Mathematics Advanced • Year 12 • Module 5 • Lesson 8
Comparing Data Sets
Build procedural fluency in z-scores, back-and-forth conversion between raw and standardised marks, and the "Centre → Spread → Shape" comparison framework.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Write the z-score formula and its inverse:
z = ____________________
x = ____________________
Q1.2 What does a z-score of 0 mean? What about z = 2? z = −1?
z = 0: ________________________________________
z = 2: ________________________________________
z = −1: _______________________________________
Q1.3 List the four headings used in the lesson's "exam-ready comparison framework":
________ → ________ → ________ → ________
2. Worked example — comparing Maths and English performance
Problem. A student scores 72% in Maths (class mean 65, SD 8) and 68% in English (class mean 60, SD 5). Which is the better performance relative to the class?
Step 1 — Recognise the situation.
Raw scores are on different scales (different means and SDs), so direct comparison is unfair. Convert each to a z-score.
Step 2 — Compute z for Maths.
z_Maths = (72 − 65) / 8 = 7/8 = 0.875
Reason: subtract the mean, divide by SD.
Step 3 — Compute z for English.
z_English = (68 − 60) / 5 = 8/5 = 1.6
Step 4 — Compare on the same standardised scale.
1.6 > 0.875 → English is further above the mean (in SD units).
Conclusion. The 68% in English is the better performance relative to its class. Despite the lower raw mark, the student outperformed English classmates by 1.6 SDs vs only 0.875 SDs in Maths.
3. Faded example — fill in the missing steps
A student wants z = 1.4 in a test with mean 65 and SD 8. What raw mark must they achieve? Fill in each blank. 3 marks
Step 1 — Recognise the situation. We are given z and asked for x, so use the inverse formula x = ______________________.
Step 2 — Substitute the values.
x = 65 + (______) × 8
Step 3 — Evaluate.
x = 65 + ______ = ____________
Conclusion. The required raw mark is ____________.
4. Graduated practice — z-scores and comparisons
Show your working. Quote z-scores to 2 decimal places.
Foundation — single-step calculations (4 questions)
| Q | Data & question | Answer |
|---|---|---|
| 4.1 1 | x = 85, x̄ = 78, s = 5. Find z. | |
| 4.2 1 | x = 62, x̄ = 70, s = 4. Find z. | |
| 4.3 1 | z = 1.4, x̄ = 65, s = 8. Find x. | |
| 4.4 1 | z = −0.6, x̄ = 100, s = 15. Find x. |
Standard — typical HSC difficulty (6 questions)
Show z-scores and one-sentence conclusions for each comparison.
4.5 Two schools sat the same exam. School X: mean 72, SD 10. School Y: mean 68, SD 6. A student scored 80 at School X and 76 at School Y. Which student performed better relative to their school? 2 marks
4.6 A test has mean 70, SD 12. State the raw marks corresponding to z = +1 and z = −1, and the raw marks corresponding to z = +2 and z = −2. 2 marks
4.7 A distribution has mean 50, median 45, mode 40. Describe its shape, and state whether mean > median, mean < median, or mean = median is expected for this shape. 2 marks
4.8 In a negatively skewed distribution, which is greater: the mean or the median? Explain in one sentence. 2 marks
4.9 Class A (median 68, IQR 12) and Class B (median 72, IQR 8). Write a two-sentence comparison using both centre and spread in context (Year 12 exam marks). 2 marks
4.10 A national test has mean 500 and SD 100. Convert the raw score 720 to a z-score. State the percentile-rank phrase you would use to describe a student with z ≈ 2.2 ("approximately the __th percentile"). 2 marks
Extension — combine concepts (2 questions)
4.11 A student's z-score in Physics is 1.2 and in Chemistry is 0.8. Without knowing means or SDs, state whether the Physics result is better relative to its class, and explain in one sentence why z-scores allow this comparison even though the raw scores are unknown. 3 marks
4.12 A test has mean 60 and SD 10. Sarah's raw mark is 75. If every raw mark in the class is then increased by 5 bonus marks (so Sarah now has 80, the class mean becomes 65, and the SD is unchanged), find Sarah's new z-score and explain in one sentence why it is the same as her old z-score. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — z-score formulas
z = (x − x̄) / s. x = x̄ + z · s.
Q1.2 — z-score meanings
z = 0: exactly average (raw mark equals the mean). z = 2: two SDs above the mean (very high, ≈ 98th percentile). z = −1: one SD below the mean (≈ 16th percentile).
Q1.3 — Comparison framework
Centre → Spread → Shape → Outliers.
Q3 — Faded example
Step 1: x = x̄ + z · s. Step 2: x = 65 + (1.4)(8). Step 3: x = 65 + 11.2 = 76.2. The required raw mark is 76.2.
Q4.1 — z for x = 85, x̄ = 78, s = 5
z = (85 − 78) / 5 = 1.40.
Q4.2 — z for x = 62, x̄ = 70, s = 4
z = (62 − 70) / 4 = −2.00.
Q4.3 — x for z = 1.4, x̄ = 65, s = 8
x = 65 + 1.4(8) = 65 + 11.2 = 76.2.
Q4.4 — x for z = −0.6, x̄ = 100, s = 15
x = 100 + (−0.6)(15) = 100 − 9 = 91.
Q4.5 — School X (80) vs School Y (76)
z_X = (80 − 72)/10 = 0.80. z_Y = (76 − 68)/6 = 1.33. School Y student performed better relative to their school (1.33 SDs above the mean vs 0.80 SDs).
Q4.6 — Raw marks at z = ±1, ±2 (mean 70, SD 12)
z = +1 → 70 + 12 = 82. z = −1 → 70 − 12 = 58. z = +2 → 70 + 24 = 94. z = −2 → 70 − 24 = 46. Roughly 95% of scores in a bell-shaped distribution fall in [46, 94].
Q4.7 — Shape with mean 50, median 45, mode 40
Mean > median > mode is characteristic of a positively skewed (right-skewed) distribution: a long tail to the right pulls the mean above the median.
Q4.8 — Negatively skewed
In a negatively skewed distribution, the long tail extends to the left (low values), pulling the mean below the median: median > mean.
Q4.9 — Class A vs Class B comparison
Class B has a higher centre (median 72 vs 68), so its typical exam mark is higher. Class B is also more consistent (IQR 8 vs 12), meaning the middle 50% of Class B's marks are clustered more tightly than Class A's.
Q4.10 — National test, x = 720
z = (720 − 500)/100 = 2.20. A student with z ≈ 2.2 sits at approximately the 99th percentile (above ≈ 98.6% of the cohort).
Q4.11 — Physics z = 1.2 vs Chemistry z = 0.8
Yes — the Physics performance is better relative to its class. z-scores are unit-free standardised values measured in standard deviations from the mean; once two raw scores are converted, they live on a common scale and can be compared even when the original means, SDs, or scoring systems are completely different.
Q4.12 — Adding 5 bonus marks to every student
Old z: (75 − 60)/10 = 1.50. New z: (80 − 65)/10 = 1.50 — unchanged. Reason: adding the same constant to every mark shifts the class mean by the same constant, so the numerator (x − x̄) is unchanged; the SD is also unchanged because all pair-wise distances are unchanged. Sarah's relative standing in the class is exactly preserved.