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:
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.