Mathematics Standard • Year 12 • Module 8 • Lesson 7
Z-Scores
Apply z-scores to real comparison problems: tests with different scales, IQ and ATAR-style admissions, and converting standardised scores back into raw values.
Problem 1 — Maths vs English (different scales)
A Sydney student, Mei, sits two tests in the same week.
Maths: Mei scores 82. Class mean = 70, SD = 8.
English: Mei scores 75. Class mean = 65, SD = 5.
Set up: What are we solving for?
(i) Calculate Mei's z-score in Maths. 1 mark
(ii) Calculate Mei's z-score in English. 1 mark
(iii) In which subject did Mei perform better relative to her class? Justify with a clear conclusion sentence. 2 marks
Stuck? Revisit lesson § Comparing Across Distributions — bigger z = better relative performance.Problem 2 — Scholarship selection (comparing schools)
A university awards a scholarship to students whose performance is at least 1.5 standard deviations above their school's mean.
Selective High: mean ATAR = 88, SD = 6. Applicant scores ATAR 96.
Comprehensive High: mean ATAR = 72, SD = 10. Applicant scores ATAR 90.
Set up: What are we solving for?
(i) Calculate the z-score for the Selective High applicant. 1 mark
(ii) Calculate the z-score for the Comprehensive High applicant. 1 mark
(iii) Which applicants (if any) meet the z ≥ 1.5 threshold? State your conclusion in one sentence. 2 marks
Stuck? Revisit lesson § Z-Score Formula — apply z = (x − mean)/SD to each applicant separately.Problem 3 — From percentile back to raw mark
A physics test has mean 64 and standard deviation 12.
Set up: What are we solving for?
(i) Convert z = −1.25 back to a raw mark on this test. 1 mark
(ii) Convert z = 1.75 back to a raw mark on this test. 1 mark
(iii) A school awards a "distinguished achiever" certificate to any student with z ≥ 2 on the test. What is the minimum raw mark required? 3 marks
Stuck? Revisit lesson § Comparing — use x = mean + z × SD; for a minimum, use the boundary value z = 2.Problem 4 — IQ scores and unusualness
IQ tests are scaled so the population mean is 100 and the standard deviation is 15. Three people sit a test.
Aiden: IQ = 130
Beth: IQ = 100
Cleo: IQ = 70
Set up: What are we solving for?
(i) Calculate the z-score for each of the three people. 2 marks
(ii) Which of the three scores would be classified as unusual using the rule |z| > 2? Justify. 2 marks
(iii) A "Gifted" program admits anyone with z > 2.0. What is the minimum IQ score needed to qualify? 1 mark
Stuck? Revisit lesson § Interpreting Z-Scores — the |z| > 2 rule means about 2.5% are below or above respectively.Problem 5 — Ranking three students across subjects
Three students each sit a different subject. The ranking committee needs to compare relative performance.
Sarah — Maths: raw mark 78, class mean 70, SD 8.
Tom — Science: raw mark 82, class mean 75, SD 6.
Emma — History: raw mark 85, class mean 80, SD 12.
Set up: What are we solving for?
(i) Calculate the z-score for each student. Show one line of substitution per student. 3 marks
(ii) Rank the three students from best to worst relative performance. 1 mark
(iii) Emma has the highest raw mark but is ranked lowest by z-score. In one or two sentences, explain why this is mathematically sensible. 2 marks
Stuck? Revisit lesson § Worked Example — Sarah/Tom/Emma. Different means and SDs mean raw marks are not directly comparable.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Mei's two tests
Set up. We are converting both raw marks to z-scores so we can compare on the same scale.
(i) Maths z = (82 − 70) ÷ 8 = 12 ÷ 8 = 1.50.
(ii) English z = (75 − 65) ÷ 5 = 10 ÷ 5 = 2.00.
(iii) English z (2.00) > Maths z (1.50), so Mei performed better in English relative to her class. Even though her raw Maths mark was higher, her position above the class mean is greater in English.
Problem 2 — Scholarship selection
Set up. We need z for each applicant, then compare against the threshold 1.5.
(i) Selective High: z = (96 − 88) ÷ 6 = 8 ÷ 6 ≈ 1.33.
(ii) Comprehensive High: z = (90 − 72) ÷ 10 = 18 ÷ 10 = 1.80.
(iii) The Comprehensive High applicant meets the threshold (z = 1.80 ≥ 1.5); the Selective High applicant does not (z ≈ 1.33). (This illustrates why z-scores are used in selection — the Comprehensive High student stood out further from their own peers than the Selective High student did from theirs.)
Problem 3 — Converting back (mean 64, SD 12)
Set up. Use x = mean + z × SD.
(i) x = 64 + (−1.25)(12) = 64 − 15 = 49.
(ii) x = 64 + (1.75)(12) = 64 + 21 = 85.
(iii) Minimum at z = 2: x = 64 + 2(12) = 64 + 24 = 88. A student needs at least 88 to be a "distinguished achiever".
Problem 4 — IQ scores
Set up. Mean = 100, SD = 15. Compute z = (x − 100)/15 for each.
(i) Aiden: z = (130 − 100)/15 = 30/15 = 2.00. Beth: z = 0/15 = 0. Cleo: z = (70 − 100)/15 = −30/15 = −2.00.
(ii) Using |z| > 2: Aiden's |z| = 2 (boundary), Cleo's |z| = 2 (boundary). Strictly > 2 means none are unusual by this rule; but both Aiden and Cleo are at the boundary, i.e. tied for "unusual". Beth's z = 0 is exactly average.
(iii) Minimum IQ for z > 2.0: x > 100 + 2.0 × 15 = 100 + 30 = 130 (so above 130).
Problem 5 — Sarah, Tom, Emma
Set up. Convert each raw mark to a z-score so the three different subjects can be compared on the same scale.
(i) Sarah: z = (78 − 70)/8 = 8/8 = 1.00. Tom: z = (82 − 75)/6 = 7/6 ≈ 1.17. Emma: z = (85 − 80)/12 = 5/12 ≈ 0.42.
(ii) Ranking: Tom (1.17) > Sarah (1.00) > Emma (0.42).
(iii) Emma's raw mark is highest, but her test had the largest SD (12), so the same raw difference of 5 above the mean is a smaller standardised gap than the differences for the other students. Her score is only about 0.42 of a standard deviation above her cohort, which is the smallest "standing" of the three.