---
title: "Sampling distribution of proportion"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:sampling-distribution-proportion-example"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/sampling-distribution-proportion-example
site: BCT Engineering
---

# Sampling distribution of proportion

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2071 Shrawan, 4 marks; 2069 Chaitra, 4 marks (asked 2 times)
- Video: [The Sampling Distribution of the Sample Proportion](https://www.youtube.com/watch?v=fuGwbG9_W1c) — jbstatistics
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/sampling-distribution-proportion-example
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

Define sampling distribution of proportion with example.

## Short answer

The sampling distribution of proportion is the distribution of sample proportions p̂ from all possible samples. Its mean equals population proportion P and standard error is sqrt(PQ/n), with finite population correction sqrt((N−n)/(N−1)). Example: P=0.05, n=100 gives SE=0.0218.

## Answer

**Sampling distribution of proportion:** if all possible samples of size *n* are drawn from a population in which the proportion having a certain attribute is P, and the sample proportion p̂ = x/n (x = number of successes in the sample) is calculated for each, the probability distribution of p̂ is the sampling distribution of the proportion.

Properties:
- Mean: E(p̂) = P
- Standard error: SE(p̂) = √(PQ/n), where Q = 1 − P
- For finite population (without replacement): SE = √(PQ/n) × √((N − n)/(N − 1))
- For large n (nP ≥ 5 and nQ ≥ 5) it is approximately normal, so z = (p̂ − P)/√(PQ/n).

#### Example
A factory knows that 5% of the items it makes are defective (P = 0.05). Samples of n = 100 items are taken.

- Mean of p̂ = 0.05
- SE = √(0.05 × 0.95/100) = √0.000475 = 0.0218

Probability that the sample proportion defective exceeds 8%:

z = (0.08 − 0.05)/0.0218 = 1.38

P(p̂ > 0.08) = P(Z > 1.38) = 0.0838

So only about 8.4% of such samples will have more than 8% defectives.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The mean of the sampling distribution of p̂ is:
   A. P
   B. P/n
   C. nP
   D. PQ
   Answer: A. E(p̂) = P, so p̂ is unbiased for P. (Key point: Mean of p̂)

2. The standard error of p̂ is:
   A. sqrt(PQ/n)
   B. PQ/n
   C. sqrt(PQ)/n
   D. sqrt(P/n)
   Answer: A. SE(p̂) = sqrt(PQ/n) with Q = 1 − P. (Key point: Standard error of p̂)

3. For P = 0.05 and n = 100, SE is about:
   A. 0.0218
   B. 0.05
   C. 0.0005
   D. 0.218
   Answer: A. sqrt(0.05 × 0.95/100) = sqrt(0.000475) = 0.0218. (Key point: Example SE)

4. p̂ is approximately normal when:
   A. nP ≥ 5 and nQ ≥ 5
   B. n < 5
   C. P = 0.5 only
   D. nP < 1
   Answer: A. A large enough sample with nP and nQ at least 5 allows z = (p̂ − P)/SE. (Key point: Normal condition)

## Related in Probability and Statistics

- [Sampling distribution of proportion; parameter vs statistic](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aparameter-vs-statistic-sampling-proportion) — Probability and Statistics, chapter 4
- [Sampling distribution of mean; standard error](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Asampling-distribution-mean-se-finite) — Probability and Statistics, chapter 4, asked 2×
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