Mathematics Advanced • Year 12 • Module 5 • Lesson 12
The Normal Distribution
Practise HSC-style writing on the empirical rule, z-scores, and cross-distribution comparison — including an extended analytical response.
1. Short-answer questions
1.1 The masses of newborn babies in a hospital are normally distributed with μ = 3.4 kg and σ = 0.5 kg. Use the empirical rule to estimate the percentage of babies with mass between 2.4 kg and 4.4 kg. 2 marks Band 3
1.2 A standardised test has scores T ~ N(500, 100²). (a) Find the z-score of a student who scored 720. (b) Use the empirical rule to estimate the percentage of students who scored above 700. 3 marks Band 3-4
1.3 Anna scores 78 on a Maths test with cohort mean 65 and SD 10. Brett scores 84 on a Chemistry test with cohort mean 70 and SD 8.
(a) Calculate each student's z-score. (b) State who performed better relative to their cohort, justifying with both z-scores. (c) Convert each z-score into the approximate percentile of the corresponding cohort using the empirical rule. 4 marks Band 4
2. Extended response
2.1 A manufacturer fills bottles with sports drink. The fill volume X is normally distributed with mean μ = 500 mL and standard deviation σ = 4 mL. The label states "500 mL minimum"; a bottle is "underfilled" if X < 488 mL and "overfilled" if X > 512 mL.
(a) Use the empirical rule to estimate the percentage of bottles that are (i) underfilled, (ii) overfilled, (iii) within the acceptable range 488 ≤ X ≤ 512.
(b) The factory operates a quality controller who weighs random bottles. A bottle measuring 495 mL is rejected. Calculate the z-score of this bottle and explain in one sentence why rejecting it is unreasonable.
(c) The marketing team asks: "If we raised μ from 500 to 504 mL (keeping σ = 4), how would the underfill rate change?" Recalculate the underfill rate at the new μ using the empirical rule and quantify the improvement.
(d) Discuss in 2-3 sentences the trade-off the company faces between increasing μ (less underfill, more product cost) and decreasing σ (tighter machine, higher capital cost). Refer to both your calculations and the empirical rule. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — identifies 488 = μ − 3σ and 512 = μ + 3σ.
• 1 mark — states (i) ≈ 0.15%, (ii) ≈ 0.15%, (iii) ≈ 99.7%.
Part (b) — 2 marks
• 1 mark — computes z = (495 − 500)/4 = −1.25.
• 1 mark — explains that |z| = 1.25 is well inside μ ± 2σ (the 95% range), so this fill is ordinary variability, not a defect.
Part (c) — 2 marks
• 1 mark — re-expresses 488 as μ_new − 4σ (since μ_new = 504, σ = 4 ⇒ 488 is 4 SDs below new mean).
• 1 mark — states the new underfill rate is well below the 0.15% (3-SD) value; the empirical rule does not give 4 SDs exactly, but P(z < −4) ≈ 0.003%, a roughly 50× reduction.
Part (d) — 1 mark
• 1 mark — names both options, quantifies one numerically (e.g. "raising μ by 1 mL costs 1 mL extra product per bottle"), and concludes that the choice depends on the marginal cost of product vs. the capital cost of a tighter machine.
Your response:
Stuck on (d)? Compare "shift the mean" (rebadge label or buy extra product) with "tighten the spread" (replace the filling machine).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Newborn masses, μ = 3.4, σ = 0.5, interval 2.4-4.4 kg (2 marks)
Sample response. 2.4 = 3.4 − 2(0.5) = μ − 2σ; 4.4 = 3.4 + 2(0.5) = μ + 2σ. By the empirical rule, P(μ − 2σ < X < μ + 2σ) ≈ 95%.
Marking notes. 1 mark — correctly expresses both endpoints as μ ± 2σ. 1 mark — quotes 95% (NOT 68% or 99.7%) and gives a clear final sentence. A response that says only "95%" with no boundary work is 1/2.
1.2 — T ~ N(500, 100²): z(720) and P(T > 700) (3 marks)
Sample response. (a) z = (720 − 500)/100 = 2.2. (b) 700 = 500 + 2(100) = μ + 2σ, so P(T > 700) ≈ (100% − 95%)/2 = 2.5%.
Marking notes. (a) 1 mark — correct z, including arithmetic. (b) 1 mark — identifies 700 as μ + 2σ; 1 mark — computes upper-tail percentage 2.5% (NOT 5%, which is the two-tailed value).
1.3 — Anna vs Brett, z-scores and percentiles (4 marks)
Sample response.
(a) z_Anna = (78 − 65)/10 = 1.3. z_Brett = (84 − 70)/8 = 14/8 = 1.75.
(b) Brett performed better relative to their cohort because 1.75 > 1.3 — Brett is further above their cohort mean in SD units.
(c) Anna's z = 1.3 lies between μ + σ (≈ 84th percentile) and μ + 2σ (≈ 97.5th percentile), so Anna is roughly at the ~90th percentile of Maths. Brett's z = 1.75 lies closer to μ + 2σ, so Brett is roughly at the ~96th percentile of Chemistry.
Marking notes. (a) 1 mark — both z-scores correct. (b) 1 mark — correct comparison with a one-sentence justification referring to z. (c) 1 mark per percentile estimate (2 marks). Common error: students confuse "raw score difference" (Brett's 84 > Anna's 78) with "relative performance" — the larger raw score is meaningless without the cohort SD.
2.1 — Bottle fill extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — empirical rule on the 488-512 range. 488 = 500 − 3(4) = μ − 3σ and 512 = 500 + 3(4) = μ + 3σ. [1 mark — boundaries identified.]
By the 68-95-99.7 rule, P(μ − 3σ < X < μ + 3σ) ≈ 99.7%, so:
(i) underfilled ≈ 0.15%, (ii) overfilled ≈ 0.15%, (iii) within range ≈ 99.7%. [1 mark.]
Part (b) — single 495 mL bottle. z = (495 − 500)/4 = −1.25. [1 mark.]
Since |z| = 1.25 lies well inside μ ± 2σ (which captures 95% of all bottles), a fill of 495 mL is well within ordinary machine variability — not a defect. Rejecting it is unreasonable because such fills occur in about 1 bottle in 5 by chance alone. [1 mark.]
Part (c) — μ raised to 504. Now 488 = 504 − 4(4) = μ_new − 4σ. [1 mark — 4-SD identification.]
The empirical rule does not give a 4-SD value directly, but extending the same logic, the proportion beyond μ − 4σ is far smaller than 0.15% — about 0.003% from tables (a ≈ 50× reduction). Practically, raising the mean by 1 SD (4 mL) wipes out underfilling almost entirely. [1 mark — quantified reduction.]
Part (d) — trade-off. The company has two levers: shift μ up (cheap to implement, but costs 1 mL extra product per bottle for every 1 mL of headroom) or shrink σ (requires capital investment in a more precise filling machine, but reduces underfill without giving away extra product). At high production volumes the per-bottle product cost dominates, so a precise machine pays for itself; at low volumes, simply raising μ by 1-2 SD is the rational choice. [1 mark — both levers named, one quantified, conclusion conditional on volume.]
Total: 7/7.
Band descriptors for marker.
Band 3: Identifies 488 and 512 as ± kσ but mismatches k (uses k = 2 instead of 3), gives part-correct percentages. ≈ 2-3 marks.
Band 4: Part (a) correct; computes the z in (b) but does not explain why rejection is unreasonable. Part (c) attempted but no quantification of the improvement. ≈ 4-5 marks.
Band 5: All numerical parts correct, (b) explains in terms of empirical rule, (c) gives a directional improvement but not a numerical estimate. Part (d) names only one lever. ≈ 5-6 marks.
Band 6: Full numerical work, both levers in (d) explicitly compared, the conclusion is conditional on production volume or includes a quantified product-vs-capital trade-off. 7/7.