Mathematics Advanced • Year 12 • Module 5 • Lesson 15
Module Synthesis
Apply Module 5's full toolkit, probability rules, summary statistics, regression, normal CDF, binomial, to integrated real-world scenarios.
Optional legacy extension. This worksheet mixes current Random Variables practice with former Statistical Analysis topics, including binomial probability and normal approximation. For the 2024 syllabus (first HSC 2027), those legacy questions are optional extension; retain them for 2017-syllabus / 2026 HSC revision.
Problem 1, Factory defects (probability + binomial)
A factory produces components. Each component has a 3% chance of being defective, independently. A quality inspector tests batches of 50 components.
Set up: What are we solving for?
(i) Name the distribution of X = number of defective components per batch and state its parameters. Justify with reference to the binomial conditions. 2 marks
(ii) Find the probability that a batch contains exactly 2 defective components. 2 marks
(iii) The factory upgrades its process and the defect rate drops to 1%. How many components must be tested in the new process so that the expected number of defectives is exactly 2? 2 marks
Problem 2, Exam cohort (data + normal)
Exam scores in a large cohort are normally distributed with μ = 68 and σ = 12.
Set up: What are we solving for?
(i) What percentage of students scored between 56 and 80? 2 marks
(ii) A student claims they are in the top 10%. Find the minimum score they need. [Use z ≈ 1.282.] 2 marks
(iii) Two students compare results: Alex scored 80 in Maths (μ = 68, σ = 12); Blake scored 74 in English (μ = 62, σ = 10). Calculate each z-score and state who performed better relative to their cohort. 2 marks
Problem 3, Study habits (regression + critical reasoning)
A study finds r = 0.75 between weekly study hours and exam score across a sample. The least-squares regression line is ŷ = 45 + 3x, where x is hours studied per week and y is the exam score.
Set up: What are we solving for?
(i) Predict the exam score for a student who studied 8 hours per week. 1 mark
(ii) A student studied 8 hours and scored 72. Calculate their residual (observed − predicted) and state in one sentence whether the student over- or under-performed relative to the line. 2 marks
(iii) A newspaper writes: "Study More, Score Higher, Causation Proven." Identify the statistical error and state in one sentence what kind of evidence would support a causal claim. 2 marks
Problem 4, Customer survey (mixed tools)
A retail chain surveys 100 customers about satisfaction (satisfied / not satisfied) and also records the dollar amount each customer spent during their visit.
Set up: What are we solving for?
(i) "Number of satisfied customers in the survey." Which Module 5 tool best fits, and what are the parameters if the long-run satisfaction rate is 70%? 2 marks
(ii) "Distribution of spending amounts is approximately bell-shaped with μ = $42 and σ = $9. What proportion spent more than $60?" Identify the tool and compute the answer. [Use P(Z < 2) ≈ 0.9772.] 2 marks
(iii) "Is customer satisfaction associated with the amount they spent?" Name the tool and state what numerical summary would quantify any association. 2 marks
Problem 5, Medical decision (independence + conditional)
A medical study investigates a new treatment. In a randomised trial, 120 patients receive the treatment and 80 recover. In a parallel control group of 120 patients receiving a placebo, 60 recover.
Set up: What are we solving for?
(i) Compute the recovery rate (proportion who recovered) for each group. 1 mark
(ii) Pool both groups into one population of 240 patients. Treat A = "received treatment" and B = "recovered". Compute P(A), P(B), and P(A ∩ B). Test whether A and B are independent using P(A ∩ B) = P(A) × P(B). 3 marks
(iii) Even if A and B are not independent in this trial, explain in 1-2 sentences why this does not by itself prove the treatment causes recovery. Name one feature of the study design that, if present, would strengthen a causal claim. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1, Factory defects
Set up. Model defective counts per batch as binomial, then compute a specific-value probability, then use E(X) = np to find the new sample size.
(i) X ~ B(50, 0.03). Conditions: fixed n = 50 ✓; independent components ✓; two outcomes (defective / not) ✓; constant p = 0.03 ✓.
(ii) P(X = 2) = C(50, 2)(0.03)²(0.97)^48 = 1225 × 0.0009 × 0.2317 ≈ 0.2555 (about 25.6%).
(iii) E(X) = np = 2 with p = 0.01 ⇒ n = 2/0.01 = 200 components.
Problem 2, Exam cohort
Set up. Use the empirical rule for the first part, the inverse normal for the percentile, and z-scores to compare two students from different normal distributions.
(i) 56 = 68 − 12 = μ − σ; 80 = 68 + 12 = μ + σ. So P(56 < X < 80) ≈ 68%.
(ii) P(X > x) = 0.10 ⇒ P(Z < z) = 0.90 ⇒ z ≈ 1.282. x = 68 + 1.282(12) = 68 + 15.38 ≈ 83.4.
(iii) z_Alex = (80 − 68)/12 = 1.0; z_Blake = (74 − 62)/10 = 1.2. Blake performed better (1.2 > 1.0).
Problem 3, Study habits
Set up. Use the regression line for prediction and residual calculation, then critique a causal claim.
(i) ŷ = 45 + 3 × 8 = 69.
(ii) residual = 72 − 69 = +3. The student over-performed the regression-line prediction by 3 marks.
(iii) Error: the article confuses correlation (which the data shows) with causation (which it does not). A confounder (e.g. motivation: motivated students study more and score higher) could fully account for the association without study causing the score. A randomised controlled experiment in which students are randomly assigned different study amounts, would be needed to support a causal claim.
Problem 4, Customer survey
Set up. Pick the right Module 5 tool for each subquestion: a count of "successes" → binomial; bell-shaped continuous measurement → normal; relationship between two variables → correlation/regression.
(i) Binomial: X ~ B(100, 0.70). E(X) = 70 satisfied customers expected.
(ii) Normal CDF: z = (60 − 42)/9 = 2. P(spend > $60) = 1 − P(Z < 2) ≈ 1 − 0.9772 = 0.0228 (about 2.3%).
(iii) Scatter plot with Pearson's r and possibly a regression line. The numerical summary that quantifies the strength of the association is r ∈ [−1, 1].
Problem 5, Medical decision
Set up. Compare recovery rates between groups; pool to compute the independence test using P(A ∩ B) = P(A) × P(B); then comment on causation.
(i) Treatment recovery rate = 80/120 ≈ 0.667 (66.7%). Placebo recovery rate = 60/120 = 0.50 (50%).
(ii) Pooled: P(A) = 120/240 = 0.5 (half received treatment); P(B) = (80 + 60)/240 = 140/240 ≈ 0.583 (recovered overall); P(A ∩ B) = 80/240 ≈ 0.333. If independent, we'd expect P(A) × P(B) = 0.5 × 0.583 = 0.292. Observed 0.333 ≠ expected 0.292, so A and B are not independent recovery and receiving treatment are associated in this data.
(iii) Even with the observed dependence, association does not prove causation: a confounder (e.g. healthier or younger patients ended up in the treatment group) could explain the difference. A randomised, double-blind assignment would strengthen the causal claim, because randomisation balances confounders between groups.