Use the normal distribution, the 68-95-99.7 rule and z-scores to describe data and compare values from different distributions.

What this dot point is asking

You must apply the 68-95-99.7 rule, compute and interpret z-scores, find proportions, and use z-scores to compare values from different distributions.

The shape of the normal distribution

A normal distribution is symmetric about its mean $\mu$, with most data clustered near the centre and the curve tailing off either side. Its spread is set by the standard deviation $\sigma$. The mean, median and mode all sit at the centre.

Z-scores: standardising a value

A z-score rescales any value so it can be compared on a common scale. It tells you how many standard deviations the value lies above (positive) or below (negative) the mean.

Comparing values from different distributions

Z-scores let you compare results measured on different scales, because they convert everything to "standard deviations from the mean."

The inverse problem: from proportion to value

Many SACE questions run the standardisation backwards: you are told a proportion or percentile and must find the corresponding value. Use the inverse normal to get the $z$-score for that area, then unstandardise with $x = \mu + z\sigma$. For example, the cut-off for the top 10% of a distribution has $z \approx 1.2816$, and substituting the mean and standard deviation gives the actual boundary value. Watch the direction: "the slowest 2%" or "the fastest 2%" both refer to a lower-tail area, while "the top 5%" is an upper-tail area, so the sign of $z$ must match the side of the distribution the question describes.

Comparing groups with different spreads

A subtle but examinable point is that a larger standard deviation puts more of a distribution into the tails. Two groups with similar means can have very different proportions beyond a high threshold if their spreads differ, with the more variable group having more extreme values at both ends. This is why a candle type with a lower mean burn time can still have a greater proportion burning beyond a long target, as in the candle question above. Comparing $z$-scores at the threshold, rather than comparing means, is the reliable way to settle such claims.

Finding proportions and percentiles

Beyond the round numbers of the empirical rule, a calculator gives the proportion below any z-score, which is the percentile. For example, $z = 1.5$ corresponds to about the 93rd percentile, meaning roughly 93% of values fall below it.