Geometric Mean Calculator Interactive tool
Variance measures the average squared distance of each value from the mean. Standard deviation is its square root — bringing the result back to the original unit. Population variance divides by n; sample variance divides by n − 1 (Bessel's correction). For dataset {4, 7, 13, 16, 20}: mean = 12, population variance = 34.8, population standard deviation = 5.90. This calculator returns both population and sample versions side by side.
Spread & dispersion
Variance measures the average squared distance of each value from the mean. Standard deviation is its square root — bringing the result back to the original unit. Population variance divides by n; sample variance divides by n − 1 (Bessel's correction). For dataset {4, 7, 13, 16, 20}: mean = 12, population variance = 34.8, population standard deviation = 5.90. This calculator returns both population and sample versions side by side.
Standard deviation quantifies the uncertainty hiding inside an average. A mean without SD is an incomplete summary.
Dataset A {10, 10, 10, 10, 10} and Dataset B {2, 6, 10, 14, 18} both have mean = 10. Dataset A has SD = 0 (perfectly consistent); Dataset B has SD = 5.66 (spread across a 16-point range). A quality-control engineer needs both numbers.
For normally distributed data: 68% of values fall within ±1 SD of the mean, 95% within ±2 SD, 99.7% within ±3 SD. For {4, 7, 13, 16, 20}: mean = 12, σ = 5.90 → 68% of values expected between 6.10 and 17.90. Three of the five values (7, 13, 16) fall in this range.
A sample tends to underestimate the full population’s spread because extreme values are less likely to be sampled. Dividing by n−1 instead of n inflates the estimate slightly to correct for this bias. For n = 5: s² = 170/4 = 42.5 versus σ² = 170/5 = 34.8. The 22% increase accounts for the missing coverage at the tails.
The choice of population or sample formula changes the numeric result. Each row shows a real situation and which formula is correct.
| Situation | Input | Best next move | Why |
|---|---|---|---|
| All scores from one class exam | 30 scores; this is the complete group | Population SD (÷ n) in Excel: =STDEV.P() | No estimation involved — you have every value in the group. |
| Survey of 100 users from 10,000 | 100 satisfaction scores | Sample SD (÷ n−1) in Excel: =STDEV.S() | The 100 scores estimate the spread of all 10,000 users. |
| Two products with the same mean defect rate | Product A: {1,1,1,1,1}, Product B: {0,0,1,2,2} | Compare SD: A = 0, B = 0.89 | Same mean (1.0) hides that Product B is inconsistent. |
| Investment return volatility | Annual returns: 6%, 8%, −35%, 10%, 12% | Pair sample SD with geometric mean (CAGR) | SD = 18.0% captures risk; CAGR = −2.0%/yr captures the compound return. |
Decide first: are these values the entire group (population) or a subset drawn from a larger group (sample)? This determines which formula applies.
Enter values separated by commas, spaces, or line breaks. At least 2 values are required; at least 3 values are needed for meaningful sample standard deviation.
Read the mean and standard deviation together. A mean of 12 with σ = 5.90 means roughly 68% of values fall between 6.10 and 17.90 (one standard deviation), assuming normal distribution.
If population σ and sample s differ significantly, your sample is small — consider collecting more data before drawing conclusions.
Values: 4, 7, 13, 16, 20 | n = 5 | mean = (4+7+13+16+20)/5 = 60/5 = 12
Squared deviations: (4−12)²=64, (7−12)²=25, (13−12)²=1, (16−12)²=16, (20−12)²=64. Sum = 170. Population variance = 170/5 = 34.8. Sample variance = 170/4 = 42.5.
Population SD = √34.8 = 5.90 | Sample SD = √42.5 = 6.52. The sample SD is 10% larger — Bessel's correction assumes the full population is more spread than the sample suggests.
SD of 5.90 on a mean of 12 means a coefficient of variation of 49% — this dataset is widely spread relative to its center. Compare with a tight dataset like {10, 11, 12, 13, 14} where SD = 1.58 and CV = 13%.
Mean tells you where data is centered. Variance and standard deviation tell you how far values typically stray from that center. Use them together — one without the other is incomplete.
| Method | Best for | Watch for | Example |
|---|---|---|---|
| Population variance (σ²) | When the dataset contains every member of the group you are describing. | Divides by n. Underestimates spread for samples — use s² instead. | All 30 students in one class: σ² = 34.8 for {4, 7, 13, 16, 20}. |
| Sample variance (s²) | When the dataset is a subset used to estimate the spread of a larger group. | Divides by n−1 (Bessel’s correction). s² is always larger than σ² for the same data. | Same data as above: s² = 170/4 = 42.5, which is 22% larger than σ² = 34.8. |
| Standard deviation (σ or s) | Reading spread in the original unit so it is directly comparable to the mean. | Sensitive to extreme values, just like the mean. A single outlier inflates both. | For {4, 7, 13, 16, 20}: mean = 12, σ = 5.90. Typical value is within 12 ± 5.90. |
| Mean (AM) | Locating the center of the data distribution. | Mean alone is misleading without SD. Mean = 10 for both {10,10,10} and {2,6,10,14,18}. | {2, 6, 10, 14, 18}: mean = 10, SD = 5.66. {10, 10, 10, 10, 10}: mean = 10, SD = 0. |
Squared deviations: (4-12)²=64, (7-12)²=25, (13-12)²=1, (16-12)²=16, (20-12)²=64. Sum = 170. σ² = 34.8, σ = 5.90.
SD = √(Σ(xᵢ − μ)² / n). Every step shown with the actual numbers.
(4+7+13+16+20)/5 = 60/5 = 12.
Differences from mean: −8, −5, +1, +4, +8.
Squared: 64, 25, 1, 16, 64. Sum = 170.
Population: 170/5 = 34.8. Sample: 170/4 = 42.5.
σ = √34.8 = 5.90. s = √42.5 = 6.52.
Use population standard deviation (σ, divides by n) when the dataset contains every member of the group you want to describe — all students in one class, all machines on one production line. Use sample standard deviation (s, divides by n − 1) when the dataset is a subset drawn to estimate the larger group — 100 survey respondents representing 10,000 users. Bessel's correction (n − 1) makes s slightly larger than σ because samples tend to underrepresent extreme values. For n = 5: dividing by 4 instead of 5 increases the estimate by 25%.
Two datasets can share the same arithmetic mean but have completely different distributions. Dataset A: {10, 10, 10, 10, 10} → mean = 10, SD = 0. Dataset B: {2, 6, 10, 14, 18} → mean = 10, SD = 5.66. Same center, completely different spread. A quality-control team watching a manufacturing process cares about both: a mean within spec but a high SD signals inconsistency that AM alone would hide.
Each mistake below produces a different number — not just a different interpretation.
For {4, 7, 13, 16, 20} as a sample: σ = 5.90 (wrong) vs s = 6.52 (correct). The 10% difference grows as the sample gets smaller. Use s² = Σ(xᵢ−x̄)² / (n−1) when estimating a larger group.
SD = 5.90 points for scores and SD = 2.1 kg for weights cannot be compared directly. Use coefficient of variation (CV = SD / mean × 100%) for unit-free comparison: CV = 5.90/12 × 100% = 49%.
Removing the 20 from {4, 7, 13, 16, 20} drops σ from 5.90 to 4.44. That is a 25% reduction from a single deletion. Keep outliers unless you have a documented data-quality reason (sensor fault, entry error).
If your dataset contains every value in the group (all employees in a company, all items produced in a shift), use population SD (÷ n). If it is a sample (e.g., 50 employees randomly chosen from 1,000), use sample SD (÷ n − 1). In Excel: =STDEV.P() for population, =STDEV.S() for sample.
Variance squares the deviations, so its unit is the square of the original (e.g., points² for test scores). Standard deviation takes the square root, returning the result to the original unit (points). For {4, 7, 13, 16, 20}: variance = 34.8 points², SD = 5.90 points — directly comparable to the mean of 12 points.
Squaring serves two purposes: (1) it prevents positive and negative deviations from canceling (|+8| and |−8| both contribute equally), and (2) it gives disproportionately more weight to large deviations — a value 8 units from the mean contributes 64 to the sum, while a value 4 units away contributes only 16. This is a deliberate feature that makes variance sensitive to large swings.
Yes, but only when every value is identical. {5, 5, 5, 5, 5}: mean = 5, all deviations = 0, σ = 0. A SD of zero means complete consistency — no variation in the data.
No. Context determines whether high spread is a problem. High SD in investment returns = more risk (bad for conservative investors). High SD in product test scores = wide customer talent range (neutral or good). Compare SD to the mean (CV) and to a relevant benchmark.
Remove values only when you have a documented, non-statistical reason: sensor malfunction, data entry error, wrong unit. Removing outliers to get a 'cleaner' SD is statistical manipulation. Instead, report the SD both with and without the outlier, and note the reason.