Use the normal distribution and the 68-95-99.7 rule, standardise to z-scores, and construct and interpret sample proportions and confidence intervals.

What this dot point is asking

You must apply the normal distribution and empirical rule, compute z-scores, and build and interpret confidence intervals for a proportion.

The normal distribution and the empirical rule

The normal distribution is a symmetric bell-shaped curve centred at its mean $\mu$, with spread set by its standard deviation $\sigma$.

Because the curve is symmetric, each interval splits evenly about the mean, so for example $34\%$ lies between the mean and $+1\sigma$. Sketching the bell curve and marking the standard-deviation lines is the safest way to read off percentages.

z-scores

A z-score standardises a value by stating how many standard deviations it is from the mean.

A positive z-score is above the mean, negative below. z-scores let you compare values from different normal distributions: a score with $z = 1.5$ in one test is more impressive than $z = 0.8$ in another, regardless of the raw marks.

Sample proportions and confidence intervals

A sample proportion $\hat{p} = \dfrac{\text{number with the trait}}{n}$ estimates the unknown population proportion $p$. Because samples vary, we report a range.

The interpretation is about the population: "we are $95\%$ confident the true population proportion lies between the lower and upper limits". A larger sample $n$ shrinks the margin of error, giving a narrower, more precise interval.

Margin of error and sample size

The margin of error is $E = z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$. Two levers change it: the confidence level (a higher level uses a larger $z$, widening the interval) and the sample size (a larger $n$ shrinks the margin). Because $n$ sits under a square root, quartering the margin requires multiplying the sample size by about $16$, so improving precision gets expensive quickly. SCSA questions often ask you to state the effect of doubling the sample size: the margin shrinks by a factor of $\sqrt{2} \approx 1.41$, not by half.

Working backwards through a normal distribution

As well as finding a percentage from a value, you may be asked for the value at a given percentile. Use the empirical rule in reverse: the value below which $97.5\%$ of a normal population lies is $\mu + 2\sigma$ (since $2.5\%$ lies above it), and the value below which $16\%$ lies is $\mu - \sigma$ (since $34\%$ lies between $\mu - \sigma$ and the mean, leaving $16\%$ below). For percentiles that do not land on whole standard deviations, you rearrange the z-score formula: $x = \mu + z\sigma$, reading $z$ from the required tail area. This connects the normal model to the confidence-interval $z$ values of $1.96$ and $2.58$.