---
title: "Standard error; population 12,19,13,16"
subject: "Probability and Statistics"
chapter: 4
key: "semester_5/probability_and_statistics:standard-error-population-12-19-13"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/standard-error-population-12-19-13
site: BCT Engineering
---

# Standard error; population 12,19,13,16

Probability and Statistics · Chapter 4: Sampling Distribution

- Asked in: 2072 Kartik, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/standard-error-population-12-19-13
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-4

## Question

Define standard error of sample mean. A population consist of the four numbers 12, 19, 13, 16.
i) Write down all possible sample size of two without replacement.
ii) Find standard error of the sample mean.

## Short answer

Standard error of the sample mean is the standard deviation of the sampling distribution of x̄. All six samples of size two are (12,19),(12,13),(12,16),(19,13),(19,16),(13,16); population σ=2.7386, so SE=(2.7386/√2)√(2/3)=1.5811.

## Answer

**Standard error of the sample mean:** the standard deviation of the sampling distribution of x̄. It is σ/√n for infinite population (or sampling with replacement) and (σ/√n)√((N − n)/(N − 1)) for sampling without replacement from a finite population of size N.

Population: 12, 19, 13, 16 (N = 4), n = 2 without replacement.

#### i) All possible samples
| No. | Sample | Sample mean |
|---|---|---|
| 1 | (12, 19) | 15.5 |
| 2 | (12, 13) | 12.5 |
| 3 | (12, 16) | 14 |
| 4 | (19, 13) | 16 |
| 5 | (19, 16) | 17.5 |
| 6 | (13, 16) | 14.5 |

#### ii) Standard error of the sample mean
Population mean and standard deviation:

μ = Σx/N = 60/4 = **15**

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

σ = √7.5 = **2.7386**

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.)

Mean of sample means = 15 = μ.

**Answer: SE of the sample mean = 1.5811.**

## Check yourself

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

1. The standard error of the sample mean is:
   A. The standard deviation of the population
   B. The mean of all sample means
   C. The standard deviation of the sampling distribution of x̄
   D. The difference between x̄ and μ in one sample
   Answer: C. SE is the standard deviation of the sampling distribution of x̄. (Key point: Definition of standard error)

2. How many samples of size 2 can be drawn without replacement from 4 numbers?
   A. 16
   B. 6
   C. 8
   D. 4
   Answer: B. 4C2 = 6 samples (12,19), (12,13), (12,16), (19,13), (19,16), (13,16). (Key point: Number of samples)

3. Why is the factor √((N − n)/(N − 1)) used here?
   A. Because the population is normal
   B. Because n is less than 30
   C. Because the sampling is without replacement from a small finite population
   D. Because σ is unknown
   Answer: C. Sampling without replacement from a finite population reduces the variation of x̄, so SE is multiplied by √((N − n)/(N − 1)). (Key point: Finite population correction)

4. With σ = 2.7386, n = 2 and N = 4, SE of the sample mean is:
   A. 1.9365
   B. 2.7386
   C. 1.5811
   D. 0.8165
   Answer: C. SE = (2.7386/√2) × √(2/3) = 1.9365 × 0.8165 = 1.5811, which matches the SD of the six sample means. (Key point: Value of SE)

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