Mathematics Standard • Year 12 • Module 8 • Lesson 7

Z-Scores

Practise HSC Mathematics Standard 2-style writing on z-scores — multi-mark short answers and one structured extended response.

Master · Past-Paper Style

1. Short-answer questions

1.1 A Year 12 Maths test has a mean of 65 and a standard deviation of 12. Jamie scored 83.
(a) Calculate Jamie's z-score.
(b) Interpret the z-score in one sentence in context. 3 marks Band 3

1.2 A driving theory test has mean 72 and SD 9. A pass requires a z-score of at least −0.5.
(a) Calculate the minimum raw mark needed to pass.
(b) A candidate scores 64. Calculate their z-score and state whether they pass. 3 marks Band 3-4

1.3 Three students sit different HSC trial subjects. Their results are:

Ana — Maths: 78 (class mean 70, SD 8).

Ben — Science: 85 (class mean 80, SD 5).

Carla — English: 82 (class mean 72, SD 10).

(a) Calculate the z-score for each student.
(b) The top 10% of standardised performance approximately corresponds to z > 1.28. State which students (if any) qualify, and write one sentence justifying your decision. 4 marks Band 4

Stuck on 1.3(b)? Compare each z-score to 1.28 directly — bigger is better.

2. Extended response

2.1 A NSW university uses z-scores from HSC subjects to decide a scholarship. The cutoff is z ≥ 1.5 within each school.

School P (selective): mean HSC mark = 86, SD = 5. Applicant Pia scored 94.

School Q (comprehensive): mean HSC mark = 70, SD = 12. Applicant Quan scored 92.

(a) Calculate Pia's z-score and state whether she meets the cutoff.
(b) Calculate Quan's z-score and state whether he meets the cutoff.
(c) Quan has a lower raw mark than Pia but a higher z-score. In a clear paragraph (4–6 sentences) explain in plain English why this is mathematically reasonable, and state which student the university would award the scholarship to (with justification). 7 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — correct substitution and value for Pia's z.

• 1 mark — explicit statement of whether she meets z ≥ 1.5.

Part (b) — 2 marks

• 1 mark — correct substitution and value for Quan's z.

• 1 mark — explicit statement of whether he meets z ≥ 1.5.

Part (c) — 3 marks

• 1 mark — refers to the different means and SDs across the two schools.

• 1 mark — explains that z measures standing within school, not absolute mark.

• 1 mark — explicit, justified award decision naming the winning student.

Your response:

Stuck on (c)? Write: "Even though Quan's raw mark of 92 is lower than Pia's 94, his z-score is higher because at School Q the typical student scored only 70 with a wider spread (SD 12), while at School P the typical student already scored 86 with little spread (SD 5). Quan therefore stood out further from his peers than Pia did from hers, so the scholarship — based on z — should go to Quan."

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Jamie's z-score (3 marks)

Sample response.
(a) z = (83 − 65) ÷ 12 = 18 ÷ 12 = 1.50.
(b) Jamie's mark is 1.5 standard deviations above the class mean, indicating clearly above-average performance.

Marking notes. 1 mark — correct substitution. 1 mark — correct value of z. 1 mark — interpretation referencing "standard deviations above the mean" in context. A bare "z = 1.5" with no interpretation scores 2/3.

1.2 — Pass mark and candidate check (3 marks)

Sample response.
(a) x = mean + z × SD = 72 + (−0.5)(9) = 72 − 4.5 = 67.5. The minimum pass mark is 67.5.
(b) Candidate: z = (64 − 72) ÷ 9 = −8 ÷ 9 ≈ −0.89. Since −0.89 < −0.5, the candidate does not pass.

Marking notes. 1 mark — correct minimum pass mark. 1 mark — correct z calculation for the candidate. 1 mark — explicit pass/fail conclusion comparing −0.89 with −0.5.

1.3 — Three students vs z = 1.28 cutoff (4 marks)

(a) Sample response. Ana: z = (78 − 70)/8 = 8/8 = 1.00. Ben: z = (85 − 80)/5 = 5/5 = 1.00. Carla: z = (82 − 72)/10 = 10/10 = 1.00.

(b) Sample response. All three students have z = 1.00, which is less than 1.28. None of the three students qualify for the top-10% threshold, because their standardised performances are all the same (one standard deviation above the mean) and the cutoff requires a slightly higher z.

Marking notes. (a) 1 mark per correct z (3 marks total — capped). (b) 1 mark — explicit decision naming all three and citing z = 1.00 < 1.28. Common error: declaring one student a winner because their raw mark is highest — z-scores are tied, so all three are equal.

2.1 — Pia vs Quan scholarship (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

(a) Pia (School P).

z_Pia = (94 − 86) ÷ 5 = 8 ÷ 5 = 1.60. [1 mark — substitution and value.]
1.60 ≥ 1.5, so Pia meets the cutoff. [1 mark — explicit cutoff check.]

(b) Quan (School Q).

z_Quan = (92 − 70) ÷ 12 = 22 ÷ 12 ≈ 1.83. [1 mark — substitution and value.]
1.83 ≥ 1.5, so Quan also meets the cutoff. [1 mark — explicit cutoff check.]

(c) Explanation and decision.

Although Pia's raw mark of 94 is higher than Quan's 92, School P had a much higher mean (86 vs 70) and a much smaller spread (SD 5 vs SD 12). This means an above-average student at School P sits much closer to the rest of the cohort than a similarly above-average student at School Q. [1 mark — refers to different means and SDs.] The z-score measures how many standard deviations each student is above their own school's mean, not their absolute mark, so it is a fair within-school comparison. [1 mark — explains the role of z as relative standing.]

Quan's z = 1.83 is higher than Pia's z = 1.60, so on the university's z-based criterion Quan should be awarded the scholarship, since he stood out further from his own peers than Pia did from hers. [1 mark — explicit award decision with justification.]

Total: 7/7.

Band descriptors for marker.

Band 3: Computes one z-score correctly but does not compare to the cutoff or makes one calculation error. ≈ 3 marks.

Band 4: Both z-scores correct, both cutoff checks correct, but no real explanation in (c) — just states an award based on raw marks. ≈ 5 marks.

Band 5: Full numerical work plus an explanation that references either the means or the SDs but not both. Award decision present. ≈ 6 marks.

Band 6: Complete, correctly substituted z-scores; explicit cutoff checks; coherent explanation referencing both the different means and the different SDs across the two schools; explicit, justified award decision that names the winning student. 7/7.