Mathematics Standard • Year 12 • Module 8 • Lesson 6
Normal Distribution — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on the 68-95-99.7 rule — interval percentages, cutoffs, and population counts.
1. Short-answer questions
1.1 A normally distributed set of test marks has μ = 60 and σ = 8.
(a) What percentage of marks lie between 52 and 76?
(b) What percentage of marks are above 76? 3 marks Band 3
1.2 The mass (g) of apples in a delivery is normally distributed with μ = 180 g and σ = 15 g. A box contains 400 apples.
(a) How many apples weigh between 150 g and 210 g?
(b) How many apples weigh less than 150 g? 3 marks Band 3-4
1.3 A factory produces light bulbs whose lifetime (hours) is normally distributed with μ = 1,200 and σ = 90.
(a) What lifetime defines the bottom 2.5% of bulbs (i.e. the cutoff below which only 2.5% of bulbs fall)?
(b) If the factory makes 20,000 bulbs per month, about how many will fail before the cutoff in part (a)?
(c) State, in one sentence, what 99.7% of bulbs' lifetimes will lie between. 4 marks Band 4
2. Extended response
2.1 A NSW catchment-management authority is testing the effectiveness of a new river-monitoring system. The system reports water-temperature readings (°C). Over a long pilot period the daily reading at 9am is normally distributed with μ = 18.5°C and σ = 1.5°C.
Decision thresholds: A reading below 15.5°C triggers a "cold-snap" alert. A reading above 21.5°C triggers a "heat-warning" alert.
Annual data: The system takes one reading every day of the year (n = 365).
(a) Express both thresholds (15.5°C and 21.5°C) as "μ ± kσ", and find the percentage of days that trigger each alert.
(b) Estimate how many cold-snap and heat-warning alerts the system will issue in one year.
(c) The authority is considering tightening the heat-warning threshold so that no more than ≈ 2.5% of days trigger the warning. What new threshold (in °C) achieves this? Explain in one sentence why this would be a safer setting for a heat-sensitive ecosystem. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — correctly converts both thresholds to μ ± kσ (μ − 2σ and μ + 2σ).
• 1 mark — correct percentage for cold-snap (2.5%).
• 1 mark — correct percentage for heat-warning (2.5%).
Part (b) — 2 marks
• 1 mark — cold-snap count: 0.025 × 365 ≈ 9 days.
• 1 mark — heat-warning count: 0.025 × 365 ≈ 9 days.
Part (c) — 2 marks
• 1 mark — recognises 2.5% threshold corresponds to μ + 2σ (= 18.5 + 3 = 21.5°C — already at 2.5%). So if the authority wants "no more than 2.5%" the threshold of 21.5°C already does that exactly; tightening to anything tighter would reduce alerts further. Acceptable alternative: realise the question may intend a smaller fraction → use μ + 3σ = 23°C for ≈ 0.15%.
• 1 mark — explains the conservation reasoning (heat-sensitive species; fewer false alarms vs more catch rate is a trade-off).
Your response:
Stuck on (c)? Notice the current 21.5°C threshold is already at μ + 2σ → already 2.5%.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Test marks μ = 60, σ = 8 (3 marks)
(a) Sample response. 52 = 60 − 8 = μ − 1σ. 76 = 60 + 16 = μ + 2σ. The interval 52 to 76 covers μ − 1σ → μ + 2σ. Percentage = 68%/2 (lower half of the inner band) + 95%/2 (upper half of the wider band) = 34% + 47.5% = 81.5%.
(b) Sample response. Above 76 = above μ + 2σ = (100 − 95)/2 = 2.5%.
Marking notes. 2 marks for (a) — students often forget to split the 68% and 95% bands into halves around the mean; one mark for the correct setup, one for the correct percentage. 1 mark for (b).
1.2 — Apples (3 marks)
(a) Sample response. 150 = 180 − 30 = μ − 2σ. 210 = 180 + 30 = μ + 2σ. Between μ ± 2σ → 95%. Count = 0.95 × 400 = 380 apples.
(b) Sample response. Below μ − 2σ → 2.5%. Count = 0.025 × 400 = 10 apples.
Marking notes. 1 mark for correctly identifying μ ± 2σ in (a). 1 mark for converting 95% to 380 apples. 1 mark for (b) (both the 2.5% and the count).
1.3 — Light bulbs (4 marks)
(a) Bottom 2.5% sits below μ − 2σ = 1,200 − 180 = 1,020 hours.
(b) 2.5% × 20,000 = 500 bulbs fail before 1,020 hours each month.
(c) 99.7% of bulb lifetimes will lie between μ − 3σ = 1,200 − 270 = 930 hours and μ + 3σ = 1,200 + 270 = 1,470 hours.
Marking notes. 1 mark for the 1,020-hour cutoff. 1 mark for the 500-bulb count with working. 1 mark for the 930-1,470 hour range. 1 mark for using "between" correctly and citing 99.7% (instead of vague "almost all").
2.1 — River-temperature monitoring (7 marks): Band-6 sample with annotations
(a) Thresholds and alert percentages.
15.5 = 18.5 − 3 = 18.5 − 2(1.5) = μ − 2σ. 21.5 = 18.5 + 3 = 18.5 + 2(1.5) = μ + 2σ. [1 mark — both thresholds converted.]
Cold-snap (below μ − 2σ) = (100 − 95)/2 = 2.5%. [1 mark.]
Heat-warning (above μ + 2σ) = (100 − 95)/2 = 2.5%. [1 mark.]
(b) Annual alert counts.
Cold-snap = 0.025 × 365 = 9.125 → about 9 days/year. [1 mark.]
Heat-warning = 0.025 × 365 = 9.125 → about 9 days/year. [1 mark.]
(c) Tightening the heat warning to ≤ 2.5%.
The current threshold of 21.5°C already sits at μ + 2σ, which by the empirical rule already triggers on only ≈ 2.5% of days. To genuinely tighten further, the authority should set the threshold at μ + 3σ = 18.5 + 4.5 = 23.0°C, which would reduce alerts to (100 − 99.7)/2 ≈ 0.15% — roughly 1 alert every 1.8 years instead of 9 per year. [1 mark — recognises current threshold is already 2.5% and identifies a meaningful tighter cutoff.]
For a heat-sensitive ecosystem, the tighter threshold means warnings are issued only for genuinely exceptional heat events, reducing alert fatigue among staff and ensuring that any warning issued is a high-confidence signal that the river is in true thermal stress. [1 mark — links the new threshold to a sensible ecological reason.]
Total: 7/7.
Band descriptors for marker.
Band 3: Converts one threshold to μ ± kσ; partial alert percentages. ≈ 2-3 marks.
Band 4: Both percentages correct (2.5%); counts attempted but not labelled with units; (c) chooses a threshold but with no reasoning. ≈ 4-5 marks.
Band 5: All numerical parts correct including the count; (c) chooses 23.0°C but justification is vague. ≈ 6 marks.
Band 6: Full numerical solution AND notices that 21.5°C is already 2.5%, then proposes a defensible alternative (23.0°C / μ + 3σ) with a one-sentence ecological justification. 7/7.