Mathematical Methods · Units 3 & 4

Sampling Distributions of Proportions

Q: What is the difference between p and p hat?

The plain p is the true proportion for the whole population, a fixed number you usually never know. The p hat is the proportion you actually measure in one sample, and it changes from sample to sample. You use p hat as your best estimate of p.

Q: Why does a bigger sample give a smaller standard deviation?

Because the sample size sits underneath a square root in the spread formula. More people in the sample means the random ups and downs average out more, so your estimate jumps around less. To halve the spread you need four times the sample, not just double it.

Q: When can I treat the sample proportion as normal?

When the sample is reasonably large, the sampling distribution of the sample proportion is close enough to a bell curve to use the normal distribution. That lets you answer probability questions about the sample proportion the same way you would for any normal variable.

Understand the sampling distribution of the sample proportion the easy way, with plain English intuition, its mean and standard deviation, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Imagine the whole country has already decided, deep down, what fraction of people will vote yes. You will never know that exact fraction, so you ring a sample of people and work out the fraction in your sample instead. Ring a different sample and you get a slightly different fraction. The sampling distribution is the picture of all those possible fractions, and the surprising part is how predictable that picture is. It sits right on the truth and its spread shrinks in a way you can calculate exactly.

The sampling distribution of the sample proportion is approximately a bell curve, centred on the true population proportion p, with a spread that shrinks as the sample size grows.

The sample proportion and why it moves

You want to know the population proportion $p$ , the fraction of the whole population with some feature. You cannot measure everyone, so you take a random sample of size $n$ , count the $X$ people in the sample who have the feature, and report the fraction:

\hat{p} = \frac{X}{n}

This $\hat{p}$ is the sample proportion, and it is your estimate of $p$ . The key idea is that $\hat{p}$ is not a fixed number. Before you collect the sample it is random, because $X$ is random. A different random sample gives a different $\hat{p}$ . So $\hat{p}$ has its own distribution, called the sampling distribution, and our whole job is to describe its centre and its spread.

The centre: the mean of p hat

Here is the reassuring fact. Although each $\hat{p}$ misses the truth a little, the sample proportion does not lean high or lean low. On average it lands exactly on the population proportion:

E(\hat{p}) = p

That is what makes $\hat{p}$ a fair estimate of $p$ . There is no built in bias pulling your samples above or below the truth, so the whole distribution is centred on the number you are trying to find. The catch is that any single sample still misses, which is why the spread matters just as much as the centre.

The spread: the standard deviation of p hat

The count $X$ in a sample of size $n$ is a binomial variable, so $\hat{p} = X/n$ inherits its spread from the binomial. Dividing by $n$ shrinks that spread, and the standard deviation of the sample proportion works out to

\mathrm{sd}(\hat{p}) = \sqrt{\frac{p(1-p)}{n}}

Read the formula as a story. The $p(1-p)$ on top is largest when $p = 0.5$ , so a population split fifty fifty is the hardest to pin down. The $n$ underneath is inside a square root, so a bigger sample shrinks the spread, just more slowly than you might expect. To halve the spread you do not double the sample, you have to multiply it by four.

Learn it, then fade the help

1 · I do — fully worked

Mean and standard deviation of $\hat{p}$ — fully worked. A population has $p = 0.3$ ; samples of size $n = 50$ are taken. The mean of $\hat{p}$ is always the population proportion: $E(\hat{p}) = p = 0.3$ . The standard deviation is $\mathrm{sd}(\hat{p}) = \sqrt{\dfrac{p(1-p)}{n}} = \sqrt{\dfrac{0.3 \times 0.7}{50}} = \sqrt{0.0042} \approx 0.0648$ (to four decimal places).

2 · We do — your turn, guided

Same steps, your turn. A population has $p = 0.6$ ; samples of size $n = 400$ are taken. Find $E(\hat{p})$ and $\mathrm{sd}(\hat{p})$ . Begin with $E(\hat{p}) = 0.6$ , then $\mathrm{sd}(\hat{p}) = \sqrt{\dfrac{0.6 \times 0.4}{400}} = \sqrt{\;?\;}$ .

Check your step

$\mathrm{sd}(\hat{p}) = \sqrt{0.0006} \approx 0.0245$ . So $E(\hat{p}) = 0.6$ and $\mathrm{sd}(\hat{p}) \approx 0.0245$ — the spread is small because the sample is large.

3 · You do — solo

No scaffolding now: a population has $p = 0.25$ and samples of size $n = 64$ are taken. Find the mean and standard deviation of $\hat{p}$ , to four decimal places.

Reveal solution

$E(\hat{p}) = p = 0.25$ , and $\mathrm{sd}(\hat{p}) = \sqrt{\dfrac{0.25 \times 0.75}{64}} = \sqrt{0.0029296\ldots} \approx 0.0541$ .

The shape: an approximately normal curve

For a large enough sample the sampling distribution of $\hat{p}$ is not just centred and spread in a known way, it is also a familiar shape. It is approximately a normal curve:

\hat{p} \approx N\!\left(p,\ \frac{p(1-p)}{n}\right)

That means once you have the mean $p$ and the standard deviation $\sqrt{p(1-p)/n}$ , you can answer probability questions about $\hat{p}$ exactly the way you handle any normal distribution. Questions like “what is the chance a sample gives a proportion below $0.15$ ” become standard normal calculations. This bell shape is also the foundation for confidence intervals, which is the next topic.

How to actually do it

Every sampling distribution question runs on the same short recipe.

Identify the population proportion $p$ and the sample size $n$ from the wording.
The mean of $\hat{p}$ is just $p$ .
The standard deviation of $\hat{p}$ is $\sqrt{p(1-p)/n}$ .
If the question asks for a probability, treat $\hat{p}$ as normal with that mean and standard deviation.

Two traps catch students every year. The first is mixing up the count and the proportion. Always divide by $n$ , so the answer is a fraction like $0.32$ , never the raw count. The second is rounding too early. Carry full accuracy through the square root and only round the final answer to the number of decimal places asked for.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

What is the sample proportion

\hat{p}

, and how does it differ from

p

What is the mean of the sampling distribution of

\hat{p}

Write the standard deviation of

\hat{p}

If you multiply the sample size

n

4

(with

p

fixed), what happens to

\mathrm{sd}(\hat{p})

For a large sample, what is the approximate distribution of

\hat{p}

Recall · The binomial distribution

Why is the count

X

in a sample binomial, and how does

\hat{p}

relate to it?

Recall · The normal distribution

How do you find a probability like

\Pr(\hat{p} < 0.15)

See this recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked Example 1Luggage labelled heavy, from a real exam

On a particular day a random sample of $35$ pieces of luggage is selected. Let $\hat{P}$ be the proportion of luggage labelled heavy in random samples of size $35$ , where the population proportion is $p = 0.234$ . (i) Find $\Pr(\hat{P} > 0.2)$ , correct to three decimal places. (ii) Find the probability that $\hat{P}$ lies within one standard deviation of its mean, correct to three decimal places. Do not use a normal approximation.

1
(i) Since $\hat{P} = X/35$ , the event $\hat{P} > 0.2$ means $X > 0.2 \times 35 = 7$ , that is $X \ge 8$ , where $X \sim \mathrm{Bi}(35, 0.234)$ .
$\Pr(X \ge 8) = 1 - \Pr(X \le 7) \approx 0.595$
2
(ii) The mean is $p = 0.234$ and $\mathrm{sd}(\hat{P}) = \sqrt{\dfrac{p(1-p)}{35}} \approx 0.0716$ . Within one standard deviation of the mean corresponds to $X$ from $6$ to $10$ .
$\Pr(6 \le X \le 10) \approx 0.684$
3
The examiner report notes the common errors were using the wrong inequality in (i) and using a normal approximation, or the wrong standard deviation, in (ii).
$\text{(i)}\ \approx 0.595, \qquad \text{(ii)}\ \approx 0.684$

Answer

\text{(i)}\ \Pr(\hat{P} > 0.2) \approx 0.595, \qquad \text{(ii)}\ \approx 0.684

VCAA 2024 Mathematical Methods Exam 2, Section B Q4d

Worked Example 2Smallest sample size for a target spread, from a real exam

A sample of $n$ people gives a sample proportion of $\hat{p} = 0.6$ . Find the smallest value of $n$ for which the standard deviation of the sample proportion is $\dfrac{\sqrt{2}}{50}$ .

1
Set the standard deviation of the sample proportion equal to the target value, with $\hat{p} = 0.6$ .
$\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.6 \times 0.4}{n}} = \frac{\sqrt{2}}{50}$
2
Square both sides to clear the square roots.
$\frac{0.24}{n} = \frac{2}{2500}$
3
Solve for $n$ . The examiner report notes that students who set up the standard-deviation equation correctly generally reached the answer.
$n = \frac{0.24 \times 2500}{2} = 300$

Answer

n = 300

VCAA 2024 Mathematical Methods Exam 1, Q5c.ii

Worked Example 3Mean and standard deviation of p hat

In a large population, the true proportion who own a pet is $p = 0.3$ . Random samples of size $n = 50$ are taken. Find the mean and the standard deviation of the sample proportion $\hat{p}$ , with the standard deviation correct to four decimal places.

1
The mean of $\hat{p}$ is always the population proportion $p$ .
$E(\hat{p}) = p = 0.3$
2
The standard deviation of $\hat{p}$ is $\sqrt{p(1-p)/n}$ . Substitute $p = 0.3$ and $n = 50$ .
$\mathrm{sd}(\hat{p}) = \sqrt{\frac{0.3 \times 0.7}{50}} = \sqrt{0.0042}$
3
Evaluate the square root and round at the very end.
$\sqrt{0.0042} \approx 0.0648$

Answer

E(\hat{p}) = 0.3, \quad \mathrm{sd}(\hat{p}) \approx 0.0648

Worked Example 4A bigger sample sharpens the spread

For a population proportion $p = 0.6$ , random samples of size $n = 400$ are taken. Find the standard deviation of the sample proportion $\hat{p}$ , correct to four decimal places.

1
Write the standard deviation formula with $p = 0.6$ and $n = 400$ .
$\mathrm{sd}(\hat{p}) = \sqrt{\frac{0.6 \times 0.4}{400}} = \sqrt{0.0006}$
2
Evaluate.
$\sqrt{0.0006} \approx 0.0245$
3
Notice the spread is small because the sample is large. The same $p$ with $n = 50$ would give a spread about three times wider.
$\mathrm{sd}(\hat{p}) \approx 0.0245$

Answer

\mathrm{sd}(\hat{p}) \approx 0.0245

Worked Example 5Using the normal shape of p hat

A population has proportion $p = 0.2$ . A random sample of size $n = 100$ is taken, so $\hat{p}$ is approximately normal. Find $\Pr(\hat{p} < 0.15)$ , correct to four decimal places.

1
The mean is $p = 0.2$ . Find the standard deviation with $p = 0.2$ and $n = 100$ .
$\mathrm{sd}(\hat{p}) = \sqrt{\frac{0.2 \times 0.8}{100}} = \sqrt{0.0016} = 0.04$
2
So $\hat{p}$ is approximately normal with mean $0.2$ and standard deviation $0.04$ .
$\hat{p} \approx N(0.2,\ 0.04^2)$
3
Find the probability below $0.15$ using the normal distribution.
$\Pr(\hat{p} < 0.15) \approx 0.1056$

Answer

\Pr(\hat{p} < 0.15) \approx 0.1056

Practice test

Try it yourself

7 questions, 10 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.The probability a driver is late on any given working day is $0.08704$ . Let $\hat{P}$ be the proportion of days the driver is late in a five-day working week. Find $\Pr(0.4 \le \hat{P} \le 0.6)$ , correct to four decimal places.

2marks

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Mathematical Methods Exam 2, Section B Q3biii

Q2.A random sample of $n = 250$ people contains $80$ who use public transport. The value of the sample proportion $\hat{p}$ is:

1mark

Need a hint?

The sample proportion is the count of successes divided by the sample size, so work out

80 \div 250

Q3.Random samples of size $n$ are taken from a population with proportion $p$ . The mean of the sample proportion $\hat{p}$ is:

1mark

Need a hint?

The sampling distribution is centred on the truth, so ask which option describes the centre rather than the spread.

Q4.For a population proportion $p = 0.25$ and samples of size $n = 64$ , the standard deviation of $\hat{p}$ is closest to:

1mark

Need a hint?

Use

\sqrt{p(1-p)/n}

with

p = 0.25

and

n = 64

, and remember the final square root.

Q5.Keeping the population proportion $p$ fixed, the sample size $n$ is multiplied by $4$ . The standard deviation of $\hat{p}$ is:

1mark

Need a hint?

\sqrt{p(1-p)/n}

the

n

sits under a square root, so think about what

\sqrt{4}

does to the spread.

Q6.A population has proportion $p = 0.5$ and samples of size $n = 100$ are taken, so $\hat{p}$ is approximately normal. Which statement is correct?

1mark

Need a hint?

The mean is just

p

; work out the standard deviation as

\sqrt{p(1-p)/n}

with

p = 0.5

and

n = 100

Q7.A population has proportion $p = 0.5$ . Random samples of size $n = 100$ are taken. Find the mean and the standard deviation of the sample proportion $\hat{p}$ . Show your working.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

What is the difference between p and p hat?

The plain p is the true proportion for the whole population, a fixed number you usually never know. The p hat is the proportion you actually measure in one sample, and it changes from sample to sample. You use p hat as your best estimate of p.

Why does a bigger sample give a smaller standard deviation?

Because the sample size sits underneath a square root in the spread formula. More people in the sample means the random ups and downs average out more, so your estimate jumps around less. To halve the spread you need four times the sample, not just double it.

When can I treat the sample proportion as normal?

When the sample is reasonably large, the sampling distribution of the sample proportion is close enough to a bell curve to use the normal distribution. That lets you answer probability questions about the sample proportion the same way you would for any normal variable.

Sampling Distributions of Proportions

Learn

The sample proportion and why it moves

The centre: the mean of p hat

The spread: the standard deviation of p hat

The shape: an approximately normal curve

How to actually do it

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions