HSC Mathematics Standard 2 the normal distribution (2026 guide)

HSCMaths Standard 2Study guide18 min readUpdated 25 May 2026

Quick answer

The normal distribution is symmetric and bell-shaped, fully specified by $\mu$ and $\sigma$ ; the empirical rule gives $68\%$ within $\pm 1\sigma$ , $95\%$ within $\pm 2\sigma$ , $99.7\%$ within $\pm 3\sigma$ ; z-scores $z = (x - \mu)/\sigma$ enable comparison across different normal distributions.

Jump to a section

What the normal distribution is
The empirical rule
z-scores
Australian context examples
Common normal-distribution mistakes
Exam strategy
Check your knowledge

Figure 1. The standard normal distribution and the empirical 68-95-99.7 rule. The variable scale (top labels) and the standard-score $z$ -scale (bottom labels) are aligned: any value $x$ converts to a $z$ -score via $z = (x - \mu)/\sigma$ . Approximately 68% of the probability mass lies within $\pm 1\sigma$ , 95% within $\pm 2\sigma$ , and 99.7% within $\pm 3\sigma$ of the mean.

The normal distribution is one of the cornerstone topics of HSC Mathematics Standard 2 statistics. Every paper since the 2017 syllabus reform has had at least one question on z-scores, and most have a separate question on the empirical rule. Across the paper, you can expect about $8$ - $12$ marks from this topic.

The good news: the normal distribution is conceptually one of the cleanest topics in the syllabus. The 68-95-99.7 rule is genuinely memorisable, the z-score formula is on the reference sheet, and the calculator does the heavy lifting for non-standard probabilities. The bad news: students lose easy marks every year through sign errors, confusing standard deviation with variance, and applying the rule to non-normal data.

What the normal distribution is

The normal distribution (or bell curve) is a continuous probability distribution fully specified by two parameters:

$\mu$ (mu): the mean.
$\sigma$ (sigma): the standard deviation.

Notation: $X \sim N(\mu, \sigma^2)$ . Note the variance, not the standard deviation, sits inside.

Properties:

Symmetric about $x = \mu$ .
Mean = median = mode = $\mu$ .
Total area under the curve is $1$ .
Higher $\sigma$ gives a flatter, wider curve. Lower $\sigma$ gives a taller, narrower curve.

The bell shape arises naturally in many contexts, especially when the quantity is the sum of many small independent random influences (heights, exam marks in large cohorts, measurement errors, manufacturing variation).

The empirical rule

For any normal distribution:

About $68\%$ of values lie within $1$ standard deviation of the mean ( $\mu \pm \sigma$ ).
About $95\%$ within $2$ standard deviations ( $\mu \pm 2\sigma$ ).
About $99.7\%$ within $3$ standard deviations ( $\mu \pm 3\sigma$ ).

By symmetry, the tail percentages above the upper threshold are half the percentage outside the band:

Above $\mu + \sigma$ : $16\%$ .
Above $\mu + 2\sigma$ : $2.5\%$ .
Above $\mu + 3\sigma$ : $0.15\%$ .

Standard region percentages:

Region	Percentage
$\mu - 1\sigma$ to $\mu + 1\sigma$	$68\%$
$\mu$ to $\mu + 1\sigma$	$34\%$
$\mu + 1\sigma$ to $\mu + 2\sigma$	$13.5\%$
$\mu + 2\sigma$ to $\mu + 3\sigma$	$2.35\%$
Above $\mu + 3\sigma$	$0.15\%$

Memorise this table; it is the heart of empirical-rule questions.

Empirical rule: heights of Year

12

boys at a Sydney school

Heights of Year 12 boys at a Sydney school are normally distributed with mean $\mu = 175$ cm and standard deviation $\sigma = 7$ cm. Find:

(a) the percentage taller than $189$ cm;
(b) the percentage between $168$ and $189$ cm.

Step 1. Convert each threshold to a number of standard deviations from the mean.

This places each value on the standard scale of the empirical rule.

$189$ cm is $\dfrac{189 - 175}{7} = 2$ SDs above the mean, so $189 = \mu + 2\sigma$ .
$168$ cm is $\dfrac{168 - 175}{7} = -1$ SD, so $168 = \mu - 1\sigma$ .

Step 2 (part a). Read the percentage above $\mu + 2\sigma$ from the empirical rule.

The rule says $95\%$ of values are within $\pm 2$ SDs, so $5\%$ are outside that band. By symmetry, half ( $2.5\%$ ) are in each tail.

Answer (a). About $2.5\%$ of the boys are taller than $189$ cm.

Step 3 (part b). Build the region $\mu - 1\sigma$ to $\mu + 2\sigma$ from the standard pieces.

Split the region at the mean and use the standard-region table:

$\mu - 1\sigma$ to $\mu$ : $34\%$ .
$\mu$ to $\mu + 1\sigma$ : $34\%$ .
$\mu + 1\sigma$ to $\mu + 2\sigma$ : $13.5\%$ .

Add these: $34 + 34 + 13.5 = 81.5\%$ .

Answer (b). About $81.5\%$ of the boys are between $168$ and $189$ cm tall.

z-scores

The z-score of 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.

To find $x$ given $z$ :

x = \mu + z \sigma.

Common percentile-to-z-score values

Percentile	z-score
50th	$0$
75th	$0.67$
90th	$1.28$
95th	$1.65$
97.5th	$1.96$
99th	$2.33$

By symmetry, lower percentiles use the negative of the upper: 10th = $-1.28$ , etc.

Using the standard normal table

The HSC Standard 2 paper provides a table of $\Phi(z) = P(Z \le z)$ , the cumulative distribution function of the standard normal.

For any value $x$ from a normal distribution:

$P(X \le x) = \Phi(z)$ where $z = (x - \mu) / \sigma$ .
$P(X > x) = 1 - \Phi(z)$ .
$P(a \le X \le b) = \Phi(z_b) - \Phi(z_a)$ .

Comparing two students from different tests

Two students sit different tests.

Anika scored $76$ on a maths test with $\mu = 68$ , $\sigma = 6$ .
Ben scored $82$ on an English essay with $\mu = 76$ , $\sigma = 8$ .

Who performed better relative to their cohort?

Step 1. Recognise that raw marks alone are misleading.

Ben's raw mark ( $82$ ) is higher than Anika's ( $76$ ), but the tests have different means and standard deviations. We need to convert both to z-scores to compare like with like.

Step 2. Compute each student's z-score.

The formula is $z = \dfrac{x - \mu}{\sigma}$ .

Anika: $z = \dfrac{76 - 68}{6} = \dfrac{8}{6} \approx 1.33$ .
Ben: $z = \dfrac{82 - 76}{8} = \dfrac{6}{8} = 0.75$ .

Step 3. Compare the z-scores.

A higher z-score means the student is further above their cohort's mean in standard-deviation units. Anika's $1.33$ is well above Ben's $0.75$ .

Answer. Anika performed better relative to her cohort. She scored about $1.33$ SDs above her test's mean, compared with Ben's $0.75$ SDs above his test's mean, even though Ben's raw mark is higher.

Standard normal table lookup: probability of failing a pre-fit test

A community gym pre-fit test has scores that are normally distributed with mean $\mu = 72$ and standard deviation $\sigma = 10$ . The pass mark is $55$ . Use the standard normal table values $\Phi(1.7) \approx 0.9554$ and $\Phi(-1.7) \approx 0.0446$ to find the probability a randomly chosen candidate fails (i.e. scores less than $55$ ).

Step 1. Convert the pass mark to a z-score using $z = (x - \mu)/\sigma$ .

We standardise so we can read the table, which is in terms of $z$ .

z = \frac{55 - 72}{10} = \frac{-17}{10} = -1.7

Step 2. Translate "fails" into a probability on the z-scale.

Failing means $X < 55$ . After standardising, this is $Z < -1.7$ , which is exactly $\Phi(-1.7)$ .

Step 3. Read $\Phi(-1.7)$ from the table.

The standard normal table gives $\Phi(-1.7) \approx 0.0446$ , which equals $1 - \Phi(1.7) = 1 - 0.9554$ by the symmetry of the standard normal.

P(X < 55) = \Phi(-1.7) \approx 0.0446

Step 4. Sanity-check with the empirical rule.

A z-score of $-1.7$ is beyond $\mu - 1\sigma$ but inside $\mu - 2\sigma$ . The empirical rule says about $2.5\%$ of values are below $\mu - 2\sigma$ and about $16\%$ below $\mu - 1\sigma$ , so a value below $z = -1.7$ should be somewhere around $4\%$ - $5\%$ . Our answer of $4.46\%$ sits in that range, so the table lookup is consistent.

Answer. About $4.46\%$ of candidates fail the pre-fit test.

Inverse normal: cut-off mark for the top

10\%

of HSC raw marks

HSC raw marks on a particular subject test are approximately normal with $\mu = 70$ and $\sigma = 13$ . The school wants to identify the top $10\%$ for a scholarship interview. Find the cut-off raw mark.

Step 1. Identify what z-score corresponds to "top $10\%$ ".

Being in the top $10\%$ means having a mark greater than the $90$ th percentile. The $90$ th-percentile z-score from the standard table (or the percentile-to-z reference table above) is $z \approx 1.28$ .

Step 2. Use $x = \mu + z\sigma$ to convert the z-score back to a raw mark.

This is the inverse of the standardising step.

x = 70 + 1.28 \times 13 = 70 + 16.64 = 86.64

Step 3. Round sensibly and state the cut-off.

Raw marks are typically integers, so we round up to be conservative (we want strictly more than the $90$ th percentile).

Answer. A raw mark of approximately $87$ or higher cuts off the top $10\%$ . Equivalently, the cut-off is about $86.6$ raw marks; students scoring $87$ or above qualify for the interview.

Bottling factory rejection rate (very small tail probability)

A bottling factory fills $600$ mL drink bottles with mean $\mu = 602$ mL and standard deviation $\sigma = 1.5$ mL. Bottles below $597$ mL are rejected and recycled. What proportion of bottles are rejected?

Step 1. Convert the rejection threshold to a z-score.

z = \frac{597 - 602}{1.5} = \frac{-5}{1.5} \approx -3.33

Step 2. Interpret the z-score using the empirical rule.

A z-score of $-3.33$ is more than $3$ standard deviations below the mean. The empirical rule says only $0.15\%$ of a normal distribution lies below $\mu - 3\sigma$ , so the proportion below $\mu - 3.33\sigma$ is even smaller.

Step 3. Refine with the standard normal table.

From the standard normal table, $\Phi(-3.33) \approx 0.00043$ , so about $0.043\%$ of bottles fall below $597$ mL.

Step 4. State the answer in factory terms.

In a production run of $100{,}000$ bottles, this is roughly $43$ rejects.

Answer. About $0.043\%$ of bottles are rejected, or about $4$ per $10{,}000$ filled. The factory has very high yield, and this is exactly why the line uses tight statistical-process-control limits at $\pm 3\sigma$ .

Australian context examples

Heights from the ABS

ABS data (Australian Health Survey, 2017-18) gives mean adult male height around $175$ cm with standard deviation about $7$ cm. The empirical rule then says:

$68\%$ of men are between $168$ and $182$ cm.
$95\%$ are between $161$ and $189$ cm.
$2.5\%$ are taller than $189$ cm.

HSC marks distribution

HSC exam marks are scaled, but a rough approximation is that raw marks in many subjects are approximately normally distributed with mean about $65$ - $70$ and standard deviation about $12$ - $15$ . The empirical rule then gives a rough sense of how rare Band 6 (typically the top $\sim 10\%$ ) is in raw terms.

For a quick estimate using $\mu = 70$ , $\sigma = 13$ : the top $10\%$ requires $z \ge 1.28$ , so $x \ge 70 + 1.28 \times 13 \approx 86.6$ raw marks for a Band 6 (matching the inverse-normal worked example above).

Common normal-distribution mistakes

Confusing $\sigma$ with $\sigma^2$: If $X \sim N(100, 25)$ , then $\sigma = 5$ (the variance is $25$ , the standard deviation is the square root).
Sign error in z-score: $z = (x - \mu) / \sigma$ . A value below the mean has a negative z-score.
Forgetting to halve: "Within $\pm 1\sigma$ " is $68\%$ . "Above $\mu + \sigma$ " is $16\%$ (half of $32\%$ ). Easy to halve the wrong number.
Using the rule on non-normal data: The empirical rule applies only to (approximately) normal data. Income, for example, is highly right-skewed; the rule does not apply.
Reading the wrong percentage from the table: $\Phi(z)$ is the area to the LEFT of $z$ . For "greater than", use $1 - \Phi(z)$ .

Exam strategy

For normal distribution questions in the HSC:

Always state the parameters. " $X \sim N(70, 13^2)$ " or "mean $70$ , SD $13$ ".
Show the z-score calculation explicitly. Even if the question is solvable by the empirical rule, marking guides usually accept either approach.
Sketch a normal curve. Shade the region in question. This catches errors about which tail you want.
State the final probability with units or percentage. " $P(X > 85) \approx 0.067$ " or " $\approx 6.7\%$ ".

In the past three years of HSC papers, normal distribution and z-score questions have appeared in every paper, contributing $6$ - $10$ marks. They are among the most reliable mark-earners in the paper for students who have drilled the empirical rule and the z-score formula.

Check your knowledge

Try these before checking the worked solutions. They cover empirical-rule reading, z-score computation, table lookup and inverse normal.

For $X \sim N(50, 8^2)$ , what is the z-score of $x = 66$ ?
For $X \sim N(50, 8^2)$ , give the value of $x$ whose z-score is $-1.25$ .
In a normal distribution, approximately what percentage of values lie between $\mu - 2\sigma$ and $\mu + 1\sigma$ ?
Marks on a test are $N(60, 10^2)$ . Use the empirical rule to estimate the percentage of students who scored above $80$ .
For $Z \sim N(0, 1)$ and $\Phi(1.2) \approx 0.8849$ , find $P(Z > 1.2)$ .
Adult female heights in Australia are approximately $N(162, 7^2)$ (cm). Using $\Phi(1.43) \approx 0.9236$ , find the proportion of women shorter than $172$ cm.
Two students sit different tests. Cara scored $74$ on a test with $\mu = 65$ , $\sigma = 9$ . Dan scored $80$ on a test with $\mu = 72$ , $\sigma = 10$ . Whose result was better, and by how many standard deviations?
ATAR-style raw marks on a paper are $N(68, 12^2)$ . Find the cut-off raw mark above which the top $5\%$ of students sit. Use $z_{0.95} \approx 1.645$ .

Solutions to check-your-knowledge questions

These eight questions cover the z-score formula, its inverse, reading empirical-rule regions, table lookups, comparing students across cohorts, and finding inverse-normal cut-offs.

Step 1: Q1. Find the z-score of the value $66$ when $\mu = 50$ , $\sigma = 8$

Apply the z-score formula $z = (x - \mu)/\sigma$ to express how many standard deviations $66$ sits above or below the mean.

z = \frac{66 - 50}{8} = \frac{16}{8} = 2

The value $66$ is exactly $2$ standard deviations above the mean.

Step 2: Q2. Find $x$ when $z = -1.25$ , $\mu = 50$ , $\sigma = 8$

Rearrange the z-score formula to the inverse form $x = \mu + z\sigma$ . A negative z-score means the value sits below the mean.

x = 50 + (-1.25)(8) = 50 - 10 = 40

The value whose z-score is $-1.25$ is $x = 40$ .

Step 3: Q3. Find the proportion of values between $\mu - 2\sigma$ and $\mu + 1\sigma$

Break the region into standard segments using the empirical rule: each segment between consecutive multiples of $\sigma$ has a known percentage.

Region from $\mu - 2\sigma$ to $\mu - 1\sigma$ : $13.5\%$ .
Region from $\mu - 1\sigma$ to $\mu$ : $34\%$ .
Region from $\mu$ to $\mu + 1\sigma$ : $34\%$ .

\text{Total} = 13.5 + 34 + 34 = 81.5\%

Step 4: Q4. Proportion above $80$ when $\mu = 60$ , $\sigma = 10$

Find how many standard deviations $80$ lies from the mean, then apply the empirical rule for the tail.

z = \frac{80 - 60}{10} = 2

The empirical rule says $95\%$ of values lie within $\pm 2$ SDs, leaving $5\%$ in both tails combined. By symmetry, $2.5\%$ lie above $\mu + 2\sigma$ .

Approximately $2.5\%$ of students scored above $80$ .

Step 5: Q5. Probability above $z = 1.2$ using the standard normal table

The table gives $\Phi(z) = P(Z < z)$ . To find the upper tail, subtract from 1.

P(Z > 1.2) = 1 - \Phi(1.2) = 1 - 0.8849 = 0.1151

About $11.5\%$ of values lie above $z = 1.2$ .

Step 6: Q6. Proportion of women shorter than $172$ cm when $\mu = 162$ , $\sigma = 7$

Standardise $172$ and look up the cumulative probability from the table.

z = \frac{172 - 162}{7} = \frac{10}{7} \approx 1.43

P(X < 172) = \Phi(1.43) \approx 0.9236

About $92.4\%$ of Australian adult women are shorter than $172$ cm.

Step 7: Q7. Compare Cara and Dan across different tests using z-scores

Z-scores place each result on the same scale, so they can be compared directly. Compute each z-score using the mean and standard deviation of that student's cohort.

Cara: $z = (74 - 65)/9 = 1.00$ . Dan: $z = (80 - 72)/10 = 0.80$ .

Cara performed better relative to her cohort, by $1.00 - 0.80 = 0.20$ standard deviations.

Step 8: Q8. Find the cut-off mark for the top $5\%$ when $\mu = 68$ , $\sigma = 12$

The top $5\%$ starts at the $95$ th percentile. Look up the z-score corresponding to $P(Z < z) = 0.95$ , which gives $z \approx 1.645$ . Then apply the inverse formula.

x = 68 + 1.645 \times 12 = 68 + 19.74 = 87.74

Final answer: the cut-off raw mark is approximately $87.7$ , so $88$ or above places a student in the top $5\%$ .

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA