---
title: "Sampling distribution: population 5,6,9,12"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:population-5-6-9-12-samples"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/population-5-6-9-12-samples
site: BCT Engineering
---

# Sampling distribution: population 5,6,9,12

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2069 Chaitra, 2×3 marks
- Video: [Sampling Distribution and Standard Error by Sanchit Sir | @utmuniquetheoryofmathemati5497](https://www.youtube.com/watch?v=iaTBYT8VECA) — UTM : UniqueTheoryOfMathematics
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/population-5-6-9-12-samples
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

A population consists of 5, 6, 9, 12. Consider all possible samples of size two which can be drawn without replacement from this population. Find
i) Population mean and population standard deviation.
ii) Mean of sampling distribution of mean.
iii) Standard error of sampling distribution of mean.

## Short answer

Population mean μ = (5+6+9+12)/4 = 8, σ = √7.5 ≈ 2.7386. Mean of sampling distribution of mean = 8. Standard error for sampling without replacement: SE = (σ/√n)√((N-n)/(N-1)) = (2.7386/√2)√(2/3) ≈ 1.5811.

## Answer

Population: 5, 6, 9, 12 (N = 4); samples of size n = 2 without replacement, ⁴C₂ = 6 samples.

#### i) Population mean and population standard deviation
μ = Σx/N = 32/4 = **8**

σ² = Σ(x − μ)²/N = (9 + 4 + 1 + 16)/4 = 30/4 = 7.5

σ = √7.5 = **2.7386**

#### ii) Mean of the sampling distribution of the mean
All samples:

| No. | Sample | Sample mean |
|---|---|---|
| 1 | (5, 6) | 5.5 |
| 2 | (5, 9) | 7 |
| 3 | (5, 12) | 8.5 |
| 4 | (6, 9) | 7.5 |
| 5 | (6, 12) | 9 |
| 6 | (9, 12) | 10.5 |

Sum of sample means = 5.5 + 7 + 8.5 + 7.5 + 9 + 10.5 = 48

Mean of sample means = 48/6 = 8

This equals the population mean μ = 8.

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

SE = (σ/√n) × √((N − n)/(N − 1)) = (2.7386/√2) × √((4 − 2)/(4 − 1))

= 1.9365 × 0.8165 = **1.5811**

(Direct check: the standard deviation of the 6 sample means is 1.5811.)

**Answer: μ = 8, σ = 2.7386; mean of sample means = 8; SE = 1.5811.**

## Check yourself

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

1. For the population 5, 6, 9, 12, the mean μ and σ are:
   A. μ = 8, σ = 2.7386
   B. μ = 8, σ = 7.5
   C. μ = 32, σ = 2.7386
   D. μ = 8, σ = 1.5811
   Answer: A. μ = 32/4 = 8; σ² = 30/4 = 7.5, so σ = √7.5 = 2.7386. (Key point: Population mean and SD)

2. How many samples of size 2 without replacement are possible?
   A. 16
   B. 4
   C. 12
   D. 6
   Answer: D. 4C2 = 6 samples. (Key point: Number of samples)

3. The mean of the sampling distribution of the mean (sum of 6 sample means = 48) is:
   A. 8, equal to the population mean
   B. 48
   C. 7.5
   D. 9.6
   Answer: A. 48/6 = 8, which equals μ because x̄ is unbiased. (Key point: Mean of sample means)

4. The standard error for sampling without replacement is:
   A. 2.7386/√2 = 1.9365
   B. 1.5811, i.e. (σ/√n) × √((N − n)/(N − 1))
   C. 2.7386 × 0.8165
   D. √7.5/2 = 1.37
   Answer: B. SE = 1.9365 × √(2/3) = 1.9365 × 0.8165 = 1.5811. (Key point: Standard error with correction)

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