# Standard deviation

A plot of a normal distribution (or bell-shaped curve) where each band has a width of 1 standard deviation – See also: 68-95-99.7 rule
Cumulative probability of a normal distribution with expected value 0 and standard deviation 1.

In statistics and probability theory, standard deviation (represented by the symbol sigma, σ) shows how much variation or dispersion exists from the average (mean), or expected value. A low standard deviation indicates that the data points tend to be very close to the mean; high standard deviation indicates that the data points are spread out over a large range of values.

The standard deviation of a random variable, statistical population, data set, or probability distribution is the square root of its variance. It is algebraically simpler though practically less robust than the average absolute deviation.[1][2] A useful property of standard deviation is that, unlike variance, it is expressed in the same units as the data. Note, however, that for measurements with percentage as unit, the standard deviation will have percentage points as unit.

In addition to expressing the variability of a population, standard deviation is commonly used to measure confidence in statistical conclusions. For example, the margin of error in polling data is determined by calculating the expected standard deviation in the results if the same poll were to be conducted multiple times. The reported margin of error is typically about twice the standard deviation ­– the radius of a 95 percent confidence interval. In science, researchers commonly report the standard deviation of experimental data, and only effects that fall far outside the range of standard deviation are considered statistically significant – normal random error or variation in the measurements is in this way distinguished from causal variation. Standard deviation is also important in finance, where the standard deviation on the rate of return on an investment is a measure of the volatility of the investment.

When only a sample of data from a population is available, the standard deviation of the population can be estimated by a modified quantity called the "sample standard deviation". When the sample is the entire population, its unmodified standard deviation is called the "population standard deviation".

## Basic examples

For a finite set of numbers, the standard deviation is found by taking the square root of the average of the squared differences of the values from their average value. For example, consider a population consisting of the following eight values:

$2,\ 4,\ 4,\ 4,\ 5,\ 5,\ 7,\ 9.$

These eight data points have the mean (average) of 5:

$\frac{2 + 4 + 4 + 4 + 5 + 5 + 7 + 9}{8} = 5.$

To calculate the population standard deviation, first compute the difference of each data point from the mean, and square the result of each:

$\begin{array}{lll} (2-5)^2 = (-3)^2 = 9 && (5-5)^2 = 0^2 = 0 \\ (4-5)^2 = (-1)^2 = 1 && (5-5)^2 = 0^2 = 0 \\ (4-5)^2 = (-1)^2 = 1 && (7-5)^2 = 2^2 = 4 \\ (4-5)^2 = (-1)^2 = 1 && (9-5)^2 = 4^2 = 16. \\ \end{array}$

Next, compute the average of these values, and take the square root:

$\sqrt{ \frac{(9 + 1 + 1 + 1 + 0 + 0 + 4 + 16)}{8} } = 2.$

This quantity is the population standard deviation, and is equal to the square root of the variance. The formula is valid only if the eight values we began with form the complete population. If the values instead were a random sample drawn from some larger parent population, then we would have divided by 7 (which is n−1) instead of 8 (which is n) in the denominator of the last formula, and then the quantity thus obtained would be called the sample standard deviation. Dividing by n−1 gives a better estimate of the population standard deviation than dividing by n.

As a slightly more complicated real-life example, the average height for adult men in the United States is about 70 in, with a standard deviation of around 3 in. This means that most men (about 68 percent, assuming a normal distribution) have a height within 3 in of the mean (67–73 in)  – one standard deviation – and almost all men (about 95%) have a height within 6 in of the mean (64–76 in) – two standard deviations. If the standard deviation were zero, then all men would be exactly 70 in tall. If the standard deviation were 20 in, then men would have much more variable heights, with a typical range of about 50–90 in. Three standard deviations account for 99.7 percent of the sample population being studied, assuming the distribution is normal (bell-shaped).

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## Definition of population values

Let X be a random variable with mean value μ:

$\operatorname{E}[X] = \mu.\,\!$

Here the operator E denotes the average or expected value of X. Then the standard deviation of X is the quantity

\begin{align} \sigma & = \sqrt{\operatorname E[(X - \mu)^2]}\\ & =\sqrt{\operatorname E[X^2] + \operatorname E[(-2 \mu X)] + \operatorname E[\mu^2]} =\sqrt{\operatorname E[X^2] -2 \mu \operatorname E[X] + \mu^2}\\ &=\sqrt{\operatorname E[X^2] -2 \mu^2 + \mu^2} =\sqrt{\operatorname E[X^2] - \mu^2}\\ & =\sqrt{\operatorname E[X^2]-(\operatorname E[X])^2}. \end{align}

(derived using the properties of expected value)

In other words the standard deviation σ (sigma) is the square root of the variance of X; i.e., it is the square root of the average value of (X − μ)2.

The standard deviation of a (univariate) probability distribution is the same as that of a random variable having that distribution. Not all random variables have a standard deviation, since these expected values need not exist. For example, the standard deviation of a random variable that follows a Cauchy distribution is undefined because its expected value μ is undefined.

### Discrete random variable

In the case where X takes random values from a finite data set x1, x2, ..., xN, with each value having the same probability, the standard deviation is

$\sigma = \sqrt{\frac{1}{N}\left[(x_1-\mu)^2 + (x_2-\mu)^2 + \cdots + (x_N - \mu)^2\right]}, {\rm \ \ where\ \ } \mu = \frac{1}{N} (x_1 + \cdots + x_N),$

or, using summation notation,

$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2}, {\rm \ \ where\ \ } \mu = \frac{1}{N} \sum_{i=1}^N x_i.$

If, instead of having equal probabilities, the values have different probabilities, let x1 have probability p1, x2 have probability p2, ..., xN have probability pN. In this case, the standard deviation will be

$\sigma = \sqrt{\sum_{i=1}^N p_i(x_i - \mu)^2} , {\rm \ \ where\ \ } \mu = \sum_{i=1}^N p_i x_i.$

### Continuous random variable

The standard deviation of a continuous real-valued random variable X with probability density function p(x) is

$\sigma = \sqrt{\int_\mathbf{X} (x-\mu)^2 \, p(x) \, dx}, {\rm \ \ where\ \ } \mu = \int_\mathbf{X} x \, p(x) \, dx,$

and where the integrals are definite integrals taken for x ranging over the set of possible values of the random variable X.

In the case of a parametric family of distributions, the standard deviation can be expressed in terms of the parameters. For example, in the case of the log-normal distribution with parameters μ and σ2, the standard deviation is [(exp(σ2) − 1)exp(2μ + σ2)]1/2.

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## Estimation

One can find the standard deviation of an entire population in cases (such as standardized testing) where every member of a population is sampled. In cases where that cannot be done, the standard deviation σ is estimated by examining a random sample taken from the population and computing a statistic of the sample, which is used as an estimate of the population standard deviation. Such a statistic is called an estimator, and the estimator (or the value of the estimator, namely the estimate) is called a sample standard deviation, and is denoted by s (possibly with modifiers). However, unlike in the case of estimating the population mean, for which the sample mean is a simple estimator with many desirable properties (unbiased, efficient, maximum likelihood), there is no single estimator for the standard deviation with all these properties, and unbiased estimation of standard deviation is a very technical involved problem. Most often, the standard deviation is estimated using the corrected sample standard deviation (using N − 1), defined below, and this is often referred to as the "sample standard deviation", without qualifiers. However, other estimators are better in other respects: the uncorrected estimator (using N) yields lower mean squared error, while using N − 1.5 (for the normal distribution) almost completely eliminates bias.

### Uncorrected sample standard deviation

Firstly, the formula for the population standard deviation (of a finite population) can be applied to the sample, using the size of the sample as the size of the population (though the actual population size from which the sample is drawn may be much larger). This estimator, denoted by sN, is known as the uncorrected sample standard deviation, or sometimes the standard deviation of the sample (considered as the entire population), and is defined as follows:[citation needed]

$s_N = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2},$

where $\scriptstyle\{x_1,\,x_2,\,\ldots,\,x_N\}$ are the observed values of the sample items and $\scriptstyle\overline{x}$ is the mean value of these observations, while the denominator N stands for the size of the sample.

This is a consistent estimator (it converges in probability to the population value as the number of samples goes to infinity), and is the maximum-likelihood estimate when the population is normally distributed.[citation needed] However, this is a biased estimator, as the estimates are generally too low. The bias decreases as sample size grows, dropping off as 1/n, and thus is most significant for small or moderate sample sizes; for $n > 75$ the bias is below 1%. Thus for very large sample sizes, the uncorrected sample standard deviation is generally acceptable. This estimator also has a uniformly smaller mean squared error than the corrected sample standard deviation.

### Corrected sample standard deviation

When discussing the bias, to be more precise, the corresponding estimator for the variance, the biased sample variance:

$s^2_N = \frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2,$

equivalently the second central moment of the sample (as the mean is the first moment), is a biased estimator of the variance (it underestimates the population variance). Taking the square root to pass to the standard deviation introduces further downward bias, by Jensen's inequality, due to the square root being a concave function. The bias in the variance is easily corrected, but the bias from the square root is more difficult to correct, and depends on the distribution in question.

An unbiased estimator for the variance is given by apply Bessel's correction, using N − 1 instead of N to yield the unbiased sample variance, denoted s2:

$s^2 = \frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2.$

This estimator is unbiased if the variance exists and the sample values are drawn independently with replacement. N − 1 corresponds to the number of degrees of freedom in the vector of residuals, $\scriptstyle(x_1-\overline{x},\; \dots,\; x_n-\overline{x}).$

Taking square roots reintroduces bias, and yields the corrected sample standard deviation, denoted by s:

$s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}.$

While s2 is an unbiased estimator for the population variance, s is a biased estimator for the population standard deviation, though markedly less biased than the uncorrected sample standard deviation. The bias is still significant for small samples (n less than 10), and also drops off as 1/n as sample size increases. This estimator is commonly used, and generally known simply as the "sample standard deviation".

### Unbiased sample standard deviation

For unbiased estimation of standard deviation, there is no formula that works across all distributions, unlike for mean and variance. Instead, s is used as a basis, and is scaled by a correction factor to produce an unbiased estimate. For the normal distribution, an unbiased estimator is given by s/c4, where the correction factor (which depends on N) is given in terms of the Gamma function, and equals:

$c_4(N)\,=\,\sqrt{\frac{2}{N-1}}\,\,\,\frac{\Gamma\left(\frac{N}{2}\right)}{\Gamma\left(\frac{N-1}{2}\right)}.$

This arises because the sampling distribution of the sample standard deviation follows a (scaled) chi distribution, and the correction factor is the mean of the chi distribution.

An approximation is given by replacing N − 1 with N − 1.5, yielding:

$\hat\sigma = \sqrt{ \frac{1}{N - 1.5} \sum_{i=1}^n (x_i - \bar{x})^2 },$

The error in this approximation decays quadratically (as 1/N2), and it is suited for all but the smallest samples or highest precision: for n = 3 the bias is equal to 1.3%, and for n = 9 the bias is already less than 0.1%.

For other distributions, the correct formula depends on the distribution, but a rule of thumb is to use the further refinement of the approximation:

$\hat\sigma = \sqrt{ \frac{1}{n - 1.5 - \tfrac14 \gamma_2} \sum_{i=1}^n (x_i - \bar{x})^2 },$

where γ2 denotes the population excess kurtosis. The excess kurtosis may be either known beforehand for certain distributions, or estimated from the data.

### Confidence interval of a sampled standard deviation

The standard deviation we obtain by sampling a distribution is itself not absolutely accurate. This is especially true if the number of samples is very low. This effect can be described by the confidence interval or CI. For example, for N=2 the 95% CI of the SD is from 0.45*SD to 31.9*SD. In other words the standard deviation of the distribution in 95% of the cases can be up to a factor of 31 larger or up to a factor 2 smaller. For N=10 the interval is 0.69*SD to 1.83*SD, the actual SD can still be almost a factor 2 higher than the sampled SD. For N=100 this is down to 0.88*SD to 1.16*SD. So to be sure the sampled SD is close to the actual SD we need to sample a large number of points.

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## Identities and mathematical properties

The standard deviation is invariant under changes in location, and scales directly with the scale of the random variable. Thus, for a constant c and random variables X and Y:

$\operatorname{stdev}(c) = 0 \,$
$\operatorname{stdev}(X + c) = \operatorname{stdev}(X), \,$
$\operatorname{stdev}(cX) = |c| \operatorname{stdev}(X). \,$

The standard deviation of the sum of two random variables can be related to their individual standard deviations and the covariance between them:

$\operatorname{stdev}(X + Y) = \sqrt{\operatorname{var}(X) + \operatorname{var}(Y) + 2 \,\operatorname{cov}(X,Y)}. \,$

where $\scriptstyle\operatorname{var} \,=\, \operatorname{stdev}^2$ and $\scriptstyle\operatorname{cov}$ stand for variance and covariance, respectively.

The calculation of the sum of squared deviations can be related to moments calculated directly from the data. The standard deviation of the sample can be computed as:

$\operatorname{stdev}(X) = \sqrt{E[(X-E(X))^2]} = \sqrt{E[X^2] - (E[X])^2}.$

The sample standard deviation can be computed as:

$\operatorname{stdev}(X) = \sqrt{\frac{N}{N-1}} \sqrt{E[(X-E(X))^2]}.$

For a finite population with equal probabilities at all points, we have

$\sqrt{\frac{1}{N}\sum_{i=1}^N(x_i-\overline{x})^2} = \sqrt{\frac{1}{N} \left(\sum_{i=1}^N x_i^2\right) - \overline{x}^2} = \sqrt{\left(\frac{1}{N} \sum_{i=1}^N x_i^2\right) - \left(\frac{1}{N} \sum_{i=1}^{N} x_i\right)^2}.$

This means that the standard deviation is equal to the square root of (the average of the squares less the square of the average). See computational formula for the variance for proof, and for an analogous result for the sample standard deviation.

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## Interpretation and application

Example of two sample populations with the same mean and different standard deviations. Red population has mean 100 and SD 10; blue population has mean 100 and SD 50.[dubious ]

A large standard deviation indicates that the data points are far from the mean and a small standard deviation indicates that they are clustered closely around the mean.

For example, each of the three populations {0, 0, 14, 14}, {0, 6, 8, 14} and {6, 6, 8, 8} has a mean of 7. Their standard deviations are 7, 5, and 1, respectively. The third population has a much smaller standard deviation than the other two because its values are all close to 7. It will have the same units as the data points themselves. If, for instance, the data set {0, 6, 8, 14} represents the ages of a population of four siblings in years, the standard deviation is 5 years. As another example, the population {1000, 1006, 1008, 1014} may represent the distances traveled by four athletes, measured in meters. It has a mean of 1007 meters, and a standard deviation of 5 meters.

Standard deviation may serve as a measure of uncertainty. In physical science, for example, the reported standard deviation of a group of repeated measurements gives the precision of those measurements. When deciding whether measurements agree with a theoretical prediction, the standard deviation of those measurements is of crucial importance: if the mean of the measurements is too far away from the prediction (with the distance measured in standard deviations), then the theory being tested probably needs to be revised. This makes sense since they fall outside the range of values that could reasonably be expected to occur, if the prediction were correct and the standard deviation appropriately quantified. See prediction interval.

While the standard deviation does measure how far typical values tend to be from the mean, other measures are available. An example is the mean absolute deviation, which might be considered a more direct measure of average distance, compared to the root mean square distance inherent in the standard deviation.

### Application examples

The practical value of understanding the standard deviation of a set of values is in appreciating how much variation there is from the average (mean).

#### Climate

As a simple example, consider the average daily maximum temperatures for two cities, one inland and one on the coast. It is helpful to understand that the range of daily maximum temperatures for cities near the coast is smaller than for cities inland. Thus, while these two cities may each have the same average maximum temperature, the standard deviation of the daily maximum temperature for the coastal city will be less than that of the inland city as, on any particular day, the actual maximum temperature is more likely to be farther from the average maximum temperature for the inland city than for the coastal one.

#### Particle physics

Particle physics uses a standard of "5 sigma" for the declaration of a discovery.[3] At five-sigma there is only one chance in nearly two million that a random fluctuation would yield the result. This level of certainty prompted the announcement that a particle consistent with the Higgs boson has been discovered in two independent experiments at CERN.[4]

#### Sports

Another way of seeing it is to consider sports teams. In any set of categories, there will be teams that rate highly at some things and poorly at others. Chances are, the teams that lead in the standings will not show such disparity but will perform well in most categories. The lower the standard deviation of their ratings in each category, the more balanced and consistent they will tend to be. Teams with a higher standard deviation, however, will be more unpredictable. For example, a team that is consistently bad in most categories will have a low standard deviation. A team that is consistently good in most categories will also have a low standard deviation. However, a team with a high standard deviation might be the type of team that scores a lot (strong offense) but also concedes a lot (weak defense), or, vice versa, that might have a poor offense but compensates by being difficult to score on.

Trying to predict which teams, on any given day, will win, may include looking at the standard deviations of the various team "stats" ratings, in which anomalies can match strengths vs. weaknesses to attempt to understand what factors may prevail as stronger indicators of eventual scoring outcomes.

In racing, a driver is timed on successive laps. A driver with a low standard deviation of lap times is more consistent than a driver with a higher standard deviation. This information can be used to help understand where opportunities might be found to reduce lap times.

#### Finance

In finance, standard deviation is often used as a measure of the risk associated with price-fluctuations of a given asset (stocks, bonds, property, etc.), or the risk of a portfolio of assets [5] (actively managed mutual funds, index mutual funds, or ETFs). Risk is an important factor in determining how to efficiently manage a portfolio of investments because it determines the variation in returns on the asset and/or portfolio and gives investors a mathematical basis for investment decisions (known as mean-variance optimization). The fundamental concept of risk is that as it increases, the expected return on an investment should increase as well, an increase known as the risk premium. In other words, investors should expect a higher return on an investment when that investment carries a higher level of risk or uncertainty. When evaluating investments, investors should estimate both the expected return and the uncertainty of future returns. Standard deviation provides a quantified estimate of the uncertainty of future returns.

For example, let's assume an investor had to choose between two stocks. Stock A over the past 20 years had an average return of 10 percent, with a standard deviation of 20 percentage points (pp) and Stock B, over the same period, had average returns of 12 percent but a higher standard deviation of 30 pp. On the basis of risk and return, an investor may decide that Stock A is the safer choice, because Stock B's additional two percentage points of return is not worth the additional 10 pp standard deviation (greater risk or uncertainty of the expected return). Stock B is likely to fall short of the initial investment (but also to exceed the initial investment) more often than Stock A under the same circumstances, and is estimated to return only two percent more on average. In this example, Stock A is expected to earn about 10 percent, plus or minus 20 pp (a range of 30 percent to −10 percent), about two-thirds of the future year returns. When considering more extreme possible returns or outcomes in future, an investor should expect results of as much as 10 percent plus or minus 60 pp, or a range from 70 percent to −50 percent, which includes outcomes for three standard deviations from the average return (about 99.7 percent of probable returns).

Calculating the average (or arithmetic mean) of the return of a security over a given period will generate the expected return of the asset. For each period, subtracting the expected return from the actual return results in the difference from the mean. Squaring the difference in each period and taking the average gives the overall variance of the return of the asset. The larger the variance, the greater risk the security carries. Finding the square root of this variance will give the standard deviation of the investment tool in question.

Population standard deviation is used to set the width of Bollinger Bands, a widely adopted technical analysis tool. For example, the upper Bollinger Band is given as x + x. The most commonly used value for n is 2; there is about a five percent chance of going outside, assuming a normal distribution of returns.

Unfortunately, financial time series are known to be non-stationary series, whereas the statistical calculations above, such as standard deviation, apply only to stationary series. Whatever apparent "predictive powers" or "forecasting ability" that may appear when applied as above is illusory. To apply the above statistical tools to non-stationary series, the series first must be transformed to a stationary series, enabling use of statistical tools that now have a valid basis from which to work.

### Geometric interpretation

To gain some geometric insights and clarification, we will start with a population of three values, x1, x2, x3. This defines a point P = (x1, x2, x3) in R3. Consider the line L = {(r, r, r) : rR}. This is the "main diagonal" going through the origin. If our three given values were all equal, then the standard deviation would be zero and P would lie on L. So it is not unreasonable to assume that the standard deviation is related to the distance of P to L. And that is indeed the case. To move orthogonally from L to the point P, one begins at the point:

$M = (\overline{x},\overline{x},\overline{x})$

whose coordinates are the mean of the values we started out with. A little algebra shows that the distance between P and M (which is the same as the orthogonal distance between P and the line L) is equal to the standard deviation of the vector x1, x2, x3, multiplied by the square root of the number of dimensions of the vector (3 in this case.)

### Chebyshev's inequality

An observation is rarely more than a few standard deviations away from the mean. Chebyshev's inequality ensures that, for all distributions for which the standard deviation is defined, the amount of data within a number of standard deviations of the mean is at least as much as given in the following table.

Minimum population Distance from mean
50% √2
75% 2
89% 3
94% 4
96% 5
97% 6
$\scriptstyle 1-\frac{1}{k^2}$[6] $\scriptstyle k$
$\scriptstyle l$ $\scriptstyle \frac{1}{\sqrt{1-l}}$

### Rules for normally distributed data

Dark blue is one standard deviation on either side of the mean. For the normal distribution, this accounts for 68.27 percent of the set; while two standard deviations from the mean (medium and dark blue) account for 95.45 percent; three standard deviations (light, medium, and dark blue) account for 99.73 percent; and four standard deviations account for 99.994 percent. The two points of the curve that are one standard deviation from the mean are also the inflection points.

The central limit theorem says that the distribution of an average of many independent, identically distributed random variables tends toward the famous bell-shaped normal distribution with a probability density function of:

$f(x;\mu,\sigma^2) = \frac{1}{\sigma\sqrt{2\pi}} e^{ -\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2 }$

where μ is the expected value of the random variables, σ equals their distribution's standard deviation divided by n1/2, and n is the number of random variables. The standard deviation therefore is simply a scaling variable that adjusts how broad the curve will be, though it also appears in the normalizing constant.

If a data distribution is approximately normal, then the proportion of data values within z standard deviations of the mean is defined by:

Proportion = $\operatorname{erf}\left(\frac{z}{\sqrt{2}}\right)$

where $\scriptstyle\operatorname{erf}$ is the error function. If a data distribution is approximately normal then about 68 percent of the data values are within one standard deviation of the mean (mathematically, μ ± σ, where μ is the arithmetic mean), about 95 percent are within two standard deviations (μ ± 2σ), and about 99.7 percent lie within three standard deviations (μ ± 3σ). This is known as the 68-95-99.7 rule, or the empirical rule.

For various values of z, the percentage of values expected to lie in and outside the symmetric interval, CI = (−), are as follows:

zσ Percentage within CI Percentage outside CI Fraction outside CI
0.674490σ 50% 50% 1 / 2
0.994458σ 68% 32% 1 / 3.125
68.2689492% 31.7310508% 1 / 3.1514872
1.281552σ 80% 20% 1 / 5
1.644854σ 90% 10% 1 / 10
1.959964σ 95% 5% 1 / 20
95.4499736% 4.5500264% 1 / 21.977895
2.575829σ 99% 1% 1 / 100
99.7300204% 0.2699796% 1 / 370.398
3.290527σ 99.9% 0.1% 1 / 1,000
3.890592σ 99.99% 0.01% 1 / 10,000
99.993666% 0.006334% 1 / 15,787
4.417173σ 99.999% 0.001% 1 / 100,000
4.891638σ 99.9999% 0.0001% 1 / 1,000,000
99.9999426697% 0.0000573303% 1 / 1,744,278
5.326724σ 99.99999% 0.00001% 1 / 10,000,000
5.730729σ 99.999999% 0.000001% 1 / 100,000,000
99.9999998027% 0.0000001973% 1 / 506,797,346
6.109410σ 99.9999999% 0.0000001% 1 / 1,000,000,000
6.466951σ 99.99999999% 0.00000001% 1 / 10,000,000,000
6.806502σ 99.999999999% 0.000000001% 1 / 100,000,000,000
99.9999999997440% 0.000000000256% 1 / 390,682,215,445
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## Relationship between standard deviation and mean

The mean and the standard deviation of a set of data are descriptive statistics usually reported together. In a certain sense, the standard deviation is a "natural" measure of statistical dispersion if the center of the data is measured about the mean. This is because the standard deviation from the mean is smaller than from any other point. The precise statement is the following: suppose x1, ..., xn are real numbers and define the function:

$\sigma(r) = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - r)^2}.$

Using calculus or by completing the square, it is possible to show that σ(r) has a unique minimum at the mean:

$r = \overline{x}.\,$

Variability can also be measured by the coefficient of variation, which is the ratio of the standard deviation to the mean. It is a dimensionless number.

### Standard deviation of the mean

Often, we want some information about the precision of the mean we obtained. We can obtain this by determining the standard deviation of the sampled mean. The standard deviation of the mean is related to the standard deviation of the distribution by:

$\sigma_{\text{mean}} = \frac{1}{\sqrt{N}}\sigma$

where N is the number of observations in the sample used to estimate the mean. This can easily be proven with (see basic properties of the variance):

\begin{align} \operatorname{var}(X) &\equiv \sigma^2_X\\ \operatorname{var}(X_1+X_2) &\equiv \operatorname{var}(X_1) + \operatorname{var}(X_2)\\ \operatorname{var}(cX_1) &\equiv c^2 \, \operatorname{var}(X_1) \end{align}

hence

\begin{align} \operatorname{var}(\text{mean}) &= \operatorname{var}\left (\frac{1}{N} \sum_{i=1}^N X_i \right) = \frac{1}{N^2}\operatorname{var}\left (\sum_{i=1}^N X_i \right ) \\ &= \frac{1}{N^2}\sum_{i=1}^N \operatorname{var}(X_i) = \frac{N}{N^2} \operatorname{var}(X) = \frac{1}{N} \operatorname{var} (X). \end{align}

Resulting in:

$\sigma_\text{mean} = \frac{\sigma}{\sqrt{N}}.$
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## Rapid calculation methods

The following two formulas can represent a running (continuous) standard deviation. A set of two power sums s1 and s2 are computed over a set of N values of x, denoted as x1, ..., xN:

$\ s_j=\sum_{k=1}^N{x_k^j}.$

Given the results of these three running summations, the values N, s1, s2 can be used at any time to compute the current value of the running standard deviation:

$\sigma = \frac{\sqrt{Ns_2-s_1^2} }{N}$

Where :$\ N=s_0=\sum_{k=1}^N{x_k^0}.$

Similarly for sample standard deviation,

$s = \sqrt{\frac{Ns_2-s_1^2}{N(N-1)}}.$

In a computer implementation, as the three sj sums become large, we need to consider round-off error, arithmetic overflow, and arithmetic underflow. The method below calculates the running sums method with reduced rounding errors.[7] This is a "one pass" algorithm for calculating variance of n samples without the need to store prior data during the calculation. Applying this method to a time series will result in successive values of standard deviation corresponding to n data points as n grows larger with each new sample, rather than a constant-width sliding window calculation.

For k = 1, ..., n:

\begin{align} A_0 &= 0\\ A_k &= A_{k-1}+\frac{x_k-A_{k-1}}{k} \end{align}

where A is the mean value.

\begin{align} Q_0 &= 0\\ Q_k &= Q_{k-1}+\frac{k-1}{k} (x_k-A_{k-1})^2 = Q_{k-1}+ (x_k-A_{k-1})(x_k-A_k) \end{align}

Sample variance:

$s^2_n=\frac{Q_n}{n-1}$

Population variance:

$\sigma^2_n=\frac{Q_n}{n}$

### Weighted calculation

When the values xi are weighted with unequal weights wi, the power sums s0, s1, s2 are each computed as:

$\ s_j=\sum_{k=1}^N{w_k x_k^j}.\,$

And the standard deviation equations remain unchanged. Note that s0 is now the sum of the weights and not the number of samples N.

The incremental method with reduced rounding errors can also be applied, with some additional complexity.

A running sum of weights must be computed for each k from 1 to n:

\begin{align} W_0 &= 0\\ W_k &= W_{k-1} + w_k \end{align}

and places where 1/n is used above must be replaced by wi/Wn:

\begin{align} A_0 &= 0\\ A_k &= A_{k-1}+\frac{w_k}{W_k}(x_k-A_{k-1})\\ Q_0 &= 0\\ Q_k &= Q _{k-1} + \frac{w_k W_{k-1}}{W_k}(x_k-A_{k-1})^2 = Q_{k-1}+w_k(x_k-A_{k-1})(x_k-A_k) \end{align}

In the final division,

$\sigma^2_n=\frac{Q_n}{W_n}\,$

and

$s^2_n = \frac{n'}{n'-1}\sigma^2_n\,$

where n is the total number of elements, and n' is the number of elements with non-zero weights. The above formulas become equal to the simpler formulas given above if weights are taken as equal to one.

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## Combining standard deviations

### Population-based statistics

The populations of sets, which may overlap, can be calculated simply as follows:

\begin{align} &&N_{X \cup Y} &= N_X + N_Y - N_{X \cap Y}\\ X \cap Y = \varnothing &\Rightarrow &N_{X \cap Y} &= 0\\ &\Rightarrow &N_{X \cup Y} &= N_X + N_Y \end{align}

Standard deviations of non-overlapping (XY = ∅) sub-populations can be aggregated as follows if the size (actual or relative to one another) and means of each are known:

\begin{align} \mu_{X \cup Y} &= \frac{ N_X \mu_X + N_Y \mu_Y }{N_X + N_Y} \\ \sigma_{X\cup Y} &= \sqrt{ \frac{N_X \sigma_X^2 + N_Y \sigma_Y^2}{N_X + N_Y} + \frac{N_X N_Y}{(N_X+N_Y)^2}(\mu_X - \mu_Y)^2 } \end{align}

For example, suppose it is known that the average American man has a mean height of 70 inches with a standard deviation of three inches and that the average American woman has a mean height of 65 inches with a standard deviation of two inches. Also assume that the number of men, N, is equal to the number of women. Then the mean and standard deviation of heights of American adults could be calculated as:

\begin{align} \mu &= \frac{N\cdot70 + N\cdot65}{N + N} = \frac{70+65}{2} = 67.5 \\ \sigma &= \sqrt{ \frac{3^2 + 2^2}{2} + \frac{(70-65)^2}{2^2} } = \sqrt{12.75} \approx 3.57 \end{align}

For the more general case of M non-overlapping populations, X1 through XM, and the aggregate population $\scriptstyle X \,=\, \bigcup_i X_i$:

\begin{align} \mu_X &= \frac{ \sum_i N_{X_i}\mu_{X_i} }{ \sum_i N_{X_i} } \\ \sigma_X &= \sqrt{ \frac{ \sum_i N_{X_i}(\sigma_{X_i}^2 + \mu_{X_i}^2) }{ \sum_i N_{X_i} } - \mu_X^2 } = \sqrt{ \frac{ \sum_i N_{X_i}\sigma_{X_i}^2 }{ \sum_i N_{X_i} } + \frac{ \sum_{i

where

$X_i \cap X_j = \varnothing, \quad \forall\ i

If the size (actual or relative to one another), mean, and standard deviation of two overlapping populations are known for the populations as well as their intersection, then the standard deviation of the overall population can still be calculated as follows:

\begin{align} \mu_{X \cup Y} &= \frac{1}{N_{X \cup Y}}\left(N_X\mu_X + N_Y\mu_Y - N_{X \cap Y}\mu_{X \cap Y}\right)\\ \sigma_{X \cup Y} &= \sqrt{\frac{1}{N_{X \cup Y}}\left(N_X[\sigma_X^2 + \mu _X^2] + N_Y[\sigma_Y^2 + \mu _Y^2] - N_{X \cap Y}[\sigma_{X \cap Y}^2 + \mu _{X \cap Y}^2]\right) - \mu_{X\cup Y}^2} \end{align}

If two or more sets of data are being added together datapoint by datapoint, the standard deviation of the result can be calculated if the standard deviation of each data set and the covariance between each pair of data sets is known:

$\sigma_X = \sqrt{\sum_i{\sigma_{X_i}^2} + \sum_{i,j}\operatorname{cov}(X_i,X_j)}$

For the special case where no correlation exists between any pair of data sets, then the relation reduces to the root-mean-square:

\begin{align} &\operatorname{cov}(X_i, X_j) = 0,\quad \forall i

### Sample-based statistics

Standard deviations of non-overlapping (XY = ∅) sub-samples can be aggregated as follows if the actual size and means of each are known:

\begin{align} \mu_{X \cup Y} &= \frac{1}{N_{X \cup Y}}\left(N_X\mu_X + N_Y\mu_Y\right)\\ \sigma_{X \cup Y} &= \sqrt{\frac{1}{N_{X \cup Y} - 1}\left([N_X - 1]\sigma_X^2 + N_X\mu_X^2 + [N_Y - 1]\sigma_Y^2 + N_Y\mu _Y^2 - [N_X + N_Y]\mu_{X \cup Y}^2\right) } \end{align}

For the more general case of M non-overlapping data sets, X1 through XM, and the aggregate data set $\scriptstyle X \,=\, \bigcup_i X_i$:

\begin{align} \mu_X &= \frac{1}{\sum_i { N_{X_i}}} \left(\sum_i { N_{X_i} \mu_{X_i}}\right)\\ \sigma_X &= \sqrt{\frac{1}{\sum_i {N_{X_i} - 1}} \left( \sum_i { \left[(N_{X_i} - 1) \sigma_{X_i}^2 + N_{X_i} \mu_{X_i}^2\right] } - \left[\sum_i {N_{X_i}}\right]\mu_X^2 \right) } \end{align}

where:

$X_i \cap X_j = \varnothing,\quad \forall i

If the size, mean, and standard deviation of two overlapping samples are known for the samples as well as their intersection, then the standard deviation of the aggregated sample can still be calculated. In general:

\begin{align} \mu_{X \cup Y} &= \frac{1}{N_{X \cup Y}}\left(N_X\mu_X + N_Y\mu_Y - N_{X\cap Y}\mu_{X\cap Y}\right)\\ \sigma_{X \cup Y} &= \sqrt{ \frac{1}{N_{X \cup Y} - 1}\left([N_X - 1]\sigma_X^2 + N_X\mu_X^2 + [N_Y - 1]\sigma_Y^2 + N_Y\mu _Y^2 - [N_{X \cap Y}-1]\sigma_{X \cap Y}^2 - N_{X \cap Y}\mu_{X \cap Y}^2 - [N_X + N_Y - N_{X \cap Y}]\mu_{X \cup Y}^2\right) } \end{align}
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## History

The term standard deviation was first used[8] in writing by Karl Pearson[9] in 1894, following his use of it in lectures. This was as a replacement for earlier alternative names for the same idea: for example, Gauss used mean error.[10] It may be worth noting in passing that the mean error is mathematically distinct from the standard deviation.

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## References

1. ^ Gauss, Carl Friedrich (1816). "Bestimmung der Genauigkeit der Beobachtungen". Zeitschrift für Astronomie und verwandt Wissenschaften 1: 187–197.
2. ^ Walker, Helen (1931). Studies in the History of the Statistical Method. Baltimore, MD: Williams & Wilkins Co. pp. 24–25.
3. ^ http://public.web.cern.ch/public/
4. ^ http://press.web.cern.ch/press/PressReleases/Releases2012/PR17.12E.html
5. ^ "What is Standard Deviation". Pristine. Retrieved 2011-10-29.
6. ^ Ghahramani, Saeed (2000). Fundamentals of Probability (2nd Edition). Prentice Hall: New Jersey. p. 438.
7. ^ Welford, BP (August 1962). "Note on a Method for Calculating Corrected Sums of Squares and Products". Technometrics 4 (3): 419–420.
8. ^ Dodge, Yadolah (2003). The Oxford Dictionary of Statistical Terms. Oxford University Press. ISBN 0-19-920613-9.
9. ^ Pearson, Karl (1894). "On the dissection of asymmetrical frequency curves". Phil. Trans. Roy. Soc. London, Series A 185: 719–810.
10. ^
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