The normal distribution and z-scores: VCE General Mathematics Unit 3 Data analysis

What this dot point is asking

Many measurements (heights, reaction times, test scores, manufacturing tolerances) form a symmetric, bell-shaped distribution. The normal model describes this shape, and it gives the standard deviation a concrete meaning. VCAA wants you to use the 68-95-99.7% rule to estimate percentages (and counts) of values in a range, to calculate standardised values (z-scores) with the formula on the formula sheet, to rearrange that formula to find a value, a mean or a standard deviation, and to use z-scores to compare data values across different distributions.

This is one of the most reliable sources of marks in the Data analysis section: 2023 Examination 1 had two such questions, and 2025 had one in each examination. The 2025 examiners also made a point about rounding (83.85%, not 84%) that applies to every normal-model question.

The answer

The normal model

A normal distribution is symmetric and bell-shaped, with a single peak at the centre. For such data:

• the mean, median and mode are all equal and sit at the centre;
• the standard deviation $s$ controls the width: a larger $s$ gives a wider, flatter bell;
• almost all values lie within three standard deviations of the mean.

Real data is never perfectly normal, but a histogram that is approximately symmetric and bell-shaped can be modelled this way.

The 68-95-99.7% rule

For an approximately normal distribution:

• about $68\%$ of values lie within 1 standard deviation of the mean ($\bar{x} \pm s$);
• about $95\%$ lie within 2 standard deviations ($\bar{x} \pm 2s$);
• about $99.7\%$ lie within 3 standard deviations ($\bar{x} \pm 3s$).

Because the curve is symmetric, these split into bands:

These are halves of the differences: $\frac{68}{2} = 34$, $\frac{95 - 68}{2} = 13.5$, $\frac{99.7 - 95}{2} = 2.35$ and $\frac{100 - 99.7}{2} = 0.15$. Almost any percentage question is answered by adding bands. Useful totals: below the mean $50\%$; below $z = -1$ is $16\%$; below $z = -2$ is $2.5\%$; above $z = 1$ is $16\%$; above $z = 2$ is $2.5\%$; between $z = -2$ and $z = 1$ is $47.5\% + 34\% = 81.5\%$.

This is also what gives the standard deviation meaning: for bell-shaped data, roughly two out of three values lie within one standard deviation of the mean.

Standardised values (z-scores)

A z-score says how many standard deviations a value lies above or below the mean:

This formula is on the formula sheet. A positive $z$ is above the mean, a negative $z$ is below it, and $z = 0$ is the mean. Rearranging gives the value from its z-score:

Heights with mean 170 cm and standard deviation 7 cm: 159.5 cm has $z = \frac{-10.5}{7} = -1.5$, and $z = 1.5$ corresponds to $170 + 1.5 \times 7 = 180.5$ cm.

Finding percentages and counts

1. Convert each end point to a z-score.
2. Add the band percentages between those z-scores.
3. For a count, multiply the percentage (as a decimal) by the total number.

Museum visits (mean 82, $s = 11$): 60 and 104 minutes are $z = -2$ and $z = 2$, so $95\%$; of 2380 visitors, $0.95 \times 2380 = 2261$ are expected in that range.

Working backwards: the mean or standard deviation

Questions often give percentages and ask for the mean and standard deviation. Turn each percentage into a z-score using the bands:

• "16% are below $a$" means $a$ is at $z = -1$;
• "2.5% are above $b$" means $b$ is at $z = 2$;
• "84% are below $c$" means $c$ is at $z = 1$ ($50\% + 34\%$);
• "0.15% are above $d$" means $d$ is at $z = 3$.

Then the gap between two such values is a known number of standard deviations. In 2025, 157.6 cm was at $z = -1$ and 178.9 cm at $z = 2$: three standard deviations span $21.3$ cm, so $s = 7.1$ cm and the mean is $157.6 + 7.1 = 164.7$ cm.

If one value and its z-score are given along with the mean, rearrange for $s$: $s = \frac{x - \bar{x}}{z}$. In 2025 Examination 2, $\frac{952\,000 - 1\,400\,000}{-1.60} = 280\,000$.

Comparing values across distributions

Raw scores from different distributions cannot be compared directly: 78 in a hard test might be a better result than 85 in an easy one. The z-score puts both on the same scale. Maths 78 (mean 65, $s = 8$) gives $z = 1.625$; English 72 (mean 60, $s = 6$) gives $z = 2$. Relative to the class, the English result is better, because it is further above its mean in standard deviations.

When the rule does not apply

The 68-95-99.7% rule is a property of the normal (bell-shaped) distribution. For clearly skewed data, or data with a very different shape, the percentages can be badly wrong. z-scores can still be calculated for any data, but their interpretation through the 68-95-99.7% rule needs an approximately normal distribution.