q8 — 32% — solve the quadratic from Σp = 1
• 2013 Exam 1 Q7bii — 32% — Pr(X ≥ E(X))
• 2013 Exam 1 Q7a — 46% — show p = 2/3 from Σp = 1
• 2008 Exam 1 Q7b — 49% — probability two independent draws give the same value
• 2009 Exam 1 Q7b — 50% — var(X) = E(X²) − μ²
• 2016 Exam 2 Section A Q19 — 15% — smallest and largest possible E(X) for a parameterised table
• 2012 Exam 2 Section A Q20 — 19% — Pr(X > 1) for Pr(X = k) = (1−p)ᵏp
• 2023 Exam 2 Section A Q12 — 29% — maximum possible mean of a pmf in k
• 2025 Exam 2 Section A Q18 — 50% — which printed pmfs have mean 3
• 2025 Exam 2 Section B Q3d — 42% — complete the distribution table for the number of red lights
• 2020 Exam 2 Section A Q19 — 15% — the pmf of the complementary count, given the pmf of the count

4.5 Binomial: computing probabilities (25)

• 2009 Exam 2 Section B Q3cii — 4% — Y ~ Bi(4, 0.1015), Pr(Y ≥ 1), with p carried from a conditional
• 2011 Exam 2 Section B Q2d — 5% — [2011] W ~ Bi(10, 23/32), Pr(W = 4)
• 2020 Exam 1 Q5b — 10% — exactly two given at least one, in the form a³/(b⁴ − c⁴)
• 2019 Exam 1 Q6b — 12% — Pr(P̂ < 1/6) for Bi(12, 1/6) in the form a(b)ⁿ
• 2020 Exam 1 Q5a — 29% — three or more of four
• 2021 Exam 1 Q6c — 29% — Pr(P̂ ≥ 0.8) for n = 5, no normal approximation
• 2007 Exam 1 Q5 — 26% — Pr(X > 2) for Bi(4, ½)
• 2025 Exam 1 Q6b — 34% — Pr(X ≥ 5) for Bi(6, ¼) in the form a/2ᵇ
• 2016 Exam 2 Section B Q3b — 37% — fewer than five given at least one
• 2007 Exam 2 Section B Q5d — 37% — Pr(X ≥ 4) for Bi(6, 7/8)
• 2024 Exam 1 Q4b — 39% — [2024 text lost] binomial expansion with powers of 0.9 and 0.1
• 2013 Exam 2 Section B Q2aii — 39% — Pr(X > 15 | X > 10)
• 2011 Exam 1 Q7a–7ii — 41% — [2011] the 3p²(1−p) term
• 2020 Exam 2 Section B Q3d — 41% — fewer than half of eight deliveries on time
• 2021 Exam 2 Section B Q4d — 40% — Pr(P̂ > 0.1) using the binomial
• 2010 Exam 2 Section A Q12 — 44% — Pr(X < 7) for Bi(15, 3/5)
• 2015 Exam 2 Section B Q3aii — 47% — Pr(Y ≥ 1) for Bi(3, 11/16)
• 2022 Exam 2 Section A Q18 — 47% — find a with Pr(X ≥ 16 | X ≥ a) = 0.9175
• 2012 Exam 2 Section B Q3aii — 48% — Pr(X ≥ 10) for Bi(20, ¼)
• 2021 Exam 2 Section A Q15 — 48% — equal heads and tails given at least one head
• 2022 Exam 2 Section B Q3aiii — 50% — Pr(X ≥ 2 | X < 5)
• 2015 Exam 2 Section B Q3di — 50% — Pr(lemons ≥ 1) for Bi(4, 0.03)
• 2020 Exam 2 Section A Q8 — 50% — Pr(X > 3) with n = 25, μ = 1.4
• 2014 Exam 2 Section B Q4fi — 50% — probability the third pot is smooth, in terms of p
• 2006 Exam 2 Section B Q2e — 35% — Pr(X = 4) for Bi(5, 0.809), with p carried from an earlier part

4.6 Binomial: finding n, finding p, optimising in p (14)

• 2011 Exam 1 Q7b–7b — 1% — [2011] solve p²(4p − 3) = 0
• 2012 Exam 2 Section B Q3d — 12% — solve two binomial equations for a and b
• 2012 Exam 2 Section B Q3c — 13% — "show that" p = 5/6
• 2007 Exam 2 Section B Q5fi — 15% — maximise q(p), give both p and q
• 2025 Exam 2 Section B Q3biv — 17% — find the integer k making "late at least once" equal 0.2
• 2019 Exam 2 Section B Q4fii — 23% — smallest n with Pr(X ≥ n) < 1%
• 2017 Exam 2 Section B Q3gi — 23% — maximum of q(p), four decimal places
• 2014 Exam 2 Section B Q4e — 23% — smallest n with Pr(Y ≥ 1) ≥ 0.95
• 2007 Exam 2 Section B Q5e — 25% — express q as a polynomial in p
• 2017 Exam 2 Section B Q3f — 32% — express q as a polynomial in p
• 2015 Exam 2 Section B Q3dii — 35% — smallest n with Pr(X ≥ 1) > 0.5
• 2017 Exam 2 Section A Q18 — 38% — smallest n with p ≤ 0.01 given μ = σ
• 2014 Exam 2 Section B Q4fii — 48% — solve 0.79 − 0.3p = 0.61
• 2025 Exam 2 Section B Q1f — 45% — show Pr(X ≥ 3) = 4p³ − 3p⁴ for Bi(4, p)

4.7 Continuous random variables: the constant, and probability by integration (10)

• 2016 Exam 1 Q8a — 17% — show by differentiation that a given expression antidifferentiates x^(k−1)logₑx
• 2016 Exam 1 Q8bi — 12% — Pr(X ≥ 1/e) using that antiderivative
• 2023 Exam 1 Q8b — 26% — E(T) for f(t) = t(16 − t²)/64
• 2006 Exam 1 Q6a — 46% — Pr(X < 3) for f(x) = x/12
• 2021 Exam 1 Q7a — 48% — show k = 2 for f(x) = k/x² on [1, 2]
• 2008 Exam 1 Q4a — 49% — show k = π/2 for k sin(πx)
• 2010 Exam 1 Q7a — 49% — find a for a(5x − x²) on [0, 5]
• 2023 Exam 1 Q8a — 43% — show k = 1/64
• 2021 Exam 1 Q7b — 39% — E(X) for f(x) = 2/x²
• 2022 Exam 2 Section B Q3bii — 34% — find a, b, c for f(h) = ah² + bh + c from three exact conditions

4.8 Continuous random variables: median, quantile, unknown terminal, and conditional probability (18)

• 2011 Exam 2 Section B Q2cii — 3% — [2011] find a with ∫f = 0.7 on a hybrid pdf
• 2016 Exam 1 Q8bii — 7% — "hence" show the median exceeds 1/e
• 2017 Exam 2 Section B Q3gii — 7% — find d from ∫_d^70 f = 0.3539
• 2025 Exam 1 Q8a — 24% — find k with Pr(X > k) = 9/16
• 2014 Exam 1 Q8b — 24% — Pr(X ≤ 1 | X ≤ m) using the median from the previous part
• 2014 Exam 2 Section B Q4d — 25% — find a with ∫₀ᵃ f = 0.15
• 2017 Exam 2 Section B Q3d — 30% — find a with Pr(T ≥ a) = 0.7
• 2006 Exam 2 Section B Q2f — 33% — median of a quartic pdf
• 2021 Exam 2 Section B Q4f — 37% — median of a hybrid pdf
• 2006 Exam 1 Q6b — 39% — find a with ∫ = 5/8, reject the negative root
• 2014 Exam 1 Q8a — 43% — median of (1/5)e^(−x/5)
• 2012 Exam 1 Q8b — 45% — find b with ∫ = 5/8, quadratic
• 2007 Exam 2 Section B Q5fii — 9% — find b with ∫_b^30 f = 1 − (2 − √2)
• 2023 Exam 1 Q8c — 11% — Pr(T > 2 | T > 1) for f(t) = t(16 − t²)/64
• 2008 Exam 1 Q4b — 27% — Pr(X ≤ ¼ | X ≤ ½) using the symmetry of the density
• 2007 Exam 2 Section B Q5c — 43% — Pr(T ≤ 15 | T ≤ 25)
• 2017 Exam 2 Section B Q3c — 49% — Pr(T ≤ 25 | T ≤ 55)
• 2018 Exam 2 Section A Q15 — 49% — which equation the median of f(x) = (8x − x³)/12 satisfies

4.9 Continuous random variables: mean, variance, sketch (11)

• 2011 Exam 2 Section A Q6 — 6% — [2011] E(X) for f(x) = logₑ(x) on [1, e]
• 2013 Exam 1 Q8 — 21% — E(X) using a supplied derivative as the antiderivative
• 2007 Exam 2 Section B Q5a — 22% — sketch a two-piece linear pdf
• 2017 Exam 2 Section B Q3a — 29% — sketch a two-piece linear pdf
• 2021 Exam 2 Section B Q4g — 32% — standard deviation of a hybrid pdf
• 2013 Exam 2 Section B Q2ci — 46% — E(X) for a hybrid pdf
• 2014 Exam 2 Section A Q16 — 46% — ∫x²p(x)dx from mean and variance
• 2015 Exam 2 Section A Q9 — 37% — E(X) for a triangular pdf with parameter a
• 2019 Exam 2 Section A Q18 — 27% — Pr(X > 0) from the average value of the density
• 2013 Exam 2 Section B Q2cii — 42% — expected count of 200 members
• 2021 Exam 2 Section B Q4e — 21% — maximum possible spin, i.e. the upper endpoint of the support

4.10 Transformations of a pdf (5)

• 2021 Exam 2 Section B Q4h — 2% — a, b in g(x) = a f(x/b) for a target median
• 2022 Exam 2 Section B Q3biii — 6% — r, s in g(d) = f(rd + s)
• 2023 Exam 2 Section B Q4j — 6% — a, b in g(w) = a f(w/b) for a target mean
• 2018 Exam 2 Section A Q20 — 20% — matrix transformation mapping one pdf to another
• 2025 Exam 1 Q8b — 27% — ∫₀⁴3h(x)dx where h = m f + n

4.11 Normal distribution (23)

• 2020 Exam 2 Section B Q3c — 6% — both values of k for a translated normal
• 2009 Exam 2 Section B Q3d — 22% — find σ from Pr = 0.99
• 2023 Exam 2 Section B Q4f — 26% — find σ so that more than 99% are grade A
• 2006 Exam 1 Q5c — 27% — Pr(X < 64 | X < 72) by standardising
• 2010 Exam 1 Q5b — 31% — Pr(X > 7) = Pr(Z < b), find b
• 2016 Exam 2 Section B Q3e — 32% — find σ from a 12% tail
• 2012 Exam 1 Q8a — 34% — Pr(94 < X < 100) in terms of q by symmetry
• 2016 Exam 2 Section B Q3d — 21% — Pr(P̂ ≥ 0.06 | P̂ ≥ 0.05), binomial not normal
• 2020 Exam 2 Section B Q3b — 41% — Pr(T ≤ 3 | T > 0)
• 2018 Exam 1 Q4b — 41% — Pr(X > 7) = Pr(Z < b), variance 4
• 2009 Exam 2 Section B Q3ci — 42% — conditional normal probability
• 2014 Exam 2 Section B Q4a — 43% — inverse normal, top 10%, to the nearest millimetre
• 2020 Exam 2 Section A Q14 — 44% — find σ when μ = 2σ and Pr(X > 5.2) = 0.9
• 2011 Exam 2 Section A Q13 — 45% — [2011] find μ from a stated tail probability
• 2006 Exam 1 Q5b — 45% — Pr(64 < X < 72) by standardising
• 2013 Exam 2 Section A Q22 — 47% — find σ from an expected count of 150 in 2000
• 2016 Exam 2 Section B Q3c — 48% — normal probability with a units trap
• 2014 Exam 2 Section B Q4b — 47% — expected count of 2000 basil plants
• 2025 Exam 2 Section B Q3cii — 48% — find σ so that a 2% tail exceeds 3.5 minutes
• 2015 Exam 2 Section B Q3c — 50% — Pr(O > 85 | O > 74)
• 2015 Exam 1 Q6a — 50% — Pr(X ≥ 2) = Pr(Z ≥ b), find b
• 2015 Exam 1 Q6b — 37% — Pr(X > 2.8 | X > 2.5) by hand, using the standardised value from part a
• 2022 Exam 2 Section A Q20 — 30% — Pr(50 sin(2Θ) > 40) where Θ is normal
• 2019 Exam 2 Section A Q18 — (counted in §4.9)

4.12 Sample proportions (10)

• 2018 Exam 2 Section B Q4ciii — 8% — least n with Pr(P̂ₙ > 1/n) > 0.99
• 2019 Exam 2 Section B Q4fiv — 19% — Pr(P̂ within one sd of 0.0527), binomial only
• 2017 Exam 1 Q4 — 31% — smallest n with sd(P̂) ≤ 1/100
• 2021 Exam 2 Section B Q4c — 33% — mean and standard deviation of P̂
• 2018 Exam 2 Section B Q4cii — 36% — Pr(P̂ > 10%) for n = 16
• 2017 Exam 2 Section A Q16 — 41% — Pr(P̂ > 0.6) given Pr(P̂ = 0) = 1/243
• 2019 Exam 2 Section B Q4fiii — 42% — E(P̂) and sd(P̂)
• 2024 Exam 1 Q5ci — 44% — [2024 text lost] interval built from sd(P̂)
• 2025 Exam 2 Section B Q3biii — 49% — Pr(0.4 ≤ P̂ ≤ 0.6) for n = 5
• 2022 Exam 2 Section B Q3bi — 38% — state the value of ∫₁.₅³ f(h)dh
• 2021 Exam 1 Q6c (29%) and 2019 Exam 1 Q6b (12%) — listed in §4.5

4.13 Confidence intervals (10)

• 2018 Exam 2 Section B Q4dii — 11% — explain why the interval suggests a difference
• 2023 Exam 2 Section B Q4g — 19% — recover the confidence level as a percentage
• 2023 Exam 1 Q6c — 23% — factor by which quadrupling n changes the width
• 2019 Exam 2 Section B Q4g — 25% — recover n from (0.0234, 0.0866)
• 2023 Exam 1 Q6b — 26% — recover n using z = 2
• 2022 Exam 2 Section B Q3ciii — 28% — flips needed to halve the width
• 2024 Exam 1 Q5cii — 33% — [2024 text lost] recover n = 300 from sd(P̂)
• 2016 Exam 2 Section B Q3g — 41% — build the 95% interval, two decimal places
• 2018 Exam 2 Section B Q4di — 45% — recover p̂ from (0.102, 0.145)
• 2017 Exam 2 Section A Q5 — 47% — recover p̂ from (0.039, 0.121)

4.14 Corpus mis-tags and unrecoverable 2024 parts (12)

These carry the statistics topic in [QJSON] but are not statistics questions, or their wording cannot be read.

• 2007 Exam 1 Q12 — 20% — [MIS-TAG] minimum distance from the origin to a line
• 2010 Exam 1 Q9b — 20% — [MIS-TAG] ∫₁³ x logₑ(x) dx by parts
• 2014 Exam 2 Section B Q1d — 29% — [MIS-TAG] fraction of the year the wombat population is below 1000
• 2016 Exam 2 Section B Q3f — 21% — first laptop with short battery life is the third one (geometric-style; legitimately statistics but no named distribution covers it)
• 2022 Exam 2 Section B Q3ciii — (counted in §4.13)
• 2024 Exam 2 Section B Q4ci — 36% — [2024 text lost] a "show that"
• 2024 Exam 2 Section B Q4cii — 11% — [2024 text lost] tree-diagram values
• 2024 Exam 2 Section B Q4di — 33% — [2024 text lost] probability, normal attempted incorrectly
• 2024 Exam 2 Section B Q4dii — 18% — [2024 text lost] standard deviation then a probability
• 2024 Exam 2 Section B Q4eii — 8% — [2024 text lost] minimum and maximum of an expression
• 2024 Exam 2 Section A Q11 — 40% — [2024 text lost] multiple choice, answer B
• 2011 Exam 1 Q5b–5b — 1% — [2011] Pr(X < 2.5 | X < 3.5) for a continuous variable
• 2011 Exam 2 Section B Q2e — 9% — [2011] Pr(Machine A | longer than 3)

4.15 Which types separate most

Ranking the types by the mean pct of their separators and by how many separators they generate:

The reports converge on four sentences that explain most of the damage:

1. 2006 [RPT]: "The denominator was usually correct but the answer to part b. was often placed in the numerator."
2. 2009 [RPT]: "Students must ensure that they work to more decimal places than what is required in the answer."
3. 2012 [RPT]: "Identifying the distribution with the correct parameters is sufficient working."
4. 2019 [RPT]: "Check that answers are reasonable… E(P̂) cannot be greater than one."

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5. What makes a hard one hard

Six mechanisms, each with the evidence.

5.1 Conditional wording that hides a set operation

The question never says "find the intersection". It says "given that they have already queued for one minute" (2023 Exam 1 Q8c, 11%), or "given that there is at least one head" (2021 Exam 2 Section A Q15, 48%), or "given that a visitor has previously visited the zoo" (2024 Exam 2 Section A Q8 (NHT)). The work is entirely in turning the English into {T > 1}, {X ≥ 1}, {N ≥ 1} and then intersecting.

The reliable move: write the two events as sets on the page before writing any formula. 2023 [RPT] diagnoses exactly this: "students had incorrectly interpreted the mathematical meaning of 'already queued for one minute'." When one event is a subset of the other, the intersection is the smaller set and the whole question collapses to one division.

5.2 Rounding rules that compound

VCAA asks for four decimal places in the answer but the intermediate value feeds the next part. 2009 Exam 2 Section B Q3cii scored 4%: students took p = 0.1015 from part c.i. rounded to four places and fed it into Bi(4, p). [RPT]: "Many students recognised that the binomial distribution was required; however, they used the wrong probability. Common incorrect answers were 0.5057 and 0.9993."

The rule: carry the full stored value between parts, round only at the printed answer. 2016 [RPT]: "Others rounded too soon and gave 0.9312 as the answer." 2020 [RPT]: "Do not round too early. Students could not be awarded marks in Questions 3 and 4 due to rounding too early."

The other half of this trap is the exact/approximate switch. 2022 [RPT]: "Be careful with questions that move between exact and approximate answers. In Questions 4ai., 4aii. and 4aiv., exact answers were required, whereas Question 4aii. required an approximate answer. This often occurred in the probability question."

5.3 Sample proportion versus probability

P̂ is a random variable with a discrete set of possible values {0, 1/n, …, 1}; p is a fixed number; p̂ is one observed value. The three get conflated constantly.

• 2019 Exam 2 Section B Q4fiii (42%): students computed E(X) = 36 × 0.0527 instead of E(P̂) = 0.0527.
• 2018 Exam 2 Section B Q4ciii (8%): [RPT]: "Many students appeared to be confused by the terminology Pr(P̂ₙ > 1/n)."
• 2016 Exam 2 Section B Q3d (21%): [RPT]: "Most students used the conditional probability formula but tried to use the normal distribution rather than the binomial distribution."
• 2022 Exam 2 Section B Q3ci (68%) is the sanity check VCAA built in: "Is the random variable P̂ discrete or continuous? Justify your answer." Answer: discrete, countable.

The operational rule: every probability statement about P̂ becomes a probability statement about X ~ Bi(n, p) by multiplying through by n. Then round the resulting bounds inwards to integers.

5.4 Integrating a pdf that carries a parameter

Three difficulties stack. (i) The unknown is in a terminal, so the integral evaluates to a polynomial or indicial expression in that unknown. (ii) The resulting equation has a root outside the support that must be rejected. (iii) On Exam 1 you must antidifferentiate a product or a shifted power by hand.

2025 Exam 1 Q8a (24%, distribution {0: 30%, 1: 23%, 2: 23%, 3: 24%}) is the model: f(x) = (3/8)(4 − 3x) on 0 ≤ x ≤ 4/3, and Pr(X > k) = 9/16. [RPT]: "Some students who chose to integrate the term as a bracketed term raised the power to 2 but then divided by 2 instead of 6. Some students were unable to form the correct quadratic equation or solve it correctly… Students who obtained the two possible solutions were mostly aware of rejecting [the invalid one]."

2016 Exam 1 Q8 is the extreme case: part a (17%) asks you to derive the antiderivative of x^(k−1)logₑ(x) by differentiating a supplied expression; part b.i (12%) uses it; part b.ii (7%) asks you to reason about the median without computing it. [RPT]: "Often the instruction 'hence' was ignored… A common misconception was to assume that since [an inequality between two numbers] then [an inequality between two probabilities]."

5.5 "At least one" complements and their inverses

Pr(≥ 1) = 1 − (1−p)ⁿ is easy forwards. Backwards it is a separator:

• 2020 Exam 2 Section B Q3ei (24%): express 1 − 0.85ⁿ in terms of n.
• 2020 Exam 2 Section B Q3eii (23%): solve 1 − 0.85ⁿ ≥ 0.95. [RPT]: "A common incorrect answer was 2, due to students solving [the wrong inequality]."
• 2014 Exam 2 Section B Q4e (23%): Pr(Y ≥ 1) ≥ 0.95 for Bi(n, 0.2), i.e. 0.8ⁿ ≤ 0.05, n = 14.
• 2025 Exam 2 Section B Q3biv (17%): find the integer k such that "late at least once in five days" is 0.2 to one decimal place — a double inversion, because k is a terminal in the integral that produces p, which then feeds 1 − (1−p)⁵.

The pattern: take logs or use solve, then round in the direction the inequality demands and state an integer.

5.6 Interpreting rather than computing

The 1-mark parts that ask for an explanation have the worst returns on the paper.

• 2018 Exam 2 Section B Q4dii: 11%. "Explain why this confidence interval suggests…"
• 2023 Exam 1 Q6c: 23%. "By what factor will the increased sample size affect the width of the confidence interval?" — distribution {0: 77%, 1: 23%}.
• 2022 Exam 2 Section B Q3biii: 6%. "Find the values of r and s." [RPT]: "This question was not answered well. Many students did not attempt it."
• 2018 Exam 2 Section B Q4bii: 44%. "Are the events H and S independent? Justify your answer." [RPT]: "A mathematical explanation was required."

Two disciplines fix most of this. First, quote the numbers you are reasoning about — the interval's endpoints, both sides of the independence test, the ratio 1/√4 = ½. Second, finish the sentence: the answer to "are they independent?" is "No, because Pr(H ∩ S) = 0.09 but Pr(H)Pr(S) = 0.046", not "No".

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6. A worked method sheet — the ten highest-yield types

Ranked by (marks available across the archive) × (separator rate). CAS steps are given generically; the key sequences are identical on TI-Nspire and ClassPad.

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M1 — Binomial probability (Exam 2)

Recognise: a fixed number of independent repetitions, constant probability, counting successes.

Write first (this is the method mark): X ~ Bi(n, p), with the numbers substituted.

CAS: binomPdf(n, p, x) for Pr(X = x); binomCdf(n, p, lower, upper) for an interval. For Pr(X ≥ k) use binomCdf(n, p, k, n) — do not use 1 − binomCdf(n, p, 0, k) unless you are sure of the boundary.

Translation table. "exactly k" → Pr(X = k). "at least k" → Pr(k ≤ X ≤ n). "more than k" → Pr(k+1 ≤ X ≤ n). "fewer than k" → Pr(0 ≤ X ≤ k−1). "at most k" → Pr(0 ≤ X ≤ k).

Exam 1 version: Pr(X = x) = ⁿCₓ pˣ(1−p)ⁿ⁻ˣ — on [FS] since 2023. Sum the required terms. Expect the answer to be demanded in a form: factor out the largest common power, then write the remaining integer. Know 2¹⁰ = 1024, 2¹² = 4096, 3⁵ = 243, 5⁴ = 625.

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M2 — Conditional probability

Formula (on [FS]): Pr(A|B) = Pr(A ∩ B)/Pr(B).

Procedure. 1. Write A and B as sets, explicitly, before anything else. 2. Write A ∩ B as a set. If A ⊆ B, then A ∩ B = A. 3. Compute numerator and denominator separately, to full stored precision. 4. Divide. Round once, at the end. 5. Check the answer is in [0, 1] and — if A ⊆ B — is larger than Pr(A).

Discrete: numerator and denominator are sums of table entries. Binomial: both are binomCdf values over the correct integer ranges. Continuous: both are definite integrals; define f once on CAS and reuse the name. Normal: both are normCdf values; watch for symmetry making the denominator exactly 0.5.

Exam 1 version: everything stays as a fraction. Pr(X < 64)/Pr(X < 72) = 0.16/0.50 = 8/25. Never convert to a decimal you then cannot simplify.

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M3 — Sample proportion probabilities

Facts (on [FS]): P̂ = X/n where X ~ Bi(n, p); E(P̂) = p; sd(P̂) = √(p(1−p)/n).

Procedure for Pr(a ≤ P̂ ≤ b). 1. Multiply the bounds by n: Pr(na ≤ X ≤ nb). 2. Round na up and nb down to integers (the possible values of X are integers). 3. binomCdf(n, p, ⌈na⌉, ⌊nb⌋).

Worked, from 2025 Exam 2 Section B Q3biii: n = 5, p = 0.08704. Pr(0.4 ≤ P̂ ≤ 0.6) = Pr(2 ≤ X ≤ 3). Worked, from 2019 Exam 2 Section B Q4fiv: p = 0.0527, n = 36, sd(P̂) = 0.0372. "Within one standard deviation of 0.0527" is 0.01546 < P̂ < 0.08993, i.e. 0.5566 < X < 3.237, i.e. Pr(1 ≤ X ≤ 3) = 0.7380.

When VCAA says "Do not use a normal approximation", this is the required method. When (2026 NHT) it says "use a normal approximation", instead treat P̂ ~ N(p, p(1−p)/n) and use normCdf.

Exam 1 version (2021 Exam 1 Q6c, 2019 Exam 1 Q6b): expand the binomial terms by hand and present in the demanded algebraic form.

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M4 — Confidence intervals

Formula (on [FS]): (p̂ − z√(p̂(1−p̂)/n), p̂ + z√(p̂(1−p̂)/n)), with z ≈ 1.96 for 95%.

Four questions, four one-line moves:

Exam 1 version: 2023 Exam 1 Q6 tells you "Use z = 2 to approximate the 95% confidence interval." With z = 2 and p̂ = 0.1, h = 2√(0.09/n) = 0.06 gives n = 100 cleanly. Expect the numbers to be built for z = 2.

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M5 — Continuous random variable: the constant, the probability, the quantile

Constant. ∫ f = 1 over the support. For a hybrid, sum the pieces. Antidifferentiate by hand on Exam 1. Write dx.

Probability. Pr(a ≤ X ≤ b) = ∫ₐᵇ f. Split at any join. Remember Pr(X = c) = 0 for a continuous variable, so < and ≤ are interchangeable — 2023 [RPT]: "Students need to be aware that with continuous probability Pr(X = a) = 0."

Quantile / median / unknown terminal. - CAS: solve(∫ₗᵏ f(x) dx = c, k), then discard roots outside the support. Better: solve(…, k) | lower < k < upper. - By hand: antidifferentiate, substitute, set equal, clear fractions, set the quadratic to zero, factorise by inspection first, reject the out-of-domain root.

Which tail. Pr(X > a) = 0.7 and Pr(X < a) = 0.7 give different a. 2017 [RPT] records a = 50.6351 as the common wrong answer to a question whose answer was 39.3649 — the mirror image.

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M6 — Continuous random variable: mean, variance, standard deviation

Formulae (all on [FS]): μ = ∫ x f(x) dx; σ² = ∫ (x − μ)² f(x) dx = ∫ x² f(x) dx − μ².

CAS: define f first. m := ∫(x·f(x), x, a, b). v := ∫(x²·f(x), x, a, b) − m². sd := √v.

The three errors the reports name every year: 1. Omitting the x — you compute 1 instead of the mean. 2. Computing the median when the mean was asked, or vice versa. 3. Stopping at the variance when the standard deviation was asked.

Exam 1 version: integration by parts is not in Methods, so VCAA supplies the antiderivative. 2013 Exam 1 Q8 hands you d/dx[x sin(πx/4)]; 2016 Exam 1 Q8a asks you to derive the antiderivative by differentiating a given expression, then use it in part b. If an Exam 1 pdf question gives you a derivative, that derivative is the antiderivative you need.

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M7 — Normal distribution: forwards and backwards

Forwards. normCdf(lower, upper, μ, σ). Use −1E99 / 1E99 for infinite tails. Enter σ, not σ². Convert units before entering — "three hours and ten minutes" is 190 minutes, not 3.1 hours.

Backwards (inverse normal). invNorm(area to the left, μ, σ). - "smallest 5%" → invNorm(0.05, μ, σ) - "largest 15%" / "longest 15%" → invNorm(0.85, μ, σ) - "larger than 90% of all" → invNorm(0.90, μ, σ)

Find σ. solve(normCdf(lower, upper, μ, s) = p, s), with a domain restriction s > 0. Or standardise: z = invNorm(p), then σ = (x − μ)/z.

Find μ and σ together. Two probability statements, two equations: solve({normCdf(...) = p₁, normCdf(...) = p₂}, {m, s}) (2025 Exam 2 Section A Q12).

Exam 1 version. Only these tools: z = (x − μ)/σ with σ = √variance; symmetry Pr(Z > c) = Pr(Z < −c); Pr(Z < 0) = 0.5; and the empirical values if supplied in the stem. Draw the curve and shade — three reports in a row name this as the difference between the students who scored and those who did not.

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M8 — Discrete random variables

Find the parameter. Σ Pr(X = x) = 1. Clear denominators, form ax² + bx + c = 0, solve, reject roots making any probability negative or greater than 1.

Mean and variance. Build a table with rows x, p(x), x·p(x), x²·p(x). μ = Σ x p(x); σ² = Σ x² p(x) − μ²; σ = √σ².

Conditional. Reduced sample space: the denominator is a partial sum.

Show that. When VCAA prints the answer (2025 Exam 1 Q4a: "Show that k = 10 or k = 15"), you must derive, not verify. Sum the probabilities, equate to 1, form and solve the quadratic, and present both roots.

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M9 — Transformations of a probability density function

The two invariants. 1. Area is 1. For g(x) = a f(x/b): ∫g = ab∫f = ab, so ab = 1. 2. Horizontal dilation scales location statistics. If g(x) = a f(x/b), then the mean, median and every quantile of g are b times those of f. The standard deviation is also b times.

Procedure (2 marks, typically the last part of the paper). 1. Write ab = 1. 2. Write the stated statistic: new mean = b × old mean, or new median = b × old median. 3. Solve for b, then a = 1/b.

Worked, from 2023 Exam 2 Section B Q4j: f has mean 30 + 3π/2 · … (exact value from part i). The new mean is 22 + 8. So b = (22 + 8)/(old mean) and a = 1/b.

If you use the integral method instead — setting up ∫ w · g(w) dw = new mean — multiply the terminals by b. 2023 [RPT]: "Many of those who attempted the second method did not multiply the terminals by b."

Exam 1 version (2025 Exam 1 Q8b): h = m f + n, find ∫₀⁴ 3h(x)dx in terms of m and n. Use linearity: 3∫(mf + n) = 3m∫f + 3n∫dx = 3m(1) + 3n(4) = 3m + 12n — because ∫₀⁴ f = 1. Do not factor m out of the whole integral.

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M10 — Finding n, and inequalities over the integers

Binomial. To find the smallest n with Pr(X ≥ 1) ≥ c: solve 1 − (1−p)ⁿ ≥ c, i.e. (1−p)ⁿ ≤ 1 − c, i.e. n ≥ logₑ(1−c)/logₑ(1−p). Round up. Report the integer.

Sample proportion. To find the smallest n with sd(P̂) ≤ c: √(p(1−p)/n) ≤ c ⟹ n ≥ p(1−p)/c². Round up. (2017 Exam 1 Q4: p = ¼, c = 1/100 gives n ≥ 1875.)

Confidence interval. n = z²p̂(1−p̂)/h² where h is the half-width.

When the inequality cannot be inverted algebraically (for example Pr(X ≥ n) < 0.01 for a fixed binomial), tabulate. Both 2019 and 2015 reports say explicitly: "Trial and error is an acceptable method." But show the two adjacent values — the one that fails and the one that works — or the working mark is lost.

Always finish with a sentence naming the integer, in the units of the question: "the minimum number of tomato plants is 14", not "n ≥ 13.4".

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Appendix — the twenty hardest statistics parts in the archive

Excluding the 2011 entries, which are unreliable, the five hardest genuinely-current statistics questions in the archive are all either a pdf transformation or a "find the parameter that makes this probability hit a target" inversion. Those two skills are worth more marks per hour of practice than anything else in the area of study.