Draw many samples of the same size from a population and compute the mean of each. Those means will not all be equal – they form their own distribution, the sampling distribution of the mean. Its standard deviation, σx̄, tells you how far a single sample mean is likely to fall from the true population mean.
The rule is simple and powerful: σx̄ = σ ÷ √n. Averages vary less than individual values, and the larger the sample the less they vary. Quadrupling the sample size halves σx̄; a hundred-fold increase shrinks it to a tenth. By the central limit theorem the sampling distribution is approximately normal for samples of about 30 or more even when the population is not, so about 95% of sample means fall within 1.96σx̄ of μ.
When you sample a sizeable share of a small population without replacement, the means vary even less. The finite population correction √((N − n) ÷ (N − 1)) accounts for this; it matters once the sample is more than about 5% of the population. When σ is unknown and estimated by the sample SD s, the same quantity is called the standard error of the mean.
How to use the Standard Deviation of the Sample Mean Calculator
- Population standard deviation (σ)
- Sample size (n)
- Population size
- Population size (N)
Results update as you type, and everything is calculated in your browser.
How it works
Enter the population standard deviation σ and the sample size n. Choose a finite population and give its size N if you sample without replacement from a small population. The calculator divides σ by √n and applies the correction if chosen.
Worked example
IQ scores have σ = 15. For random samples of n = 36 people, σx̄ = 15 ÷ √36 = 2.5 points, so 95% of sample means fall within ±1.96 × 2.5 = ±4.9 points of the population mean. If the 36 are drawn without replacement from a school of 500 students, the correction √(464 ÷ 499) = 0.9643 reduces σx̄ to 2.41. To halve σx̄ to 1.25 you would need 144 people.
How sample size shrinks σx̄ (σ = 15)
| n | σx̄ | 95% range of sample means |
|---|---|---|
| 1 | 15.0 | ±29.4 |
| 4 | 7.5 | ±14.7 |
| 25 | 3.0 | ±5.9 |
| 100 | 1.5 | ±2.9 |
| 400 | 0.75 | ±1.5 |
Inputs explained
| Field | What to enter |
|---|---|
| Population standard deviation (σ) | A number |
| Sample size (n) | A number |
| Population size | Choose one: Very large or sampling with replacement, Finite – apply the correction |
| Population size (N) | A number |
What each result means
- Standard deviation of the sample mean σx̄ = [Population standard deviation (σ)]÷sqrt([Sample size (n)])×[Finite population correction factor]
- Variance of the sample mean σx̄² = [Standard deviation of the sample mean σx̄]^2
- 95% of sample means within ± (1.96σx̄) = 1.959964×[Standard deviation of the sample mean σx̄]
- 99% of sample means within ± (2.576σx̄) = 2.575829×[Standard deviation of the sample mean σx̄]
- Reduction compared with one observation = [Standard deviation of the sample mean σx̄]÷[Population standard deviation (σ)]×100
- Sample size needed to halve σx̄ = 4×[Sample size (n)]
Formula
σx̄ = σ ÷ √n · with a finite population of N: σx̄ = (σ ÷ √n) × √((N − n) ÷ (N − 1)) · 95% of sample means lie within μ ± 1.96σx̄
Worked example
Inputs — Population standard deviation (σ): 15; Sample size (n): 36; Population size: Very large or sampling with replacement.
Results — Standard deviation of the sample mean σx̄ = 2.5; Variance of the sample mean σx̄² = 6.25; 95% of sample means within ± (1.96σx̄) = 4.89991; 99% of sample means within ± (2.576σx̄) = 6.439573; Reduction compared with one observation = 16.67 %; Sample size needed to halve σx̄ = 144.
Key terms
- Sampling distribution
- The distribution of a statistic, such as the mean, over all possible samples of a given size.
- Central limit theorem
- The result that sample means are approximately normal for large samples, whatever the population shape.
- Finite population correction
- The factor √((N − n)/(N − 1)) that reduces the spread when sampling without replacement.
Frequently asked questions
What is the standard deviation of the sample mean?
The standard deviation of all possible sample means of size n: σx̄ = σ ÷ √n. With σ = 7.1 cm and n = 100 it is 0.71 cm.
Is it the same as the standard error?
Conceptually yes. "Standard deviation of the sample mean" usually uses the known population σ; the "standard error" uses the sample SD s as an estimate.
Why does a larger sample give a smaller value?
Random highs and lows cancel out more in a larger average. Because of the square root, four times the sample size is needed to halve σx̄.
When do I need the finite population correction?
When sampling without replacement and n is more than about 5% of N. Sampling 36 from 500 gives a factor of 0.964, reducing σx̄ from 2.5 to 2.41.
What is the difference between the sample distribution and the sampling distribution?
The sample distribution is the spread of individual values in one sample; the sampling distribution is the spread of a statistic (such as the mean) across many samples.
Does the population have to be normal?
No. For n of about 30 or more the central limit theorem makes the sampling distribution approximately normal; for very skewed populations a larger n is safer.