---
title: "Sampling distribution: population 4,8,12,16,20"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:population-4-8-12-16-20"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/population-4-8-12-16-20
site: BCT Engineering
---

# Sampling distribution: population 4,8,12,16,20

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2076 Chaitra, 5 marks
- Video: [8.18: Sampling Distribution of the Mean | Population Mean and Standard Deviation | Walpole](https://www.youtube.com/watch?v=CRKk-6DhYe0) — Engineering Tutor
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/population-4-8-12-16-20
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

A population consists of five numbers 4, 8, 12, 16 and 20. If a random sample of size 2 is drawn without replacement.
a) Find the population mean and population standard deviation.
b) List all possible sample and find their sample mean.
c) Show the mean of sample mean is equal to the population mean.
d) Find the standard error of sample mean.

## Short answer

Population mean is 12 and standard deviation is 5.6569. All 10 samples without replacement have means summing to 120, so mean of sample means is 12, equal to μ. Standard error is (5.6569/√2)√(3/4)=3.4641.

## Answer

Population: 4, 8, 12, 16, 20 (N = 5); sample size n = 2, without replacement. Number of samples = ⁵C₂ = 10.

#### a) Population mean and standard deviation
μ = Σx/N = 60/5 = **12**

σ² = Σ(x − μ)²/N = (64 + 16 + 0 + 16 + 64)/5 = 160/5 = 32

σ = √32 = **5.6569**

#### b) All possible samples and their means
| No. | Sample | Sample mean |
|---|---|---|
| 1 | (4, 8) | 6 |
| 2 | (4, 12) | 8 |
| 3 | (4, 16) | 10 |
| 4 | (4, 20) | 12 |
| 5 | (8, 12) | 10 |
| 6 | (8, 16) | 12 |
| 7 | (8, 20) | 14 |
| 8 | (12, 16) | 14 |
| 9 | (12, 20) | 16 |
| 10 | (16, 20) | 18 |

#### c) Mean of sample means equals population mean
Sum of sample means = 6 + 8 + 10 + 12 + 10 + 12 + 14 + 14 + 16 + 18 = 120

Mean of sample means = 120/10 = 12

This is equal to the population mean μ = 12. Hence E(x̄) = μ is verified.

#### d) Standard error of the sample mean
With sampling without replacement from a finite population:

SE = (σ/√n) × √((N − n)/(N − 1)) = (5.6569/√2) × √((5 − 2)/(5 − 1))

= 4.0000 × 0.8660 = **3.4641**

(Direct check: the standard deviation of the 10 sample means is 3.4641.)

**Answer: μ = 12, σ = 5.6569; mean of sample means = 12; SE = 3.4641.**

## Check yourself

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

1. How many samples of size 2 without replacement come from 5 numbers?
   A. 25
   B. 10
   C. 20
   D. 5
   Answer: B. 5C2 = 10. (Key point: Number of samples)

2. For 4, 8, 12, 16, 20, μ and σ are:
   A. 12 and sqrt(32) = 5.657
   B. 12 and 32
   C. 10 and 5.657
   D. 12 and 160
   Answer: A. μ = 60/5 = 12 and σ^2 = 160/5 = 32. (Key point: Population mean and SD)

3. The mean of the 10 sample means is:
   A. 12, equal to μ
   B. 10
   C. 120
   D. 6
   Answer: A. Sum 120 divided by 10 = 12 = μ, verifying E(x̄) = μ. (Key point: Mean of sample means)

4. Which calculation gives the standard error of the sample mean?
   A. (σ/√n) × sqrt((N − n)/(N − 1)) = 3.4641
   B. σ/√n = 4.0
   C. σ × sqrt(n) = 8
   D. σ/n = 2.83
   Answer: A. Sampling without replacement from a finite population needs the correction: 4.0 × 0.8660 = 3.4641. (Key point: SE with finite correction)

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