---
title: "Samples of 2,8,14,20 with standard error"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:population-2-8-14-20-with"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/population-2-8-14-20-with
site: BCT Engineering
---

# Samples of 2,8,14,20 with standard error

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2073 Shrawan, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/population-2-8-14-20-with
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

A population consists of the four number 2, 8, 14, 20:
i) Write down all possible sample size of two without replacement
ii) Verify that the population mean is equal to the mean of the sample mean
iii) Calculate the standard error of the sampling distribution of the sample mean

## Short answer

All six samples of size two without replacement are (2,8),(2,14),(2,20),(8,14),(8,20),(14,20), with means 5,8,11,11,14,17. Population mean and mean of sample means both equal 11; SE=(6.7082/√2)√(2/3)=3.8730.

## Answer

Population: 2, 8, 14, 20 (N = 4), n = 2, without replacement.

#### i) All possible samples
There are ⁴C₂ = 6 samples.

| No. | Sample | Sample mean |
|---|---|---|
| 1 | (2, 8) | 5 |
| 2 | (2, 14) | 8 |
| 3 | (2, 20) | 11 |
| 4 | (8, 14) | 11 |
| 5 | (8, 20) | 14 |
| 6 | (14, 20) | 17 |

#### ii) Population mean equals mean of sample means
μ = (2 + 8 + 14 + 20)/4 = 11

Sum of sample means = 5 + 8 + 11 + 11 + 14 + 17 = 66

Mean of sample means = 66/6 = 11

Both equal 11, so verified.

#### iii) Standard error of the sample mean
First the population standard deviation:

μ = Σx/N = 44/4 = **11**

σ² = Σ(x − μ)²/N = (81 + 9 + 9 + 81)/4 = 180/4 = 45

σ = √45 = **6.7082**

With sampling without replacement from a finite population:

SE = (σ/√n) × √((N − n)/(N − 1)) = (6.7082/√2) × √((4 − 2)/(4 − 1))

= 4.7434 × 0.8165 = **3.8730**

(Direct check: the standard deviation of the 6 sample means is 3.8730.)

**Answer: SE of sample mean = 3.8730.**

## Check yourself

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

1. For 2, 8, 14, 20 (μ = 11), σ is:
   A. sqrt(45) = 6.708
   B. 45
   C. sqrt(180) = 13.42
   D. 4.743
   Answer: A. σ^2 = (81 + 9 + 9 + 81)/4 = 45. (Key point: Population SD)

2. Sampling without replacement from N = 4 uses:
   A. σ/√n only
   B. (σ/√n) × sqrt((N − n)/(N − 1))
   C. σ × sqrt(n)
   D. σ/n
   Answer: B. A finite population sampled without replacement needs the finite population correction. (Key point: Which SE formula)

3. The standard error of the sample mean is:
   A. 4.7434
   B. 3.8730
   C. 6.7082
   D. 2.5820
   Answer: B. 4.7434 × sqrt(2/3) = 4.7434 × 0.8165 = 3.8730. (Key point: SE value)

4. The mean of the six sample means is 11, which shows:
   A. E(x̄) = μ
   B. SE = μ
   C. σ = 11
   D. the correction factor is 1
   Answer: A. The sampling distribution of x̄ is centred on the population mean. (Key point: Verification)

## Close variants set in other exams

- [Sampling distribution: population 2,8,14,20](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apopulation-2-8-14-20-samples) — Probability and Statistics, chapter 4, asked 2×

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