Standard deviation

Standard deviation is a widely used measure of variability or diversity used in statistics and probability theory. It 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, whereas high standard deviation indicates that the data points are spread out over a large range of values.

The standard deviation of a 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.

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 population standard deviation can be estimated by a modified quantity called the sample standard deviation, explained below.

Basic examples

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:

$\frac{2 %2B 4 %2B 4 %2B 4 %2B 5 %2B 5 %2B 7 %2B 9}{8} = 5$

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:

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

This quantity is the population standard deviation; it 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 they instead were a random sample, drawn from some larger, "parent" population, then we should have used 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 have been called the sample standard deviation. See the section Estimation below for more details.

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

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

$\sigma = \sqrt{\operatorname{E}\left[(X - \mu)^2\right]}.$

That is, 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, …, x_N, with each value having the same probability, the standard deviation is

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

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, ..., x_N have probability p_N. 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.

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. Some estimators are given below:

With standard deviation of the sample

An estimator for σ sometimes used is the standard deviation of the sample, denoted by s_N and defined as follows:

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

This estimator has a uniformly smaller mean squared error than the sample standard deviation (see below), and is the maximum-likelihood estimate when the population is normally distributed. But this estimator, when applied to a small or moderately sized sample, tends to be too low: it is a biased estimator.

The standard deviation of the sample is the same as the population standard deviation of a discrete random variable that can assume precisely the values from the data set, where the probability for each value is proportional to its multiplicity in the data set.

With sample standard deviation

The most common estimator for σ used is an adjusted version, the sample standard deviation, denoted by s and defined as follows:

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

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. This correction (the use of N − 1 instead of N) is known as Bessel's correction. The reason for this correction is that s² is an unbiased estimator for the variance σ² of the underlying population, if that variance exists and the sample values are drawn independently with replacement. However, s is not an unbiased estimator for the standard deviation σ; it tends to overestimate the population standard deviation.

The term standard deviation of the sample is used for the uncorrected estimator (using N) while the term sample standard deviation is used for the corrected estimator (using N − 1). The denominator N − 1 is the number of degrees of freedom in the vector of residuals, $\scriptstyle(x_1-\overline{x},\; \dots,\; x_n-\overline{x})$ .

Other estimators

Further information: Unbiased estimation of standard deviation and Bias of an estimator

Although an unbiased estimator for σ is known when the random variable is normally distributed, the formula is complicated and amounts to a minor correction. Moreover, unbiasedness (in this sense of the word) is not always desirable.

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.

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:

$\operatorname{stdev}(c) = 0 \,$

$\operatorname{stdev}(X %2B c) = \operatorname{stdev}(X), \,$

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

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

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

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

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

Thus, 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 a proof of this fact, and for an analogous result for the sample standard deviation.

Interpretation and application

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. In a loose sense, the standard deviation tells us how far from the mean the data points tend to be. 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 should give 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.

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.

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 a representation of the risk associated with price-fluctuations of a given asset (stocks, bonds, property, etc.), or the risk of a portfolio of assets ^[3] (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 + nσ_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.

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) : r ∈ R}. 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

Main article: 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
$\scriptstyle 1-\frac{1}{k^2}$ ^[4]	$\scriptstyle k$
$\scriptstyle l$	$\scriptstyle \frac{1}{\sqrt{1-l}}$

Rules for normally distributed data

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:

$\frac{1}{\sigma\sqrt{2\pi}} \exp\!\left(-\frac{[x-\mu]^2}{2\sigma^2} \right)$

where μ is the expected value of the random variables, σ equals their distribution's standard deviation divided by n^1/2, 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 = (−zσ, zσ), 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
1σ	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
2σ	95.4499736%	4.5500264%	1 / 21.977895
2.575829σ	99%	1%	1 / 100
3σ	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
4σ	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
5σ	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
6σ	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
7σ	99.9999999997440%	0.0000000002560%	1 / 390,682,215,445

Relationship between standard deviation and mean

The mean and the standard deviation of a set of data are 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, ..., x_n 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.

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 observation in the sample used to estimate the mean. This can easily be proven with:

$\begin{align} \operatorname{var}(X) &\equiv \sigma^2_X\\ \operatorname{var}(X_1%2BX_2) &\equiv \operatorname{var}(X_1) %2B \operatorname{var}(X_2)\\ \operatorname{var}(cX_1) &\equiv c^2 \, \operatorname{var}(X_1) \end{align}$

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:

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

Rapid calculation methods

The following two formulas can represent a running (continuous) standard deviation. A set of three power sums s₀, s₁, s₂ are each computed over a set of N values of x, denoted as x₁, ..., x_N:

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

Note that s₀ raises x to the zero power, and since x⁰ is always 1, s₀ evaluates to N.

Given the results of these three running summations, the values s₀, s₁, s₂ can be used at any time to compute the current value of the running standard deviation:

$\sigma = \frac{\sqrt{s_0s_2-s_1^2} }{s_0}$

Similarly for sample standard deviation,

$s = \sqrt{\frac{s_0s_2-s_1^2}{s_0(s_0-1)}}.$

In a computer implementation, as the three s_j 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:

$\begin{align} A_0 &= 0\\ A_i &= A_{i-1}%2B\frac{x_i-A_{i-1}}{i} \end{align}$

where A is the mean value.

$\begin{align} Q_0 &= 0\\ Q_i &= Q_{i-1}%2B\frac{i-1}{i} (x_i-A_{i-1})^2 = Q_{i-1}%2B (x_i-A_{i-1})(x_i-A_i) \end{align}$

Sample variance:

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

Standard variance:

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

Weighted calculation

When the values x_i are weighted with unequal weights w_i, 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:

$\begin{align} W_0 &= 0\\ W_i &= W_{i-1} %2B w_i \end{align}$

and places where 1/i is used above must be replaced by w_i/W_i:

$\begin{align} A_0 &= 0\\ A_i &= A_{i-1}%2B\frac{w_i}{W_i}(x_i-A_{i-1})\\ Q_0 &= 0\\ Q_i &= Q _{i-1} %2B \frac{w_i W_{i-1}}{W_i}(x_i-A_{i-1})^2 = Q_{i-1}%2Bw_i(x_i-A_{i-1})(x_i-A_i) \end{align}$

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.

Combining standard deviations

Population-based statistics

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

$\begin{align} &&N_{X \cup Y} &= N_X %2B N_Y - N_{X \cap Y}\\ X \cap Y = \varnothing &\Rightarrow &N_{X \cap Y} &= 0\\ &\Rightarrow &N_{X \cup Y} &= N_X %2B N_Y \end{align}$

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

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

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:

$\begin{align} \mu &= \frac{N\cdot70 %2B N\cdot65}{N %2B N} = \frac{70%2B65}{2} = 67.5 \\ \sigma &= \sqrt{ \frac{3^2 %2B 2^2}{2} %2B \frac{(70-65)^2}{2^2} } = \sqrt{12.75} \approx 3.57 \end{align}$

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

$\begin{align} \mu_X &= \frac{ \sum_i N_{X_i}\mu_{X_i} }{ \sum_i N_{X_i} } \\ \sigma_X &= \sqrt{ \frac{ \sum_i N_{X_i}(\sigma_{X_i}^2 %2B \mu_{X_i}^2) }{ \sum_i N_{X_i} } - \mu_X^2 } = \sqrt{ \frac{ \sum_i N_{X_i}\sigma_{X_i}^2 }{ \sum_i N_{X_i} } %2B \frac{ \sum_{i<j} N_{X_i}N_{X_j} (\mu_{X_i}-\mu_{X_j})^2 }{\big(\sum_i N_{X_i}\big)^2} } \end{align}$

where

$X_i \cap X_j = \varnothing, \quad \forall\ i<j.$

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:

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

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:

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

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

$\begin{align} &\operatorname{cov}(X_i, X_j) = 0,\quad \forall i<j\\ \Rightarrow &\;\sigma_X = \sqrt{\sum_i {\sigma_{X_i}^2}}. \end{align}$

Sample-based statistics

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

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

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

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

where:

$X_i \cap X_j = \varnothing,\quad \forall i<j.$

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:

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

History

The term standard deviation was first used^[5] in writing by Karl Pearson^[6] 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.^[7]

References

^ Gauss, Carl Friedrich (1816). "Bestimmung der Genauigkeit der Beobachtungen". Zeitschrift für Astronomie und verwandt Wissenschaften 1: 187–197.
^ Walker, Helen (1931). Studies in the History of the Statistical Method. Baltimore, MD: Williams & Wilkins Co. pp. 24–25.
^ "What is Standard Deviation". Pristine. http://www.edupristine.com/blog/what-is-standard-deviation. Retrieved 2011-10-29.
^ Ghahramani, Saeed (2000). Fundamentals of Probability (2nd Edition). Prentice Hall: New Jersey. p. 438.
^ Dodge, Yadolah (2003). The Oxford Dictionary of Statistical Terms. Oxford University Press. ISBN 0-19-920613-9.
^ Pearson, Karl (1894). "On the dissection of asymmetrical frequency curves". Phil. Trans. Roy. Soc. London, Series A 185: 719–810.
^ Miller, Jeff. "Earliest Known Uses of Some of the Words of Mathematics". http://jeff560.tripod.com/mathword.html.

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