Sampling distribution of a sample proportion | Year 12

01 / LEARN

Understand the idea.

Scale a binomial count

Under $X \sim Bin (n, p)$ , the sample proportion is the scaled count $\hat{p} = \frac{X}{n}$ . Scaling gives $E (\hat{p}) = p$ and $Var (\hat{p}) = \frac{p (1 - p)}{n}$ . Take the square root for standard deviation. Exact possible proportions remain discrete, separated by $\frac{1}{n}$ .

A normal approximation is more credible when expected successes and failures are both reasonably large. Check $n p \geq 10$ and $n (1 - p) \geq 10$ as a screening rule, alongside the binomial assumptions. These checks support an approximation rather than guaranteeing a particular accuracy.

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

Keep model probabilities separate from interval estimates

For a probability question under a stated population model, use the specified $p$ in both the centre and spread. Do not substitute an observed $\hat{p}$ : that substitution belongs to estimating uncertainty when the population proportion is unknown.

A normal curve is continuous, whereas a binomial count is discrete. Exact count probabilities can require an exact binomial calculation or a stated continuity correction. If the question requests an uncorrected continuous proportion approximation, follow that instruction and label the result approximate.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Centre and spread

Assume independent trials with p = 0.4 and n = 600. Find sample-proportion mean and SD.

Worked solution

1
Keep the model proportion as the centre.
$E (\hat{p}) = p = 0.4$
A sample proportion from independent trials with common success probability estimates that probability without bias.
2
Calculate the variance.
$Var (\hat{p}) = \frac{0.4 (0.6)}{600} = 0.0004$
Scaling the binomial count by sample size gives variance divided by the square of that size.
3
Take the square root.
$SD (\hat{p}) = \sqrt{0.0004} = 0.02$
Standard deviation is the square root of variance, not the variance itself.
4
Check the expected counts.
$\begin{aligned} n p & = 600 (0.4) = 240 \\ n (1 - p) & = 600 (0.6) = 360 \end{aligned}$
Both counts comfortably pass the usual screening rule if a normal approximation is subsequently used.

Answer

Mean 0.4; SD 0.02.

EXAMPLE 02

Approximate range probability

For the same model, approximate $P (0.36 < \hat{p} < 0.44)$ without continuity correction.

Worked solution

1
Use the sample-proportion normal approximation.
$\hat{p} \approx N (0.4, {0.02}^{2})$
The independent-trial model has sufficiently large expected success and failure counts for the requested approximation.
2
Standardise the endpoints.
$\begin{aligned} z_{L} & = \frac{0.36 - 0.4}{0.02} = - 2 \\ z_{U} & = \frac{0.44 - 0.4}{0.02} = 2 \end{aligned}$
The sampling standard deviation is 0.02; the bounds lie two such standard deviations from the centre.
3
Calculate the central area.
$\begin{aligned} P (0.36 < \hat{p} < 0.44) & \approx Φ (2) - Φ (- 2) \\ \approx 0.9545 \end{aligned}$
This uses the continuous normal approximation without continuity correction, as requested, rather than the exact discrete binomial probability.

Answer

0.9545 approximately.

03 / YOUR TURN

Try it yourself.

Check your understanding

For p = 0.5 and n = 100, find the sample-proportion SD.

Show answer & reasoning

0.05.

Substitute the model probability and sample size: $SD (\hat{p}) = \sqrt{\frac{0.5 (0.5)}{100}} = \sqrt{0.0025} = 0.05$ . The proportion has no measurement unit; this spread equals five percentage points.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

For $p = 0.3$ , $n = 400$ , find sample-proportion mean and variance.
2 marks
For $p = 0.25$ , $n = 300$ , find sample-proportion SD.
2 marks
Check normal expected-count conditions for $p = 0.02$ , $n = 100$ .
2 marks
For $p = 0.5$ , $n = 100$ , approximate $P (0.4 < \hat{p} < 0.6)$ without continuity correction.
2 marks
Why cannot a sample of 20 have sample proportion exactly 0.333?
2 marks
Compare sample-proportion SDs for $n = 150$ and $n = 600$ with $p$ fixed.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For $p = 0.3$ , $n = 400$ , find sample-proportion mean and variance.
2 marks
Variance $= \frac{0.3 \times 0.7}{400}$ .
0.3 and 0.000525.
For $p = 0.25$ , $n = 300$ , find sample-proportion SD.
2 marks
$\sqrt{\frac{0.25 (0.75)}{300}} = 0.025$ .
$0.025$ .
Check normal expected-count conditions for $p = 0.02$ , $n = 100$ .
2 marks
Expected failures are 98; both counts must be adequately large.
They fail: expected successes $= 2$ .
For $p = 0.5$ , $n = 100$ , approximate $P (0.4 < \hat{p} < 0.6)$ without continuity correction.
2 marks
SD $= 0.05$ ; standardised bounds are $- 2$ and $2$ .
0.9545 approximately.
Why cannot a sample of 20 have sample proportion exactly 0.333?
2 marks
The numerator is an integer and the denominator is 20.
Possible values are integer multiples of 0.05.
Compare sample-proportion SDs for $n = 150$ and $n = 600$ with $p$ fixed.
2 marks
Quadrupling $n$ divides the SD by two.
The second is half the first.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: sampling distributions and confidence intervals for means and proportions. Mean tasks distinguish known population spread from sufficiently large-sample estimated-spread normal intervals. Repeated-interval computational simulation, approximation conditions, design and interpretation are included.

Official curriculum source ↗

In this lesson

Understand the idea.

Scale a binomial count

Keep model probabilities separate from interval estimates

See the method in action.

Centre and spread

Keep the model proportion as the centre.

Calculate the variance.

Take the square root.

Check the expected counts.

Approximate range probability

Use the sample-proportion normal approximation.

Standardise the endpoints.

Calculate the central area.

Try it yourself.

Check your understanding

Keep practising