NSWMaths Standard 2Syllabus dot point

What is a z-score, and how is it used to compare observations from different normal distributions?

Calculate z-scores and use them to compare values from different normal distributions and find probabilities

A focused answer to the HSC Maths Standard 2 dot point on z-scores. The formula $z = (x - \mu)/\sigma$ , interpreting z-scores as standard-deviation distances, comparing observations from different normal distributions, and worked examples from exam scores and Australian salary data.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to compute a z-score for any value from a normal distribution, interpret it as "how many standard deviations above (or below) the mean", and use z-scores to compare observations from different normal distributions.

The answer

The z-score formula

For a value $x$ from a normal distribution with mean $\mu$ and standard deviation $\sigma$ :

z = \frac{x - \mu}{\sigma}.

The z-score is the number of standard deviations $x$ is from the mean. Negative if $x$ is below the mean, positive if above.

Reverse: finding $x$ from IMATH_11

Rearrange to find $x$ given a z-score:

x = \mu + z \sigma.

This is useful when you want to find the value at a particular percentile.

Interpretation

IMATH_13 : value equals the mean.
IMATH_14 : value is one standard deviation above the mean.
IMATH_15 : value is two standard deviations below the mean.
IMATH_16 : a very extreme value (less than $0.3\%$ of data are this far from the mean).

Comparing across different distributions

Two observations from different distributions can be compared on a common scale by converting both to z-scores. The one with the higher z-score is further above its own mean (relative to its own standard deviation).

This is useful for comparing exam results from different tests, salaries in different industries, or athletic performance against different reference populations.

Z-scores and percentiles

Common percentile-to-z-score values (from the standard normal table):

Percentile	z-score
50th	IMATH_18
75th	IMATH_19
90th	IMATH_20
95th	IMATH_21
97.5th	IMATH_22
99th	IMATH_23
99.5th	IMATH_24

By symmetry, lower percentiles use the negative of the upper z-score: the 10th percentile is at $z = -1.28$ , the 5th at $z = -1.65$ , etc.

Z-scores and probabilities

To find the probability that an observation is less than some value $x$ :

Compute $z = (x - \mu) / \sigma$ .
Look up $\Phi(z)$ in the standard normal table (provided in the HSC exam paper).
IMATH_30 .

For "greater than" probabilities: $P(X > x) = 1 - \Phi(z)$ .

For probabilities between two values: $P(a \le X \le b) = \Phi(z_b) - \Phi(z_a)$ .

For empirical-rule-friendly endpoints (integer SDs from the mean), you can skip the table and use the empirical rule directly.

Worked example

Computing a z-score

A student scores $85$ on a test with $\mu = 70$ , $\sigma = 12$ .

$z = \frac{85 - 70}{12} = \frac{15}{12} = 1.25$ .

The student is $1.25$ standard deviations above the mean. Looking up $\Phi(1.25) \approx 0.8944$ , so $P(X < 85) \approx 0.8944$ , meaning about $89.4\%$ of students scored lower than $85$ . The student is at roughly the $89$ th percentile.

Cross-distribution comparison

A Sydney accountant earns $\$ 135000 $in an industry where the mean is$ \ $110000$ and SD is $\$ 20000 $. A Melbourne consultant earns$ \ $120000$ in an industry where the mean is $\$ 95000 $and SD is$ \ $15000$ .

Accountant z: $z = \frac{135000 - 110000}{20000} = 1.25$ .

Consultant z: $z = \frac{120000 - 95000}{15000} \approx 1.67$ .

The consultant earns less in dollars but ranks higher relative to peers (closer to the $95$ th percentile vs the $89$ th percentile).

Finding a percentile

For exam marks normally distributed with $\mu = 65$ , $\sigma = 12$ , find the cutoff for the top $5\%$ .

Top $5\%$ corresponds to the $95$ th percentile, $z = 1.65$ .

$x = 65 + 1.65 \times 12 = 65 + 19.8 = 84.8$ .

So the top $5\%$ of students scored above $84.8$ .

Australian salary example (ABS-style 2024 data)

Full-time adult average weekly earnings (ABS, August 2024) are approximately $\$ 1830 $. Assume normal with SD$ \ $500$ .

A worker earning $\$ 2500 $per week has z$ = \frac{2500 - 1830}{500} = 1.34 $. About$ \Phi(1.34) \approx 0.9099 $, so$ 91% $of workers earn less. The worker is in the top$ 9%$.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q184 marksTwo students sit different tests. Anika scores

76

on a test with

\mu = 68

and

\sigma = 6

. Ben scores

82

on a test with

\mu = 76

and

\sigma = 8

. Which student performed better relative to their cohort?

Show worked answer →

Compute each z-score: $z = \frac{x - \mu}{\sigma}$ .

Anika: $z = \frac{76 - 68}{6} = \frac{8}{6} \approx 1.33$ .

Ben: $z = \frac{82 - 76}{8} = \frac{6}{8} = 0.75$ .

Anika's z-score is higher, so she performed better relative to her cohort (further above the mean in standard-deviation units), even though Ben's raw score is higher.

Markers reward both z-scores, comparison, and the final conclusion with the reason ("higher z-score means further above the mean").

2023 HSC Q194 marksHeights of Year 12 girls at a school are normally distributed with mean

164

cm and standard deviation

6

cm. Find the height that corresponds to (a) the 90th percentile, and (b) the 10th percentile. (Use

z = 1.28

for the 90th percentile.)

Show worked answer →

(a) For the 90th percentile, $z = 1.28$ . Use $x = \mu + z\sigma$ .

$x = 164 + 1.28 \times 6 = 164 + 7.68 = 171.68$ cm. Round to $171.7$ cm.

(b) By symmetry, the 10th percentile has $z = -1.28$ .

$x = 164 + (-1.28) \times 6 = 164 - 7.68 = 156.32$ cm. Round to $156.3$ cm.

Markers reward the formula rearrangement, the symmetric z-score for the lower tail, and answers to one decimal place.