Sampling distribution of a sample proportion | Year 12

01 / LEARN

The idea, step by step

Scale a binomial count

For $X \sim Bin (n, p)$ , $\hat{p} = X / n$ has mean p and variance $p (1 - p) / n$ . Its standard deviation is the square root of that variance. Its possible values are separated by 1/n, so the exact distribution is discrete.

A normal approximation can be useful with sufficiently many expected successes and failures. The checks np ≥ 10 and n(1−p) ≥ 10 are a screening rule, not a universal accuracy guarantee.

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

Keep model probabilities separate from interval estimates

A sampling probability under a stated model uses its specified p in the spread. A confidence interval uses data to estimate an unknown p. These are different tasks.

Normal approximation is continuous. For count probabilities near integer boundaries use an exact binomial model or a stated continuity correction. When simply using a continuous approximate proportion range, identify it as approximate.

A common trap

The sample-proportion SD is p(1−p)/n.

Instead: That expression is the variance; take its square root.

02 / WORKED EXAMPLES

See the thinking

Example 1

Centre and spread

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

Variance $= 0.4 (0.6) / 600 = 0.0004$ .
Take its square root. Expected counts are 240 and 360.

Mean 0.4; SD 0.02.

Example 2

Approximate range probability

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

Bounds standardise to −2 and 2.
$Φ (2) - Φ (- 2) = 0.954499 \dots$ .

0.9545 approximately.

Check your understanding

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

Show answer

0.05.

$\sqrt{0.25 / 100} = 0.05$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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 = 0.3 × 0.7 / 400.
0.3 and 0.000525.
For p = 0.25, n = 300, find sample-proportion SD.
2 marks
$\sqrt{0.25 (0.75) / 300} = 0.025$ .
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 ↗

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