Mathematics Advanced • Year 12 • Module 5 • Lesson 12
The Normal Distribution
Apply z-scores and the empirical rule to real contexts: apple weights, bolt tolerances, IQ classifications, exam comparison, and outlier detection.
Problem 1 — Orchard apples (empirical rule)
The weights of apples from an orchard are normally distributed with μ = 150 g and σ = 15 g.
Set up: What are we solving for?
(i) Calculate the z-score for an apple weighing 180 g, and state in one sentence what it tells you about that apple. 1 mark
(ii) Use the empirical rule to estimate the percentage of apples weighing between 120 g and 180 g. 2 marks
(iii) A grocer rejects apples weighing more than 195 g (oversize). Use the empirical rule to estimate the percentage rejected. 2 marks
Stuck on (iii)? 195 = μ + 3σ; the upper tail beyond μ + 3σ is half of (100% − 99.7%) = 0.15%.Problem 2 — Comparing two students (cross-distribution)
Alex scores 72 in Mathematics, where the cohort distribution is N(60, 12²). Blake scores 68 in English, where the cohort distribution is N(55, 10²).
Set up: What are we solving for?
(i) Calculate the z-score for each student. 2 marks
(ii) State which student performed better relative to their cohort, and justify in one sentence. 1 mark
(iii) Use the empirical rule to estimate the percentage of the Maths cohort that scored below Alex. 2 marks
Problem 3 — Bolt tolerance (manufacturing inverse problem)
A manufacturer produces bolts with lengths normally distributed with μ = 50 mm and σ = 0.5 mm. Bolts are rejected if their length differs from the target by more than 1 mm.
Set up: What are we solving for?
(i) Find the z-scores for the upper and lower rejection boundaries. 2 marks
(ii) Use the empirical rule to estimate the percentage of bolts rejected. 2 marks
(iii) The manufacturer wants to reduce the rejection rate to approximately 0.3% (the empirical-rule cut-off for 3 SDs). Find the new σ required (keeping μ = 50 mm). 2 marks
Stuck on (iii)? Set 3σ_new = 1 (the half-width tolerance).Problem 4 — IQ classification bands
Adult IQ scores are I ~ N(100, 15²). A psychologist classifies the population as follows:
• "Above average": I > 115
• "Gifted": I > 130
• "Highly gifted": I > 145
Set up: What are we solving for?
(i) Express each cut-off as μ + kσ for an integer k, and use the empirical rule to estimate the percentage in each category. 3 marks
(ii) In a town of 50 000 adults, estimate how many are "highly gifted" (I > 145). 2 marks
(iii) Explain in one sentence why the term "above average" (z > 1) refers to substantially more people than "gifted" (z > 2), even though both are above the mean. 1 mark
Problem 5 — Outlier detection (|z| > 3)
A laboratory measures reaction times of 4 000 trials. The times are approximately normal with μ = 1.20 s and σ = 0.18 s. The lab flags any measurement with |z| > 3 as a probable outlier (instrumentation error or distracted subject).
Set up: What are we solving for?
(i) Convert the |z| > 3 cut-off back to raw reaction times. Express the result as t < ____ or t > ____. 2 marks
(ii) Use the empirical rule to estimate the expected number of measurements flagged out of 4 000. 2 marks
(iii) A junior researcher flags every reading with |z| > 1 instead. Estimate the number flagged under this stricter rule, and explain in one sentence why this rule is unhelpful for detecting genuine errors. 2 marks
Stuck on (iii)? About 32% of normal data has |z| > 1 by chance — that's not "errors", that's normal variability.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Orchard apples
Set up. Convert apple weights to z-scores and use the empirical rule to estimate proportions inside and outside chosen intervals.
(i) z = (180 − 150)/15 = 30/15 = 2. The apple is 2 standard deviations above the mean — heavier than most.
(ii) 120 = 150 − 2(15) = μ − 2σ; 180 = μ + 2σ. So P(120 < X < 180) ≈ 95%.
(iii) 195 = 150 + 3(15) = μ + 3σ. P(X > μ + 3σ) ≈ (100% − 99.7%)/2 = 0.15%. About 1 apple in 670 is rejected as oversize.
Problem 2 — Comparing two students
Set up. Use z-scores to put scores from two different normal distributions on the same scale and to read off where each score sits.
(i) z_Alex = (72 − 60)/12 = 1.0. z_Blake = (68 − 55)/10 = 13/10 = 1.3.
(ii) Blake performed better relative to their cohort because 1.3 > 1.0 (further above the cohort mean in SD units).
(iii) Alex's z = 1, i.e. x = μ + σ. By symmetry, P(X < μ + σ) ≈ 50% + 34% = 84% of the Maths cohort scored below Alex.
Problem 3 — Bolt tolerance
Set up. Convert the ± 1 mm tolerance to z-scores, read the rejection rate from the empirical rule, then invert to find a tighter σ.
(i) Upper: z = (51 − 50)/0.5 = +2. Lower: z = (49 − 50)/0.5 = −2.
(ii) Rejection means |z| > 2, i.e. outside μ ± 2σ. By the empirical rule, P(outside μ ± 2σ) ≈ 100% − 95% = 5% (about 2.5% in each tail).
(iii) Want rejection ≈ 0.3%, i.e. tolerance = μ ± 3σ_new. So 3σ_new = 1 ⇒ σ_new = 1/3 ≈ 0.33 mm. The machine must be tightened from σ = 0.5 mm down to about 0.33 mm.
Problem 4 — IQ classification bands
Set up. Express each cut-off as μ + kσ, read the upper-tail proportion from the empirical rule, then scale to a real population.
(i) 115 = 100 + 15 = μ + σ ⇒ P(I > 115) ≈ 16% (above-average). 130 = μ + 2σ ⇒ P(I > 130) ≈ 2.5% (gifted). 145 = μ + 3σ ⇒ P(I > 145) ≈ 0.15% (highly gifted).
(ii) Expected count ≈ 0.0015 × 50 000 = 75 adults.
(iii) The normal curve has its highest density near the mean, so the "thin" tail beyond μ + 2σ contains far fewer people than the wider region between μ + σ and μ + 2σ — moving the cut-off out one more SD slashes the proportion roughly tenfold (16% → 2.5%).
Problem 5 — Outlier detection
Set up. Convert the |z| > 3 outlier rule to raw seconds, scale to 4 000 trials, then compare with a too-strict |z| > 1 rule.
(i) |z| > 3 ⇔ t < μ − 3σ or t > μ + 3σ. μ − 3σ = 1.20 − 0.54 = 0.66; μ + 3σ = 1.20 + 0.54 = 1.74. So flagged: t < 0.66 s or t > 1.74 s.
(ii) P(|z| > 3) ≈ 100% − 99.7% = 0.3%. Expected flags ≈ 0.003 × 4 000 = 12 measurements.
(iii) P(|z| > 1) ≈ 100% − 68% = 32%. Expected flags ≈ 0.32 × 4 000 = 1 280 measurements. This is unhelpful because ~32% of perfectly normal data lies outside μ ± σ by ordinary variability — the rule would flag almost a third of all readings as "outliers", drowning genuine errors in noise.