Random sampling and the distribution of sample proportions (TCE Mathematics Methods, Tasmania)

Syllabus dot point

“Understand random sampling and describe the distribution of the sample proportion, including its mean and standard deviation.”

8 min readUpdated 2 Jun 2026

Quick answer

A sample proportion $\hat{p}$ is itself a random variable because a different random sample gives a different value. For a random sample of size $n$ from a population with true proportion $p$ , the sampling distribution of $\hat{p}$ has mean $p$ and standard deviation $\sqrt{\dfrac{p(1-p)}{n}}$ , and is approximately normal when $n$ is large.

What this dot point is asking

Before you can build a confidence interval, you need to understand how a sample proportion behaves. This dot point is the bridge between the binomial work of Unit 3 and the interval estimation that follows.

Why random sampling matters

A sample is random when every member of the population has a known, nonzero chance of being chosen and the selection is free of bias. Random sampling is what justifies using the sample to make claims about the whole population. A convenience or self-selected sample can be systematically unrepresentative, so the methods of this topic simply do not apply to it.

Population proportion versus sample proportion

The population proportion $p$ is a fixed but usually unknown number describing everyone. The sample proportion $\hat{p} = \dfrac{\text{number of successes in the sample}}{n}$ is computed from your data and changes with each new sample. We use the variable $\hat{p}$ to estimate the constant $p$ .

The sample proportion as a random variable

Imagine taking many independent random samples of the same size $n$ and recording $\hat{p}$ each time. The collection of those values has its own distribution, called the sampling distribution of the sample proportion. Because the count of successes is binomial, dividing by $n$ rescales it, and the centre and spread follow directly.

Mean and standard deviation of the sample proportion

E(\hat{p}) = p, \qquad \text{standard deviation of } \hat{p} = \sqrt{\frac{p(1-p)}{n}}.

The standard deviation is also called the standard error. It shrinks as $n$ grows, in proportion to $\dfrac{1}{\sqrt{n}}$ .

Because the mean of $\hat{p}$ equals $p$ , the sample proportion is an unbiased estimator: on average across many samples it lands on the true value.

Approximate normality

The normal approximation

For a large random sample, the sampling distribution of $\hat{p}$ is approximately normal, centred at $p$ with standard deviation $\sqrt{p(1-p)/n}$ . This follows because the underlying binomial count is approximately normal when $n$ is large, and it is what lets us use $z$ -values later.

Simulating to see the variability

A useful way to understand this is simulation. If you repeatedly generate samples from a population with a known $p$ and plot all the resulting $\hat{p}$ values in a histogram, you see a roughly bell-shaped cluster centred on $p$ . Larger samples produce a tighter, taller cluster, visually confirming that the standard error falls as $n$ rises.

Describing a sampling distribution

Problem

A factory produces components of which a true proportion $p = 0.2$ are defective. A quality inspector takes random samples of $n = 100$ components. Describe the sampling distribution of the sample proportion of defectives.

Solution

The mean of the sampling distribution is the true proportion:

E(\hat{p}) = p = 0.2.

The standard deviation (standard error) is

\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2 \times 0.8}{100}} = \sqrt{\frac{0.16}{100}} = \sqrt{0.0016} = 0.04.

So $\hat{p}$ is approximately normal with mean $0.2$ and standard deviation $0.04$ . Most samples will give a defective proportion within about two standard deviations of $0.2$ , that is roughly between $0.12$ and $0.28$ .

Using n as the number of samples instead of the sample size

In the standard error $\sqrt{p(1-p)/n}$ , the value $n$ is the size of a single sample, not how many samples you took. Taking one sample of $400$ uses $n=400$ . The number of repeated samples affects how well a histogram reveals the distribution, but it never appears in the formula for the standard error.

When is the sample large enough?

The normal approximation only works when the sample is large enough that the binomial count is not bunched against $0$ or $n$ . A standard rule of thumb requires both $np \ge 10$ and $n(1-p) \ge 10$ (some texts use $5$ ). For a true proportion near $0.5$ a sample of a few dozen already behaves normally, but for a rare event such as $p = 0.02$ you need a much larger $n$ before the bell shape emerges, because most samples would otherwise record zero successes. Checking this condition is a quick way to justify, in an exam answer, that you may use $z$ -values at all.

Comparing two sample sizes

Problem

A true proportion is $p = 0.4$ . Compare the standard error of $\hat{p}$ for $n = 100$ and for $n = 400$ .

Standard error for n = 100

\sqrt{\frac{0.4\times 0.6}{100}} = \sqrt{0.0024} = 0.04899.

Standard error for n = 400

\sqrt{\frac{0.4\times 0.6}{400}} = \sqrt{0.0006} = 0.02449.

Interpretation

Quadrupling the sample size from $100$ to $400$ halves the standard error, because the spread scales as $\dfrac{1}{\sqrt{n}}$ and $\sqrt{4} = 2$ . To halve the margin of error again you would need to quadruple the sample once more, to $1600$ . This diminishing return is why very precise polls are expensive.

Summary

A random sample lets the sample proportion stand in for the population proportion. Treat $\hat{p}$ as a random variable whose sampling distribution is centred at the true $p$ , with standard error $\sqrt{p(1-p)/n}$ , and is approximately normal for large $n$ . This unbiased, shrinking-spread behaviour is exactly what makes the confidence interval in the next dot point work.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks

Calculator-assumed. A reporter surveys

10

people and finds

6

prefer product A, giving a

95\%

confidence interval of

(29.6\%, 90.4\%)

. The editor asks for

1000

people instead, a

100

-times increase. Assuming the larger survey still finds

60\%

prefer product A, what is the new

95\%

confidence interval?

Show worked answer →

The standard error of the sample proportion is $\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}$ . With $\hat{p} = 0.6$ and the new $n = 1000$ :

\sqrt{\frac{0.6\times 0.4}{1000}} = \sqrt{0.00024} = 0.015492.

The $95\%$ margin of error is $1.96\times 0.015492 = 0.0304$ (about $3.04\%$ ).

So the new interval is $0.6 \pm 0.0304 = (0.5696, 0.6304)$ , i.e. about $(56.96\%, 63.04\%)$ .

The key idea: multiplying the sample size by $100$ divides the standard error (and the margin) by $\sqrt{100} = 10$ , shrinking the old margin from about $0.304$ to about $0.0304$ . Larger samples give narrower intervals.

Practice3 marks

Calculator-free. In a large population the true proportion supporting a policy is

p = 0.25

. Random samples of size

n = 75

are taken. (a) State the mean of the sampling distribution of

\hat{p}

. (b) Find the standard deviation of

\hat{p}

, to four decimal places. (c) Within what range would about

95\%

of sample proportions lie?

Show worked answer →

(a) The sampling distribution is unbiased, so $E(\hat{p}) = p = 0.25$ .

(b) The standard error is

\sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.25\times 0.75}{75}} = \sqrt{0.0025} = 0.05 = 0.0500.

(c) About $95\%$ of values lie within two standard deviations of the mean: $0.25 \pm 2(0.05) = 0.25 \pm 0.10$ , i.e. from $0.15$ to $0.35$ . One mark each for the mean, the standard error, and the $\pm 2\sigma$ range.

Practise this

Sources & how we know this

Mathematics Methods Level 4 (MTM415117) course document (2017 onwards) — TASC