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.