Why do we divide by n-1 instead of n for sample standard deviation?

Using n-1 corrects for the fact that the sample mean is already calculated from your data, which uses up one degree of freedom. This makes the sample standard deviation an unbiased estimator of the true population standard deviation—without it, sample values would systematically underestimate population spread.

What is the difference between standard deviation and variance?

Variance is the average of squared deviations, while standard deviation is the square root of variance. Standard deviation is usually preferred because it is in the same units as the original data, making it directly comparable to values in your dataset.

Can standard deviation be negative?

No, standard deviation is always non-negative. Since it is a square root of a sum of squared values, the result is zero only when all data points are identical, and positive whenever there is any spread.

When should I use standard deviation versus range?

Use standard deviation when you want to account for all data points and their distances from the mean; it is stable and informative. Use range (max - min) for a quick rough estimate when you only care about the span between extremes.

Standard Deviation

Standard deviation measures how spread out data points are from the average—use it to understand variability in a dataset.

s = \sqrt{\frac{1}{n-1}\sum (x_i - \bar{x})^2}

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What each symbol means

What Standard Deviation takes

Symbol	Meaning
$s$	The sample standard deviation, a measure of spread in the same units as the original data; if you interpret it as variance instead, you get squared units and lose interpretability.
$\bar{x}$	The mean (arithmetic average) of the dataset; if you read it as the median, mode, or another center measure, you'll calculate deviations from the wrong point.
$x_i$	The $i$ -th individual data point in the dataset; misidentifying which values to include changes the calculation fundamentally.
$n$	The number of data points in the sample; dividing by $n$ instead of $n-1$ produces a biased estimate that underestimates population spread.

When to use it

Use standard deviation when you need to understand how much individual values in a dataset deviate from the average.

Level

Usually taught in: Statistics · Appears on: AP Statistics

Worked examples

1. Find standard deviation of a simple dataset

Problem

Find the standard deviation of the dataset

\{2, 4, 6, 8, 10\}

$\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6$
Calculate the mean by summing all values and dividing by the count $n = 5$ .
$(2-6)^2 + (4-6)^2 + (6-6)^2 + (8-6)^2 + (10-6)^2 = 16 + 4 + 0 + 4 + 16 = 40$
Subtract the mean from each value, square the result, and sum—this captures total squared deviation from the center.
$\frac{40}{5-1} = \frac{40}{4} = 10$
Divide by $n-1 = 4$ (not $n$ ) to get the sample variance, which corrects for using the sample mean.
$s = \sqrt{10} \approx 3.162$
Take the square root of the variance to obtain standard deviation in the original units.

Answer: $s = \sqrt{10} \approx 3.162$

This straightforward example uses the formula directly: find the mean, calculate squared deviations from it, sum them, divide by $n-1$ , and take the square root.

2. Standard deviation with negative values and fractions

Problem

Find the standard deviation of

\{-2, 0, 1, 3\}

$\bar{x} = \frac{-2 + 0 + 1 + 3}{4} = \frac{2}{4} = 0.5$
Sum all values including negatives carefully, then divide by the count $n = 4$ to find the mean.
$(-2 - 0.5)^2 + (0 - 0.5)^2 + (1 - 0.5)^2 + (3 - 0.5)^2 = 6.25 + 0.25 + 0.25 + 6.25 = 13$
Each deviation is squared, which converts negatives to positives—this is crucial so that negative and positive deviations do not cancel out.
$s = \sqrt{\frac{13}{4-1}} = \sqrt{\frac{13}{3}} \approx 2.08$
Divide by $n-1 = 3$ because this is sample standard deviation, then take the square root.

Answer: $s = \sqrt{\frac{13}{3}} \approx 2.08$

With negative values and a non-integer mean, care is needed when computing deviations—always divide by $n-1$ for sample data, and square each deviation before summing.

3. Standard deviation of temperature measurements in a lab experiment

Problem

A chemistry student measures the temperature (in °C) of a reaction at five times: 20, 22, 21, 23, 19. Calculate the sample standard deviation.

$\bar{x} = \frac{20 + 22 + 21 + 23 + 19}{5} = \frac{105}{5} = 21$
The average temperature across all five measurements is 21°C.
$20 - 21 = -1, \quad 22 - 21 = 1, \quad 21 - 21 = 0, \quad 23 - 21 = 2, \quad 19 - 21 = -2$
Each measurement's deviation from the mean shows how far it is from the center; some are above (positive) and some below (negative).
$(-1)^2 + 1^2 + 0^2 + 2^2 + (-2)^2 = 1 + 1 + 0 + 4 + 4 = 10$
Squaring converts all deviations to positive values and emphasizes larger differences from the mean.
$s = \sqrt{\frac{10}{5-1}} = \sqrt{\frac{10}{4}} = \sqrt{2.5} \approx 1.58 \text{ °C}$
Divide by $n-1 = 4$ for sample data, then take the square root to return to the original units (°C).

Answer: $s = \sqrt{2.5} \approx 1.58 \text{ °C}$

Standard deviation quantifies how much the reaction temperature varied—a value of about 1.58°C tells us that individual measurements typically deviated from 21°C by roughly ±1.6°C.

Common mistakes

Where Standard Deviation usually goes wrong

The mistake	Why it is wrong	The fix
Using $n$ instead of $n-1$ in the denominator.	Dividing by $n$ instead of $n-1$ produces a biased estimator—sample standard deviations will systematically underestimate the true population spread.	Always use $n-1$ for sample data: $s = \sqrt{\frac{1}{n-1}\sum (x_i - \bar{x})^2}$ .
Reporting the variance as the standard deviation without taking the square root.	Variance and standard deviation have different units: variance is in squared units, while standard deviation is in the original units, making them incomparable.	Always complete the formula with a square root: after dividing by $n-1$ , compute $s = \sqrt{\text{that result}}$ .
Adding the deviations $(x_i - \bar{x})$ without squaring them first.	Positive and negative deviations cancel each other out, producing a sum near zero regardless of actual spread—this masks variability.	Square each deviation before summing: calculate $\sum(x_i - \bar{x})^2$ , not $\sum(x_i - \bar{x})$ .

Tips and when to use something else

If you have data for an entire population (not a sample), use Population Variance with $n$ in the denominator instead of $n-1$ .
Standard deviation is sensitive to outliers—one extreme value can inflate it significantly; use the Interquartile Range for highly skewed data.
The Empirical Rule states roughly 68% of data lies within one standard deviation of the mean for normally distributed data; use this as a reasonableness check.
For skewed or non-normal distributions, pair the Interquartile Range with the median instead of standard deviation with the mean.

Related formulas

Arithmetic MeanArithmetic Mean is the average value found by adding all numbers in a dataset and dividing by how many numbers there are.Sample VarianceSample variance measures how spread out data points are around the mean, using n-1 instead of n to account for sample uncertainty.Population VariancePopulation variance quantifies how spread data points are around the mean across an entire population, used instead of sample variance when you have all data.Empirical RuleThe Empirical Rule describes where approximately 68%, 95%, and 99.7% of normally distributed data fall, measured in standard deviations from the mean.Normal DistributionThe normal distribution describes the probability density of a continuous variable, used to model measurements and natural phenomena.

Frequently asked questions

Why do we divide by n-1 instead of n for sample standard deviation?: Using $n-1$ corrects for the fact that the sample mean is already calculated from your data, which uses up one degree of freedom. This makes the sample standard deviation an unbiased estimator of the true population standard deviation—without it, sample values would systematically underestimate population spread.
What is the difference between standard deviation and variance?: Variance is the average of squared deviations, while standard deviation is the square root of variance. Standard deviation is usually preferred because it is in the same units as the original data, making it directly comparable to values in your dataset.
Can standard deviation be negative?: No, standard deviation is always non-negative. Since it is a square root of a sum of squared values, the result is zero only when all data points are identical, and positive whenever there is any spread.
When should I use standard deviation versus range?: Use standard deviation when you want to account for all data points and their distances from the mean; it is stable and informative. Use range (max - min) for a quick rough estimate when you only care about the span between extremes.

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Reviewed 2026-09-18