Use the normal distribution and the 68-95-99.7 rule to estimate the percentage of values within a number of standard deviations of the mean, and standardise a value to a z-score to compare values from different distributions

What this dot point is asking

VCAA wants you to work with the normal distribution, the symmetric bell-shaped model that fits many real measurements such as heights, test scores and birth weights. You use the 68-95-99.7 rule to estimate the proportion of values lying within one, two or three standard deviations of the mean, you convert a raw value into a standardised z-score, and you use that z-score to compare values that come from different distributions. This material underpins the way Data analysis talks about "typical" and "unusual" values.

The 68-95-99.7 rule

When data are approximately normal, the percentages of values within whole numbers of standard deviations of the mean are fixed:

• about $68\%$ lie within $1$ standard deviation, that is between $\bar{x} - s$ and $\bar{x} + s$,
• about $95\%$ lie within $2$ standard deviations,
• about $99.7\%$ lie within $3$ standard deviations.

Because the curve is symmetric, you can split these. For example, half of the central $68\%$, that is $34\%$, lies between the mean and one standard deviation above it. The tail beyond $\bar{x} + 2s$ holds about $(100 - 95)/2 = 2.5\%$ of values.

Standardising to a z-score

To compare a value with the rest of its distribution, or with a value from a different distribution, convert it to a z-score:

A z-score of $+1.5$ means the value is $1.5$ standard deviations above the mean. To reverse the process and recover the raw value, rearrange to $x = \bar{x} + z\,s$.

Comparing values from different distributions

The whole point of standardising is fairness. Two raw marks from exams with different means and spreads cannot be compared directly. Their z-scores can, because each z-score measures position within its own distribution in the same unit: standard deviations from the mean. This is exactly how VCE study scores are placed on a common scale.

Why this matters for the exams

Normal distribution questions appear in both exams and are usually quick marks if you know the rule and the z-score formula. Examiners like to test the tails (percentage above or below a marker) and direct comparison of values across distributions. Always state whether a value is above or below the mean and by how many standard deviations, since that sentence is often worth a mark in the written-response paper.