Mathematics Standard • Year 12 • Module 8 • Lesson 6
Normal Distribution — Problem Set
Apply the 68-95-99.7 rule to realistic Australian contexts — manufacturing, exams, weights, heights and recruitment thresholds.
Problem 1 — Coffee bag fill (manufacturing)
A coffee roaster fills 250 g bags. Measurements over a long production run show the actual fill mass is normally distributed with μ = 252 g and σ = 3 g.
Set up: What are we solving for?
(i) What percentage of bags contain between 246 g and 258 g of coffee? 2 marks
(ii) Customer policy is to refund any bag containing under 246 g. In a batch of 5,000 bags, about how many refunds should the company expect? 2 marks
(iii) The factory manager says "We could save money by lowering the target μ to 250 g". State, in 1-2 sentences, what would happen to refund rates and why σ matters here. 2 marks
Stuck? Lowering μ moves the whole curve left without changing its shape — what then falls into the "under 246 g" region?Problem 2 — HSC exam marks (band cutoffs)
A particular HSC subject has exam marks normally distributed with μ = 70 and σ = 12.
Set up: What are we solving for?
(i) The top 2.5% of students receive Band 6. What is the cutoff mark? 1 mark
(ii) What percentage of students score between 58 and 82? 2 marks
(iii) In a state-wide cohort of 8,000 students, estimate how many score between 82 and 94. 2 marks
Stuck on (iii)? 82 to 94 = μ + 1σ to μ + 2σ → one "shell" on the right.Problem 3 — Cattle weights (livestock)
An Australian beef producer finds that the live weight of Angus steers ready for sale is normally distributed with μ = 540 kg and σ = 40 kg. A herd has 600 steers.
Set up: What are we solving for?
(i) How many steers weigh between 460 kg and 620 kg? 2 marks
(ii) A premium buyer pays a bonus for steers above 620 kg. About how many steers in the herd qualify? 2 marks
(iii) The producer's accountant predicts "about 50% of the herd will weigh more than 540 kg". Verify or correct this claim in one sentence, explaining briefly why. 2 marks
Stuck? In a symmetric normal distribution, what percentage sits above the mean exactly?Problem 4 — Recruitment cutoff (police academy)
An adult-male timed run is normally distributed with μ = 14.0 min and σ = 1.2 min. To enter the academy, candidates must finish in under 12.8 minutes.
Set up: What are we solving for?
(i) Express 12.8 min in terms of μ and σ. 1 mark
(ii) Use the empirical rule to estimate the percentage of candidates who currently meet the cutoff. 2 marks
(iii) The academy needs to recruit roughly 2.5% of candidates only. What new cutoff (in minutes) would achieve this? Show your reasoning. 2 marks
Stuck on (iii)? "Top 2.5% fastest" means a time below μ − 2σ.Problem 5 — Height and uniform sizes (school)
Heights of Year 12 boys at a NSW school are normally distributed with μ = 178 cm and σ = 6 cm. The cohort has 240 boys. The uniform supplier offers four sizes: Small (≤ 166 cm), Medium (166-178 cm), Large (178-190 cm), X-Large (> 190 cm).
Set up: What are we solving for?
(i) Calculate the percentage of boys in each of the four size brackets. 2 marks
(ii) Translate those percentages into actual student counts (round to whole boys). 2 marks
(iii) The supplier wants a single-paragraph order list for the school. Write the recommendation in one sentence, ordered Small / Medium / Large / X-Large, with quantities. 2 marks
Stuck on (i)? 166 = μ − 2σ; 190 = μ + 2σ; 178 = μ. Sketch the curve and shade.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Coffee bag fill
(i) 246 = 252 − 6 = μ − 2σ. 258 = 252 + 6 = μ + 2σ. Between μ ± 2σ → 95%.
(ii) Below 246 = below μ − 2σ → (100 − 95)/2 = 2.5%. In 5,000 bags: 0.025 × 5,000 = 125 refunds.
(iii) Lowering μ from 252 to 250 g shifts the whole curve left. With σ unchanged at 3 g, more of the curve now sits below 246 g — refund rates would jump (a 250 g mean puts 246 g at only μ − 1.33σ, so about 9% of bags would be under-filled instead of 2.5%). σ matters because it controls how wide the curve is — a smaller σ would let you safely lower μ without much extra risk.
Problem 2 — HSC exam marks
(i) Top 2.5% sits above μ + 2σ = 70 + 24 = 94 marks.
(ii) 58 = 70 − 12 = μ − 1σ. 82 = 70 + 12 = μ + 1σ. Between μ ± 1σ → 68%.
(iii) 82 to 94 = μ + 1σ to μ + 2σ → one shell on the right = (95 − 68)/2 = 13.5%. In 8,000 students: 0.135 × 8,000 = 1,080 students.
Problem 3 — Cattle weights
(i) 460 = 540 − 80 = μ − 2σ. 620 = 540 + 80 = μ + 2σ. Between μ ± 2σ → 95%. In 600 steers: 0.95 × 600 = 570 steers.
(ii) Above 620 = above μ + 2σ → 2.5%. In 600 steers: 0.025 × 600 = 15 steers qualify for the bonus.
(iii) The accountant is essentially correct: in a normal (symmetric) distribution, exactly 50% sit above the mean, so about 300 of the 600 steers weigh over 540 kg.
Problem 4 — Police academy run
(i) 12.8 = 14.0 − 1.2 = μ − 1σ.
(ii) "Finish in under 12.8" = below μ − 1σ = (100 − 68)/2 = 16% of candidates currently meet the cutoff.
(iii) Top 2.5% fastest = below μ − 2σ = 14.0 − 2(1.2) = 11.6 min. Set the new cutoff at 11.6 minutes to narrow the intake to about 2.5%.
Problem 5 — Year 12 uniform sizes
(i) 166 = μ − 2σ, 178 = μ, 190 = μ + 2σ.
Small (< μ − 2σ) → 2.5%. Medium (μ − 2σ to μ) → (95% − 0%)/2 = 47.5%. Large (μ to μ + 2σ) → 47.5%. X-Large (> μ + 2σ) → 2.5%.
(ii) Cohort = 240. Small = 0.025 × 240 = 6. Medium = 0.475 × 240 = 114. Large = 0.475 × 240 = 114. X-Large = 0.025 × 240 = 6.
(iii) "Order 6 Small, 114 Medium, 114 Large and 6 X-Large uniforms for the Year 12 cohort of 240 boys."