Use the sample proportion $\hat{p} = X/n$ as a random variable, with mean $p$, variance $pq/n$ and standard deviation $\sqrt{pq/n}$, and apply its normal approximation $\hat{p} \sim N(p,\, pq/n)$

What this dot point is asking

NESA wants you to treat the sample proportion $\hat{p} = X/n$ as a random variable in its own right, know its mean $p$, its variance $\dfrac{pq}{n}$ and its standard deviation $\sqrt{\dfrac{pq}{n}}$, and use the normal approximation $\hat{p} \sim N\!\left(p,\, \dfrac{pq}{n}\right)$ to compute probabilities about how close a survey's estimate is likely to be to the true value. This is the one slice of the statistics chapter that is uniquely Extension 1, and it is where binomial theory turns into real polling and quality-control reasoning.

The answer

What a sample proportion is

Run a binomial experiment: $n$ independent Bernoulli trials, each a success with probability $p$. Let $X \sim B(n, p)$ count the successes. The population proportion $p$ is a fixed (usually unknown) number, for example the true fraction of Australians who support a policy. The sample proportion

is the fraction of your sample that were successes. If you poll $n = 2000$ voters and $X = 700$ say they support a party, your sample proportion is $\hat{p} = \dfrac{700}{2000} = 0.35$, an estimate of the true $p$.

The crucial idea is that $\hat{p}$ is itself a random variable. Run the survey again with a fresh random sample and you get a different $X$, hence a different $\hat{p}$. So $\hat{p}$ has its own distribution, mean and spread, and those are exactly what tell us how trustworthy a single survey is.

Why the mean is $p$ and the variance is $pq/n$

These are not new results to memorise blindly; they fall straight out of the binomial mean and variance you already know, $E(X) = np$ and $\operatorname{Var}(X) = npq$, using the scaling rules $E(aX) = aE(X)$ and $\operatorname{Var}(aX) = a^2 \operatorname{Var}(X)$ with the constant $a = \dfrac{1}{n}$:

The mean result, $E(\hat{p}) = p$, is the whole reason surveys work: on average the sample proportion equals the population proportion, so $\hat{p}$ is an unbiased estimator of $p$. The variance result is the whole reason large samples are better: the spread carries a $\dfrac{1}{n}$, so it shrinks as $n$ grows. Note the dimension check too, $\hat{p}$ is a proportion (a number between $0$ and $1$), so its SD $\sqrt{pq/n}$ is also a small fraction, never the large count-sized spread $\sqrt{npq}$ that $X$ has.

The distribution of $\hat{p}$ is the binomial, restretched

Because $\hat{p} = X/n$, each value of $\hat{p}$ corresponds to exactly one value of $X$ and carries the same probability:

So the probability graph of $\hat{p}$ is just the probability graph of $X$ with the horizontal axis relabelled (squashed from the integers $0,1,\ldots,n$ onto the fractions $0, \tfrac1n, \ldots, 1$). Nothing about the probabilities changes, only the scale of the axis. That is why every binomial tool you have still applies: to find a probability about $\hat{p}$ exactly, convert it into the matching statement about $X$ and sum binomial terms; to find it quickly for large $n$, use the normal approximation below.

The normal approximation $\hat{p} \sim N(p,\, pq/n)$

For large $n$ the binomial is well approximated by a normal distribution (the central limit theorem). Dividing through by $n$ carries that approximation over to $\hat{p}$: it becomes approximately normal with the same mean and variance we just derived,

The validity conditions are the same $np \ge 5$, $nq \ge 5$ rule used for approximating $X$ itself, because $\hat{p}$ and $X$ have identical probabilities. The picture below is the sampling distribution of $\hat{p}$ for a national poll with true support $p = 0.4$ and $n = 1000$: a bell curve centred exactly on the true value $p$, with standard deviation $\sqrt{pq/n} \approx 0.0155$. The shaded band is the central $\pm 2$ standard deviations, the "expected range" almost every poll will fall inside.

The effect of sample size $n$

The mean of $\hat{p}$ is always $p$, no matter the sample size, so a bigger sample does not move the centre, it sharpens it. Since

the spread is inversely proportional to $\sqrt{n}$. To halve the standard deviation you must quadruple the sample. The overlay below fixes $p = 0.5$ and stacks the sampling distributions for $n = 100$, $n = 400$ and $n = 1600$: each fourfold jump in $n$ halves the standard deviation (from $0.05$ to $0.025$ to $0.0125$) and the curve becomes correspondingly taller and tighter around $0.5$.

This is the link to confidence-style "expected range" reasoning. Because $\hat{p}$ is roughly normal, about $95\%$ of the time a single survey's $\hat{p}$ lands within $\pm 2$ standard deviations of $p$ (more precisely $\pm 1.96$ SD). Read backwards, that interval is the "margin of error": if a poll of $n = 1000$ reports $\hat{p} = 0.40$, you can say the true value is very likely within $2 \times 0.0155 \approx 0.031$, i.e. roughly $0.40 \pm 0.03$. The smaller you want that margin, the larger $n$ has to be, and because of the $\sqrt{n}$, shrinking the margin by half costs four times the sample.

Exact versus approximate

For small $n$, do not reach for the normal curve, the values of $\hat{p}$ are too few and chunky. Instead convert the question about $\hat{p}$ into the matching question about $X = n\hat{p}$ and sum binomial terms exactly. For large $n$ (both $np \ge 5$ and $nq \ge 5$), the normal approximation $N(p, pq/n)$ replaces a long sum with one or two $z$-lookups. A continuity correction of $\pm \dfrac{1}{2n}$ (the half-step between adjacent $\hat{p}$ values is $\dfrac1n$) can be applied, but for the large samples typical of polling it is negligible and is usually dropped, which is the convention this page follows.

How exam questions ask about sample proportions

The wording is the tell. Map the phrase to the move:

• "Write down / find the sample proportion": just compute $\hat{p} = X/n$ from the given count. A $1$ mark opener.
• "State the mean and standard deviation of $\hat{p}$": quote $E(\hat{p}) = p$ and $\text{SD}(\hat{p}) = \sqrt{pq/n}$ (note the square root, and that the SD is a small fraction, not $\sqrt{npq}$).
• "Show that $E(\hat{p}) = p$ / $\operatorname{Var}(\hat{p}) = pq/n$": derive from $E(X) = np$, $\operatorname{Var}(X) = npq$ using $\hat{p} = X/n$ and the scaling rules $E(aX)=aE(X)$, $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$.
• "Use the normal approximation to find $P(\hat{p} \ldots)$" or "estimate the probability the sample proportion is between ...": confirm $np \ge 5$ and $nq \ge 5$, write $\hat{p} \sim N(p, pq/n)$, standardise with $z = \dfrac{\hat{p} - p}{\sqrt{pq/n}}$, look up.
• "within $0.0X$ of the true value": this is $P(|\hat{p} - p| \le 0.0X)$, a symmetric interval $p \pm 0.0X$; standardise both ends and use $2\,P(Z \le z) - 1$.
• "What sample size is needed so the estimate is within ... with probability ...": set $z \sqrt{pq/n} \le \text{margin}$ and solve for $n$, using the worst case $p = 0.5$ if $p$ is unknown.