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Study GuideYear 12Level 3 · ChallengeHSCVCE

Free Year 12 The Normal Distribution and z-scores

An HSC/VCE-style study guide to the bell curve: calculating z-scores, comparing results from different tests, using the 68–95–99.7 rule, reading probabilities from a standard normal table or calculator and working backwards from a percentage to a raw score. Includes a six-question self-check with answers.

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Year
Year 12
Subject
Mathematics
Topic
Statistics
Difficulty
Level 3 · Challenge
Estimated time
25 minutes
Curriculum
Australian Curriculum
Answers
Not applicable
Format
PDF (A4) + print

Students will practise

  • calculating z-scores with z = (x − μ) ÷ σ and converting back to raw scores
  • comparing scores from different distributions using z-scores
  • applying the 68–95–99.7 (empirical) rule
  • finding probabilities and percentiles for a normal distribution using a table or calculator

Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: The Normal Distribution and z-scores

More resources for this topic are coming soon.

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practise with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

Year 12 students (typically ages 17–18) working on statistics. It is pitched at level 3 · challenge.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 25 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

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Year 12 · Mathematics · Statistics

The Normal Distribution and z-scores

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What you need to know

Many measurements — heights, test scores, measurement errors — follow a normal distribution: a symmetric bell-shaped curve centred on the mean μ, with spread measured by the standard deviation σ. The mean, median and mode are all equal, and the total area under the curve is 1 (100%). A z-score tells you how many standard deviations a value is from the mean: z = [frac x − μ/σ]. Positive z is above the mean, negative z is below, and z = 0 is exactly the mean.

Converting to z-scores lets you compare values from different distributions on the same scale, and lets you use one standard normal table (μ = 0, σ = 1) for every normal distribution. The table or calculator gives P(Z < z); use symmetry and the complement for other regions: P(Z > z) = 1 − P(Z < z), and P(Z < −z) = P(Z > z).

IntervalApproximate percentage of data (empirical rule)Useful halves
μ − σ to μ + σ (z from −1 to 1)68%34% each side of the mean
μ − 2σ to μ + 2σ (z from −2 to 2)95%47.5% each side; 2.5% in each tail
μ − 3σ to μ + 3σ (z from −3 to 3)99.7%49.85% each side; 0.15% in each tail
between 1σ and 2σ above the mean13.5%47.5% − 34%

Test scores have μ = 65 and σ = 8. Find the z-scores for 77 and 53.

  1. z = = = 1.5 (1.5 standard deviations above the mean)
  2. z = = = −1.5 (1.5 standard deviations below the mean)

Which result is better: 82 in Maths (μ = 70, σ = 8) or 75 in English (μ = 62, σ = 10)?

  1. Maths: z = = 1.5
  2. English: z = = 1.3
  3. The Maths result is better relative to its class, because it is further above the mean in standard-deviation units.

Heights are normal with μ = 170 cm and σ = 6 cm. Use the empirical rule.

  1. Percentage between 164 cm and 182 cm: 164 is z = −1 and 182 is z = 2. Area = 34% + 47.5% = 81.5%
  2. Percentage above 188 cm: 188 is z = 3. Only 0.15% lies above z = 3, so 0.15%.
  3. Height with z = −2: x = μ + zσ = 170 + (−2)(6) = 158 cm. About 2.5% of people are shorter than this.

Using a standard normal table with μ = 65 and σ = 8

  1. P(X < 77) = P(Z < 1.5) = 0.9332, so about 93.3% of scores are below 77.
  2. P(X > 77) = 1 − 0.9332 = 0.0668
  3. P(57 < X < 77) = P(−1 < Z < 1.5) = P(Z < 1.5) − P(Z < −1) = 0.9332 − 0.1587 = 0.7745

Working backwards: what score is needed to be in the top 10% (μ = 65, σ = 8)?

  1. Top 10% means P(Z > z) = 0.10, so P(Z < z) = 0.90.
  2. From the table (or inverse normal on a calculator), z ≈ 1.28.
  3. x = μ + zσ = 65 + 1.28 × 8 = 75.24…, so a score of about 75.3 (in practice, 76 if scores are whole numbers).
Common mistake: dropping the negative sign on a z-score below the mean, or reading the table as if P(Z < −1.5) were 0.9332. For a negative z, use symmetry: P(Z < −1.5) = P(Z > 1.5) = 1 − 0.9332 = 0.0668.
Common mistake: using the empirical rule for values that are not whole numbers of standard deviations. 68–95–99.7 only applies at z = ±1, ±2, ±3. For z = 1.5 you must use a table or calculator.

Check yourself

  1. A distribution has μ = 50 and σ = 5. Find the z-score of 60.
  2. Using μ = 50 and σ = 5, find the z-score of 42.
  3. What percentage of a normal distribution lies within one standard deviation of the mean?
  4. Using μ = 50 and σ = 5, what percentage of values are below 40?
  5. Using μ = 50 and σ = 5, find the value with a z-score of 1.5.
  6. Alex scored 60 in a test with μ = 50 and σ = 5. Bea scored 85 in a test with μ = 70 and σ = 10. Who did better relative to their class?

Answers: 1. 2 2. −1.6 3. 68% 4. 40 is z = −2, so 2.5% 5. 50 + 1.5 × 5 = 57.5 6. Alex (z = 2) did better than Bea (z = 1.5)

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Last reviewed
1 October 2026
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