Mathematics Advanced • Year 12 • Module 5 • Lesson 13
Normal Distribution Applications
Practise HSC-style writing on normal CDF, inverse normal, and quality-control reasoning — including an extended applied response.
1. Short-answer questions
1.1 Reaction times in a driving test are T ~ N(0.85, 0.12²) seconds. Find P(T > 1.00). [Use P(Z < 1.25) ≈ 0.8944.] 2 marks Band 3
1.2 A university accepts the top 25% of applicants on a test S ~ N(60, 64). (a) Find the minimum acceptance score. [Use z ≈ 0.674.] (b) State the probability that a randomly chosen applicant scores between 60 and the cut-off. 3 marks Band 3-4
1.3 Battery lives are L ~ N(420, 25²) hours. The manufacturer offers a refund to any customer whose battery dies before the "guaranteed minimum life" stated on the packet. They want the refund rate to be at most 2.5%.
(a) Find the largest guaranteed minimum life they can claim. [Use z ≈ −1.96.]
(b) The marketing team wants to round the guarantee to the nearest 10 hours. State the rounded value and recompute the actual refund rate under the rounded value.
4 marks Band 4
2. Extended response
2.1 A cereal manufacturer dispenses boxes of cereal that are normally distributed in weight. The label states "at least 375 g". The company wants fewer than 0.5% of boxes to be underweight.
(a) If the machine is set to μ = 380 g, find the largest allowable σ (to 1 decimal place). [Use z = −2.576 for P(Z < z) = 0.005.]
(b) An inspector suggests setting μ = 385 g with σ = 4 g instead. Calculate the percentage of underweight boxes under this proposal and state whether it meets the 0.5% target.
(c) The marketing team wants to be 99% confident that no more than 1 box in 200 (i.e. 0.5%) is underweight. They consider two practical changes: option X (raise μ from 380 to 382 g, keep σ = 1.9 g) or option Y (keep μ = 380 g but reduce σ to 1.5 g through a machine upgrade). Find the underweight rate under each option and recommend which option to choose if extra raw material costs 3% per extra average gram and the upgrade costs a one-off $40 000.
(d) The CEO insists on guaranteeing 380 g on the label (rather than 375 g) while keeping the underweight rate at 0.5%. Calculate the minimum μ required if σ stays at 1.9 g, and comment in one sentence on whether this is a sensible business decision.
8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — sets up 375 = 380 + (−2.576)σ.
• 1 mark — σ = 5/2.576 ≈ 1.94 g; reports 1.9 g to 1 d.p.
Part (b) — 2 marks
• 1 mark — z = (375 − 385)/4 = −2.5; P(Z < −2.5) ≈ 0.0062.
• 1 mark — states underweight rate ≈ 0.62% > 0.5%, so it fails the target.
Part (c) — 3 marks
• 1 mark — option X: z = (375 − 382)/1.9 ≈ −3.68; P(Z < −3.68) ≈ 0.0001 = 0.01% (well within target).
• 1 mark — option Y: z = (375 − 380)/1.5 ≈ −3.33; P(Z < −3.33) ≈ 0.0004 = 0.04% (also within target).
• 1 mark — cost comparison: option X has recurring cost ≈ 2 g × 3% × (volume) per box, option Y has a one-off $40 000. Recommendation depends on production volume; at high volume the recurring cost dominates, so option Y wins.
Part (d) — 1 mark
• 1 mark — needs 380 = μ + (−2.576)(1.9), so μ = 380 + 4.89 = 384.9 g; commentary: this gives away 4.9 g of free cereal per box vs the "375 g" label, a significant ongoing cost — sensible only if competitors can't match or if the brand premium justifies it.
Your response:
Stuck on (c)? Compute the z-score for each option separately, then compare cost vs benefit.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Reaction time T ~ N(0.85, 0.0144), P(T > 1.00) (2 marks)
Sample response. z = (1.00 − 0.85)/0.12 = 1.25. P(T > 1.00) = 1 − P(Z < 1.25) ≈ 1 − 0.8944 = 0.1056 (about 10.6%).
Marking notes. 1 mark — correct z. 1 mark — correct right-tail probability (uses 1 − the table value, not the table value itself). Common error: students quote 0.8944 directly as the answer.
1.2 — Top 25% cut-off for S ~ N(60, 64) (3 marks)
Sample response. (a) P(S > s) = 0.25 ⇒ P(Z < z) = 0.75 ⇒ z ≈ 0.674. s = 60 + 0.674(8) = 60 + 5.39 = 65.39 ≈ 65.4. (b) P(60 < S < 65.4) = 0.75 − 0.5 = 0.25 (25%) — by symmetry, the range from the mean to the 75th percentile is exactly a quarter of the distribution.
Marking notes. (a) 1 mark — correct cumulative probability 0.75; 1 mark — correct s. (b) 1 mark — explanation OR direct probability with reasoning (a bare 0.25 with no justification is 0.5).
1.3 — Battery guaranteed minimum life (4 marks)
Sample response.
(a) P(L < l_guarantee) = 0.025 ⇒ z = −1.96. l = 420 + (−1.96)(25) = 420 − 49 = 371 hours.
(b) Rounded to nearest 10 = 370 hours. Recompute: z = (370 − 420)/25 = −2. P(Z < −2) ≈ 0.0228 = 2.28%. Refund rate falls slightly (from 2.5% to 2.28%), so rounding down to 370 is safe for the company.
Marking notes. (a) 1 mark — correct setup with z = −1.96; 1 mark — correct l. (b) 1 mark — correct rounding direction (down, not up — rounding up would raise the refund rate above 2.5%); 1 mark — correct recomputed refund rate.
2.1 — Cereal box extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — μ = 380, find max σ. P(X < 375) = 0.005 ⇒ z = −2.576. [1 mark — set-up.]
375 = 380 + (−2.576)σ ⇒ σ = 5/2.576 ≈ 1.94 g, i.e. 1.9 g to 1 d.p. [1 mark.]
Part (b) — μ = 385, σ = 4. z = (375 − 385)/4 = −2.5. P(Z < −2.5) ≈ 0.0062. [1 mark.]
Underweight rate ≈ 0.62%, which is greater than the 0.5% target — this setting fails. [1 mark.]
Part (c) — option X vs option Y.
Option X (μ = 382, σ = 1.9): z = (375 − 382)/1.9 ≈ −3.68, so underweight rate ≈ P(Z < −3.68) ≈ 0.012% ≪ 0.5%. [1 mark.]
Option Y (μ = 380, σ = 1.5): z = (375 − 380)/1.5 ≈ −3.33, so underweight rate ≈ P(Z < −3.33) ≈ 0.04% ≪ 0.5%. [1 mark.]
Both options easily meet the target. Option X dispenses an extra 2 g of product per box at 3% per gram = 6% extra raw-material cost forever. Option Y costs a one-off $40 000 but no recurring per-box cost. For high production volumes, option Y wins; the upgrade pays for itself once the cumulative 6% premium exceeds $40 000. [1 mark — comparison + conditional recommendation.]
Part (d) — guarantee 380 g instead of 375. P(X < 380) = 0.005 ⇒ z = −2.576. 380 = μ + (−2.576)(1.9) ⇒ μ = 380 + 4.89 = 384.9 g. Giving away ≈ 4.9 g of free cereal per box is a substantial ongoing cost; this is sensible only if marketing analysis shows the "380 g" claim drives enough premium pricing or extra sales to cover the increased raw-material cost. [1 mark — μ and one-sentence business commentary.]
Total: 8/8.
Band descriptors for marker.
Band 3: Parts (a) and (b) attempted; arithmetic errors in z but correct method. ≈ 3-4 marks.
Band 4: Parts (a)-(b) correct, (c) computes one option only or both with no cost comparison. ≈ 4-5 marks.
Band 5: All four parts attempted, (c) recommends an option but without quantitative cost analysis. ≈ 6-7 marks.
Band 6: Full numerical and contextual work, (c) compares cost structures and gives a conditional recommendation, (d) closes with sensible business commentary. 8/8.