Mathematics Advanced • Year 12 • Module 5 • Lesson 13
Normal Distribution Applications
Apply CDF and inverse-normal techniques to real contexts: ball bearings, employee commutes, cereal box weights, school timetables, and medical screening.
Problem 1 — Ball-bearing quality control
Ball-bearing diameters D ~ N(12.00 mm, 0.05²). The acceptance window is 11.90 mm ≤ D ≤ 12.10 mm.
Set up: What are we solving for?
(i) Find the z-scores for the lower and upper acceptance limits. 2 marks
(ii) Find the percentage of bearings that are acceptable. [Use P(Z < 2) ≈ 0.9772.] 2 marks
(iii) Find the percentage that are too large (D > 12.10 mm). 1 mark
Stuck on (ii)? P(−2 < Z < 2) = P(Z < 2) − P(Z < −2) = 0.9772 − 0.0228.Problem 2 — Commute times and a work-from-home policy
Employees' commute times are T ~ N(35, 64) minutes (so σ = 8).
Set up: What are we solving for?
(i) What percentage of employees take more than 45 minutes? [Use P(Z < 1.25) ≈ 0.8944.] 2 marks
(ii) The company introduces a policy: anyone whose commute exceeds the 90th percentile can work from home. Find this threshold in minutes. [Use z ≈ 1.282.] 2 marks
(iii) If 500 employees are affected by this policy, estimate the company's total headcount. 2 marks
Problem 3 — Cereal box "at least 375 g"
A factory dispenses cereal X ~ N(μ, σ²) g per box. The label states "at least 375 g". The company wants fewer than 0.5% of boxes to be underweight.
Set up: What are we solving for?
(i) If the machine is set to μ = 380 g, find the maximum allowable σ (to 1 d.p.). [Use z = −2.576 for P(Z < z) = 0.005.] 2 marks
(ii) An inspector suggests μ = 385 g with σ = 4 g instead. Calculate the underweight rate under this proposal and state whether it meets the 0.5% target. 2 marks
(iii) Comment in one or two sentences on which setting is preferable, given that every extra average gram costs 3% more in raw material. 2 marks
Stuck on (i)? 375 = 380 + (−2.576)σ ⇒ σ = (380 − 375)/2.576.Problem 4 — Setting exam band boundaries
A statewide exam has scores S ~ N(70, 12²). The examination board sets band boundaries at percentiles: Band 4 begins at the 60th percentile, Band 5 at the 85th, Band 6 at the 95th.
Set up: What are we solving for?
(i) Find the score that marks the start of each band. [Use z values 0.253, 1.036, 1.645.] 3 marks
(ii) In a cohort of 65 000 students, estimate the number of Band 6 awards. 1 mark
(iii) A teacher complains that "the Band 6 cut-off should be raised to the 99th percentile". Find the new cut-off and the number of Band 6 awards under this change. [Use z ≈ 2.326.] 2 marks
Problem 5 — Medical screening trade-off
Resting heart rate of healthy adults is H ~ N(72, 10²) bpm. A clinic uses two thresholds in a screening rule:
• "Slightly elevated" if H exceeds the 90th percentile.
• "Significantly elevated" if H exceeds the 99th percentile.
Set up: What are we solving for?
(i) Find the two thresholds (in bpm). [Use z values 1.282 and 2.326.] 2 marks
(ii) If the clinic screens 1 200 healthy adults, estimate how many are flagged at each level. 2 marks
(iii) Explain in 1-2 sentences why the "90th percentile" cut-off has a much higher rate of false positives in healthy adults than the "99th percentile" cut-off, and what trade-off this implies for screening sensitivity. 2 marks
Stuck on (iii)? By definition, the 90th percentile flags 10% of healthy adults; the 99th percentile flags 1%.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ball bearings
Set up. Convert tolerance limits to z-scores, then read the acceptance probability from the standard normal.
(i) z_lower = (11.90 − 12.00)/0.05 = −2; z_upper = (12.10 − 12.00)/0.05 = +2.
(ii) P(−2 < Z < 2) = 0.9772 − 0.0228 ≈ 0.9544, so about 95.4% of bearings are acceptable.
(iii) P(D > 12.10) = P(Z > 2) ≈ 1 − 0.9772 = 0.0228 ≈ 2.3% too large.
Problem 2 — Commute times
Set up. Right-tail probability for the >45-min question; inverse normal for the 90th-percentile cut-off.
(i) z = (45 − 35)/8 = 1.25. P(T > 45) = 1 − 0.8944 = 0.1056 (≈ 10.6%).
(ii) P(T < t) = 0.90 ⇒ z = 1.282; t = 35 + 1.282(8) = 35 + 10.26 = 45.26 min.
(iii) 500 affected ≈ 10% of headcount. Total ≈ 500 / 0.10 = 5 000 employees.
Problem 3 — Cereal boxes
Set up. Use the inverse normal to find σ from the 0.5% underweight requirement, then test an alternative (μ, σ) configuration and compare cost.
(i) P(X < 375) = 0.005 ⇒ z = −2.576. 375 = 380 + (−2.576)σ ⇒ σ = 5/2.576 ≈ 1.94 ≈ 1.9 g. (To 1 d.p.)
(ii) z = (375 − 385)/4 = −2.5. P(Z < −2.5) ≈ 0.0062, i.e. underweight rate ≈ 0.62%. This fails the 0.5% target.
(iii) Setting (a) at μ = 380 g, σ ≤ 1.9 g does meet the target and uses 5 g less average product per box than setting (b). At 3% cost per gram, setting (a) is both cheaper and compliant — preferable to (b), which fails the target and also uses more cereal.
Problem 4 — Exam band cut-offs
Set up. Use the inverse normal at the 60th, 85th, 95th and 99th percentiles to convert percentile cut-offs into raw scores.
(i) Band 4 start: x = 70 + 0.253(12) ≈ 73.0. Band 5 start: x = 70 + 1.036(12) ≈ 70 + 12.43 = 82.4. Band 6 start: x = 70 + 1.645(12) ≈ 70 + 19.74 = 89.7.
(ii) 5% of 65 000 = 3 250 Band 6 awards.
(iii) 99th percentile: x = 70 + 2.326(12) ≈ 70 + 27.91 = 97.9. Number of Band 6 awards ≈ 1% × 65 000 = 650 students — about one-fifth of the previous figure.
Problem 5 — Heart-rate screening trade-off
Set up. Inverse normal to find both bpm thresholds, then estimate counts of healthy people flagged at each level.
(i) 90th percentile: H = 72 + 1.282(10) = 84.82 bpm. 99th percentile: H = 72 + 2.326(10) = 95.26 bpm.
(ii) "Slightly elevated" flags ≈ 10% × 1 200 = 120 healthy adults. "Significantly elevated" flags ≈ 1% × 1 200 = 12 healthy adults.
(iii) The 90th-percentile rule misclassifies ten times as many healthy adults as the 99th-percentile rule (by construction: 10% vs 1%). The trade-off is between sensitivity (catching every sick patient: easier with a low cut-off) and specificity (avoiding false alarms: easier with a high cut-off). Choosing the threshold is a clinical decision, not a purely mathematical one.