Как обозначается ошибка в математике

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«Invalid proof» redirects here. For any type of invalid proof besides mathematics, see Fallacy.

«0 = 1» redirects here. For the algebraic structure where this equality holds, see Null ring.

In mathematics, certain kinds of mistaken proof are often exhibited, and sometimes collected, as illustrations of a concept called mathematical fallacy. There is a distinction between a simple mistake and a mathematical fallacy in a proof, in that a mistake in a proof leads to an invalid proof while in the best-known examples of mathematical fallacies there is some element of concealment or deception in the presentation of the proof.

For example, the reason why validity fails may be attributed to a division by zero that is hidden by algebraic notation. There is a certain quality of the mathematical fallacy: as typically presented, it leads not only to an absurd result, but does so in a crafty or clever way.^[1] Therefore, these fallacies, for pedagogic reasons, usually take the form of spurious proofs of obvious contradictions. Although the proofs are flawed, the errors, usually by design, are comparatively subtle, or designed to show that certain steps are conditional, and are not applicable in the cases that are the exceptions to the rules.

The traditional way of presenting a mathematical fallacy is to give an invalid step of deduction mixed in with valid steps, so that the meaning of fallacy is here slightly different from the logical fallacy. The latter usually applies to a form of argument that does not comply with the valid inference rules of logic, whereas the problematic mathematical step is typically a correct rule applied with a tacit wrong assumption. Beyond pedagogy, the resolution of a fallacy can lead to deeper insights into a subject (e.g., the introduction of Pasch’s axiom of Euclidean geometry,^[2] the five colour theorem of graph theory). Pseudaria, an ancient lost book of false proofs, is attributed to Euclid.^[3]

Mathematical fallacies exist in many branches of mathematics. In elementary algebra, typical examples may involve a step where division by zero is performed, where a root is incorrectly extracted or, more generally, where different values of a multiple valued function are equated. Well-known fallacies also exist in elementary Euclidean geometry and calculus.^[4]^[5]

Howlers[edit]

${displaystyle {begin{array}{l};;;{dfrac {d}{dx}}{dfrac {1}{x}}\={dfrac {d}{d}}{dfrac {1}{x^{2}}}\={dfrac {d!!!backslash }{d!!!backslash }}{dfrac {1}{x^{2}}}\=-{dfrac {1}{x^{2}}}end{array}}}$

Anomalous cancellation in calculus

Examples exist of mathematically correct results derived by incorrect lines of reasoning. Such an argument, however true the conclusion appears to be, is mathematically invalid and is commonly known as a howler. The following is an example of a howler involving anomalous cancellation:

${displaystyle {frac {16}{64}}={frac {16!!!/}{6!!!/4}}={frac {1}{4}}.}$

Here, although the conclusion 16/64 = 1/4 is correct, there is a fallacious, invalid cancellation in the middle step.^{[note 1]} Another classical example of a howler is proving the Cayley–Hamilton theorem by simply substituting the scalar variables of the characteristic polynomial by the matrix.

Bogus proofs, calculations, or derivations constructed to produce a correct result in spite of incorrect logic or operations were termed «howlers» by Maxwell.^[2] Outside the field of mathematics the term howler has various meanings, generally less specific.

Division by zero[edit]

The division-by-zero fallacy has many variants. The following example uses a disguised division by zero to «prove» that 2 = 1, but can be modified to prove that any number equals any other number.

Let a and b be equal, nonzero quantities
Multiply by a

$a^{2}=ab$
Subtract b²

$a^{2}-b^{2}=ab-b^{2}$
Factor both sides: the left factors as a difference of squares, the right is factored by extracting b from both terms
Divide out (a − b)
Use the fact that a = b
Combine like terms on the left
Divide by the non-zero b

Q.E.D.^[6]

The fallacy is in line 5: the progression from line 4 to line 5 involves division by a − b, which is zero since a = b. Since division by zero is undefined, the argument is invalid.

Analysis[edit]

Mathematical analysis as the mathematical study of change and limits can lead to mathematical fallacies — if the properties of integrals and differentials are ignored. For instance, a naive use of integration by parts can be used to give a false proof that 0 = 1.^[7] Letting u = 1/log x and dv = dx/x, we may write:

$int {frac {1}{x,log x}},dx=1+int {frac {1}{x,log x}},dx$

after which the antiderivatives may be cancelled yielding 0 = 1. The problem is that antiderivatives are only defined up to a constant and shifting them by 1 or indeed any number is allowed. The error really comes to light when we introduce arbitrary integration limits a and b.

${displaystyle int _{a}^{b}{frac {1}{x,log x}},dx=1|_{a}^{b}+int _{a}^{b}{frac {1}{x,log x}},dx=0+int _{a}^{b}{frac {1}{xlog x}},dx=int _{a}^{b}{frac {1}{xlog x}},dx}$

Since the difference between two values of a constant function vanishes, the same definite integral appears on both sides of the equation.

Multivalued functions[edit]

Many functions do not have a unique inverse. For instance, while squaring a number gives a unique value, there are two possible square roots of a positive number. The square root is multivalued. One value can be chosen by convention as the principal value; in the case of the square root the non-negative value is the principal value, but there is no guarantee that the square root given as the principal value of the square of a number will be equal to the original number (e.g. the principal square root of the square of −2 is 2). This remains true for nth roots.

Positive and negative roots[edit]

Care must be taken when taking the square root of both sides of an equality. Failing to do so results in a «proof» of^[8] 5 = 4.

Proof:

Start from

Write this as

Rewrite as

${displaystyle 5^{2}-5times 9=4^{2}-4times 9}$

Add 81/4 on both sides:

${displaystyle 5^{2}-5times 9+{frac {81}{4}}=4^{2}-4times 9+{frac {81}{4}}}$

These are perfect squares:

${displaystyle left(5-{frac {9}{2}}right)^{2}=left(4-{frac {9}{2}}right)^{2}}$

Take the square root of both sides:

${displaystyle 5-{frac {9}{2}}=4-{frac {9}{2}}}$

Add 9/2 on both sides:

Q.E.D.

The fallacy is in the second to last line, where the square root of both sides is taken: a² = b² only implies a = b if a and b have the same sign, which is not the case here. In this case, it implies that a = –b, so the equation should read

${displaystyle 5-{frac {9}{2}}=-left(4-{frac {9}{2}}right)}$

which, by adding 9/2 on both sides, correctly reduces to 5 = 5.

Another example illustrating the danger of taking the square root of both sides of an equation involves the following fundamental identity^[9]

$cos ^{2}x=1-sin ^{2}x$

which holds as a consequence of the Pythagorean theorem. Then, by taking a square root,

${displaystyle cos x={sqrt {1-sin ^{2}x}}}$

Evaluating this when x = π , we get that

which is incorrect.

The error in each of these examples fundamentally lies in the fact that any equation of the form

$x^{2}=a^{2}$

where aneq 0 , has two solutions:

and it is essential to check which of these solutions is relevant to the problem at hand.^[10] In the above fallacy, the square root that allowed the second equation to be deduced from the first is valid only when cos x is positive. In particular, when x is set to π, the second equation is rendered invalid.

Square roots of negative numbers[edit]

Invalid proofs utilizing powers and roots are often of the following kind:

1={sqrt {1}}={sqrt {(-1)(-1)}}={sqrt {-1}}{sqrt {-1}}=icdot i=-1.

The fallacy is that the rule is generally valid only if at least one of and is non-negative (when dealing with real numbers), which is not the case here.^[11]

Alternatively, imaginary roots are obfuscated in the following:

${displaystyle i={sqrt {-1}}=left(-1right)^{frac {2}{4}}=left(left(-1right)^{2}right)^{frac {1}{4}}=1^{frac {1}{4}}=1}$

The error here lies in the third equality, as the rule ${displaystyle a^{bc}=(a^{b})^{c}}$ only holds for positive real a and real b, c.

Complex exponents[edit]

When a number is raised to a complex power, the result is not uniquely defined (see Exponentiation § Failure of power and logarithm identities). If this property is not recognized, then errors such as the following can result:

${displaystyle {begin{aligned}e^{2pi i}&=1\left(e^{2pi i}right)^{i}&=1^{i}\e^{-2pi }&=1\end{aligned}}}$

The error here is that the rule of multiplying exponents as when going to the third line does not apply unmodified with complex exponents, even if when putting both sides to the power i only the principal value is chosen. When treated as multivalued functions, both sides produce the same set of values, being {e^2πn | n ∈ ℤ}.

Geometry[edit]

Many mathematical fallacies in geometry arise from using an additive equality involving oriented quantities (such as adding vectors along a given line or adding oriented angles in the plane) to a valid identity, but which fixes only the absolute value of (one of) these quantities. This quantity is then incorporated into the equation with the wrong orientation, so as to produce an absurd conclusion. This wrong orientation is usually suggested implicitly by supplying an imprecise diagram of the situation, where relative positions of points or lines are chosen in a way that is actually impossible under the hypotheses of the argument, but non-obviously so.

In general, such a fallacy is easy to expose by drawing a precise picture of the situation, in which some relative positions will be different from those in the provided diagram. In order to avoid such fallacies, a correct geometric argument using addition or subtraction of distances or angles should always prove that quantities are being incorporated with their correct orientation.

Fallacy of the isosceles triangle[edit]

The fallacy of the isosceles triangle, from (Maxwell 1959, Chapter II, § 1), purports to show that every triangle is isosceles, meaning that two sides of the triangle are congruent. This fallacy was known to Lewis Carroll and may have been discovered by him. It was published in 1899.^[12]^[13]

Given a triangle △ABC, prove that AB = AC:

Draw a line bisecting ∠A.
Draw the perpendicular bisector of segment BC, which bisects BC at a point D.
Let these two lines meet at a point O.
Draw line OR perpendicular to AB, line OQ perpendicular to AC.
Draw lines OB and OC.
By AAS, △RAO ≅ △QAO (∠ORA = ∠OQA = 90°; ∠RAO = ∠QAO; AO = AO (common side)).
By RHS,^{[note 2]} △ROB ≅ △QOC (∠BRO = ∠CQO = 90°; BO = OC (hypotenuse); RO = OQ (leg)).
Thus, AR = AQ, RB = QC, and AB = AR + RB = AQ + QC = AC.

Q.E.D.

As a corollary, one can show that all triangles are equilateral, by showing that AB = BC and AC = BC in the same way.

The error in the proof is the assumption in the diagram that the point O is inside the triangle. In fact, O always lies on the circumcircle of the △ABC (except for isosceles and equilateral triangles where AO and OD coincide). Furthermore, it can be shown that, if AB is longer than AC, then R will lie within AB, while Q will lie outside of AC, and vice versa (in fact, any diagram drawn with sufficiently accurate instruments will verify the above two facts). Because of this, AB is still AR + RB, but AC is actually AQ − QC; and thus the lengths are not necessarily the same.

Proof by induction[edit]

There exist several fallacious proofs by induction in which one of the components, basis case or inductive step, is incorrect. Intuitively, proofs by induction work by arguing that if a statement is true in one case, it is true in the next case, and hence by repeatedly applying this, it can be shown to be true for all cases. The following «proof» shows that all horses are the same colour.^[14]^{[note 3]}

Let us say that any group of N horses is all of the same colour.
If we remove a horse from the group, we have a group of N − 1 horses of the same colour. If we add another horse, we have another group of N horses. By our previous assumption, all the horses are of the same colour in this new group, since it is a group of N horses.
Thus we have constructed two groups of N horses all of the same colour, with N − 1 horses in common. Since these two groups have some horses in common, the two groups must be of the same colour as each other.
Therefore, combining all the horses used, we have a group of N + 1 horses of the same colour.
Thus if any N horses are all the same colour, any N + 1 horses are the same colour.
This is clearly true for N = 1 (i.e., one horse is a group where all the horses are the same colour). Thus, by induction, N horses are the same colour for any positive integer N, and so all horses are the same colour.

The fallacy in this proof arises in line 3. For N = 1, the two groups of horses have N − 1 = 0 horses in common, and thus are not necessarily the same colour as each other, so the group of N + 1 = 2 horses is not necessarily all of the same colour. The implication «every N horses are of the same colour, then N + 1 horses are of the same colour» works for any N > 1, but fails to be true when N = 1. The basis case is correct, but the induction step has a fundamental flaw.

Notes[edit]

^ The same fallacy also applies to the following:

${displaystyle {begin{aligned}{frac {19}{95}}={frac {19!!!/}{9!!!/5}}&={frac {1}{5}}\{frac {26}{65}}={frac {26!!!/}{6!!!/5}}&={frac {2}{5}}\{frac {49}{98}}={frac {49!!!/}{9!!!/8}}&={frac {4}{8}}={frac {1}{2}}end{aligned}}}$
^ Hypotenuse–leg congruence
^ George Pólya’s original «proof» was that any n girls have the same colour eyes.

References[edit]

^ Maxwell 1959, p. 9
^ ^a ^b Maxwell 1959
^ Heath & Heiberg 1908, Chapter II, §I
^ Barbeau, Ed (1991). «Fallacies, Flaws, and Flimflam» (PDF). The College Mathematics Journal. 22 (5). ISSN 0746-8342.
^ «soft question – Best Fake Proofs? (A M.SE April Fools Day collection)». Mathematics Stack Exchange. Retrieved 2019-10-24.
^ Heuser, Harro (1989), Lehrbuch der Analysis – Teil 1 (6th ed.), Teubner, p. 51, ISBN 978-3-8351-0131-9
^ Barbeau, Ed (1990), «Fallacies, Flaws and Flimflam #19: Dolt’s Theorem», The College Mathematics Journal, 21 (3): 216–218, doi:10.1080/07468342.1990.11973308
^ Frohlichstein, Jack (1967). Mathematical Fun, Games and Puzzles (illustrated ed.). Courier Corporation. p. 207. ISBN 0-486-20789-7. Extract of page 207
^ Maxwell 1959, Chapter VI, §I.1
^ Maxwell 1959, Chapter VI, §II
^ Nahin, Paul J. (2010). An Imaginary Tale: The Story of «i«. Princeton University Press. p. 12. ISBN 978-1-4008-3029-9. Extract of page 12
^ S.D.Collingwood, ed. (1899), The Lewis Carroll Picture Book, Collins, pp. 190–191
^ Robin Wilson (2008), Lewis Carroll in Numberland, Penguin Books, pp. 169–170, ISBN 978-0-14-101610-8
^ Pólya, George (1954). Induction and Analogy in Mathematics. Mathematics and plausible reasoning. Vol. 1. Princeton. p. 120.

Barbeau, Edward J. (2000), Mathematical fallacies, flaws, and flimflam, MAA Spectrum, Mathematical Association of America, ISBN 978-0-88385-529-4, MR 1725831.
Bunch, Bryan (1997), Mathematical fallacies and paradoxes, New York: Dover Publications, ISBN 978-0-486-29664-7, MR 1461270.
Heath, Sir Thomas Little; Heiberg, Johan Ludvig (1908), The thirteen books of Euclid’s Elements, Volume 1, The University Press.
Maxwell, E. A. (1959), Fallacies in mathematics, Cambridge University Press, ISBN 0-521-05700-0, MR 0099907.

External links[edit]

Invalid proofs at Cut-the-knot (including literature references)
Classic fallacies with some discussion
More invalid proofs from AhaJokes.com
Math jokes including an invalid proof

Источник

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Стандартной ошибкой называется величина, которая характеризует стандартное (среднеквадратическое) отклонение выборочного среднего. Другими словами, эту величину можно использовать для оценки точности выборочного среднего. Множество областей применения стандартной ошибки по умолчанию предполагают нормальное распределение. Если вам нужно рассчитать стандартную ошибку, перейдите к шагу 1.

1

Запомните определение среднеквадратического отклонения. Среднеквадратическое отклонение выборки – это мера рассеянности значения. Среднеквадратическое отклонение выборки обычно обозначается буквой s. Математическая формула среднеквадратического отклонения приведена выше.
2

Узнайте, что такое истинное среднее значение. Истинное среднее является средним группы чисел, включающим все числа всей группы – другими словами, это среднее всей группы чисел, а не выборки.
3

Научитесь рассчитывать среднеарифметическое значение. Среднеаримфетическое означает попросту среднее: сумму значений собранных данных, разделенную на количество значений этих данных.
4

Узнайте, что такое выборочное среднее. Когда среднеарифметическое значение основано на серии наблюдений, полученных в результате выборок из статистической совокупности, оно называется “выборочным средним”. Это среднее выборки чисел, которое описывает среднее значение лишь части чисел из всей группы. Его обозначают как:
5

Усвойте понятие нормального распределения. Нормальные распределения, которые используются чаще других распределений, являются симметричными, с единичным максимумом в центре – на среднем значении данных. Форма кривой подобна очертаниям колокола, при этом график равномерно опускается по обе стороны от среднего. Пятьдесят процентов распределения лежит слева от среднего, а другие пятьдесят процентов – справа от него. Рассеянность значений нормального распределения описывается стандартным отклонением.
6

Запомните основную формулу. Формула для вычисления стандартной ошибки приведена выше.

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1
Рассчитайте выборочное среднее. Чтобы найти стандартную ошибку, сначала нужно определить среднеквадратическое отклонение (поскольку среднеквадратическое отклонение s входит в формулу для вычисления стандартной ошибки). Начните с нахождения средних значений. Выборочное среднее выражается как среднее арифметическое измерений x1, x2, . . . , xn. Его рассчитывают по формуле, приведенной выше.
- Допустим, например, что вам нужно рассчитать стандартную ошибку выборочного среднего результатов измерения массы пяти монет, указанных в таблице:
  Вы сможете рассчитать выборочное среднее, подставив значения массы в формулу:
2
Вычтите выборочное среднее из каждого измерения и возведите полученное значение в квадрат. Как только вы получите выборочное среднее, вы можете расширить вашу таблицу, вычтя его из каждого измерения и возведя результат в квадрат.
- Для нашего примера расширенная таблица будет иметь следующий вид:
3
Найдите суммарное отклонение ваших измерений от выборочного среднего. Общее отклонение – это сумма возведенных в квадрат разностей от выборочного среднего. Чтобы определить его, сложите ваши новые значения.
- В нашем примере нужно будет выполнить следующий расчет:
  Это уравнение дает сумму квадратов отклонений измерений от выборочного среднего.
4
Рассчитайте среднеквадратическое отклонение ваших измерений от выборочного среднего. Как только вы будете знать суммарное отклонение, вы сможете найти среднее отклонение, разделив ответ на n -1. Обратите внимание, что n равно числу измерений.
- В нашем примере было сделано 5 измерений, следовательно n – 1 будет равно 4. Расчет нужно вести следующим образом:
5
Найдите среднеквадратичное отклонение. Сейчас у вас есть все необходимые значения для того, чтобы воспользоваться формулой для нахождения среднеквадратичного отклонения s.
- В нашем примере вы будете рассчитывать среднеквадратичное отклонение следующим образом:
  Следовательно, среднеквадратичное отклонение равно 0,0071624.
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1
Чтобы вычислить стандартную ошибку, воспользуйтесь базовой формулой со среднеквадратическим отклонением.
- В нашем примере вы сможете рассчитать стандартную ошибку следующим образом:
  Таким образом в нашем примере стандартная ошибка (среднеквадратическое отклонение выборочного среднего) составляет 0,0032031 грамма.

Советы

Стандартную ошибку и среднеквадратическое отклонение часто путают. Обратите внимание, что стандартная ошибка описывает среднеквадратическое отклонение выборочного распределения статистических данных, а не распределения отдельных значений
В научных журналах понятия стандартной ошибки и среднеквадратического отклонения несколько размыты. Для объединения двух величин используется знак ±.

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Стандартное отклонение (англ. Standard Deviation) — простыми словами это мера того, насколько разбросан набор данных.

Вычисляя его, можно узнать, являются ли числа близкими к среднему значению или далеки от него. Если точки данных находятся далеко от среднего значения, то в наборе данных имеется большое отклонение; таким образом, чем больше разброс данных, тем выше стандартное отклонение.

Стандартное отклонение обозначается буквой σ (греческая буква сигма).

Стандартное отклонение также называется:

среднеквадратическое отклонение,
среднее квадратическое отклонение,
среднеквадратичное отклонение,
квадратичное отклонение,
стандартный разброс.

Использование и интерпретация величины среднеквадратического отклонения

Стандартное отклонение используется:

в финансах в качестве меры волатильности,
в социологии в опросах общественного мнения — оно помогает в расчёте погрешности.

Пример:

Рассмотрим два малых предприятия, у нас есть данные о запасе какого-то товара на их складах.

	День 1	День 2	День 3	День 4
Пред.А	19	21	19	21
Пред.Б	15	26	15	24

В обеих компаниях среднее количество товара составляет 20 единиц:

А -> (19 + 21 + 19+ 21) / 4 = 20
Б -> (15 + 26 + 15+ 24) / 4 = 20

Однако, глядя на цифры, можно заметить:

в компании A количество товара всех четырёх дней очень близко находится к этому среднему значению 20 (колеблется лишь между 19 ед. и 21 ед.),
в компании Б существует большая разница со средним количеством товара (колеблется между 15 ед. и 26 ед.).

Если рассчитать стандартное отклонение каждой компании, оно покажет, что

стандартное отклонение компании A = 1,
стандартное отклонение компании Б ≈ 5.

Стандартное отклонение показывает эту волатильность данных — то, с каким размахом они меняются; т.е. как сильно этот запас товара на складах компаний колеблется (поднимается и опускается).

Расчет среднеквадратичного (стандартного) отклонения

Формулы вычисления стандартного отклонения

Где:
σ — стандартное отклонение,
xi — величина отдельного значения выборки,
μ — среднее арифметическое выборки,
n — размер выборки.
Эта формула применяется, когда анализируются все значения выборки.

стандартное отклонение формула, среднее квадратичное отклонение формула, среднеквадратическое отклонение формула, среднее квадратическое отклонение формула

Где:
S — стандартное отклонение,
n — размер выборки,
xi — величина отдельного значения выборки,
xср — среднее арифметическое выборки.
Эта формула применяется, когда присутствует очень большой размер выборки, поэтому на анализ обычно берётся только её часть.
Единственная разница с предыдущей формулой: “n — 1” вместо “n”, и обозначение «xср» вместо «μ».

Разница между формулами S и σ («n» и «n–1»)

Состоит в том, что мы анализируем — всю выборку или только её часть:

только её часть – используется формула S (с «n–1»),
полностью все данные – используется формула σ (с «n»).

Как рассчитать стандартное отклонение?

Пример 1 (с σ)

Рассмотрим данные о запасе какого-то товара на складах Предприятия Б.

	День 1	День 2	День 3	День 4
Пред.Б	15	26	15	24

Если значений выборки немного (небольшое n, здесь он равен 4) и анализируются все значения, то применяется эта формула:

Применяем эти шаги:

1. Найти среднее арифметическое выборки:

μ = (15 + 26 + 15+ 24) / 4 = 20

2. От каждого значения выборки отнять среднее арифметическое:

x1 — μ = 15 — 20 = -5

x2 — μ = 26 — 20 = 6

x3 — μ = 15 — 20 = -5

x4 — μ = 24 — 20 = 4

3. Каждую полученную разницу возвести в квадрат:

(x1 — μ)² = (-5)² = 25

(x2 — μ)² = 6² = 36

(x3 — μ)² = (-5)² = 25

(x4 — μ)² = 4² = 16

4. Сделать сумму полученных значений:

Σ (xi — μ)² = 25 + 36+ 25+ 16 = 102

5. Поделить на размер выборки (т.е. на n):

(Σ (xi — μ)²)/n = 102 / 4 = 25,5

6. Найти квадратный корень:

√((Σ (xi — μ)²)/n) = √ 25,5 ≈ 5,0498

Пример 2 (с S)

Задача усложняется, когда существуют сотни, тысячи или даже миллионы данных. В этом случае берётся только часть этих данных и анализируется методом выборки.

У Андрея 20 яблонь, но он посчитал яблоки только на 6 из них.

Популяция — это все 20 яблонь, а выборка — 6 яблонь, это деревья, которые Андрей посчитал.

Яблоня 1	Яблоня 2	Яблоня 3	Яблоня 4	Яблоня 5	Яблоня 6
9	2	5	4	12	7

Так как мы используем только выборку в качестве оценки всей популяции, то нужно применить эту формулу:

Применяем эти шаги:

1. Найти среднее арифметическое выборки:

μ = (15 + 26 + 15+ 24) / 4 = 20

2. От каждого значения выборки отнять среднее арифметическое:

x1 — μ = 15 — 20 = -5

x2 — μ = 26 — 20 = 6

x3 — μ = 15 — 20 = -5

x4 — μ = 24 — 20 = 4

3. Каждую полученную разницу возвести в квадрат:

(x1 — μ)² = (-5)² = 25

(x2 — μ)² = 6² = 36

(x3 — μ)² = (-5)² = 25

(x4 — μ)² = 4² = 16

4. Сделать сумму полученных значений:

Σ (xi — μ)² = 25 + 36+ 25+ 16 = 102

5. Поделить на размер выборки (т.е. на n):

(Σ (xi — μ)²)/n = 102 / 4 = 25,5

6. Найти квадратный корень:

√((Σ (xi — μ)²)/n) = √ 25,5 ≈ 5,0498

Пример 2 (с S)

У Андрея 20 яблонь, но он посчитал яблоки только на 6 из них.

Популяция — это все 20 яблонь, а выборка — 6 яблонь, это деревья, которые Андрей посчитал.

Яблоня 1	Яблоня 2	Яблоня 3	Яблоня 4	Яблоня 5	Яблоня 6
9	2	5	4	12	7

Так как мы используем только выборку в качестве оценки всей популяции, то нужно применить эту формулу:

Математически она отличается от предыдущей формулы только тем, что от n нужно будет вычесть 1. Формально нужно будет также вместо μ (среднее арифметическое) написать X ср.

Применяем практически те же шаги:

1. Найти среднее арифметическое выборки:

Xср = (9 + 2 + 5 + 4 + 12 + 7) / 6 = 39 / 6 = 6,5

2. От каждого значения выборки отнять среднее арифметическое:

X1 – Xср = 9 – 6,5 = 2,5

X2 – Xср = 2 – 6,5 = –4,5

X3 – Xср = 5 – 6,5 = –1,5

X4 – Xср = 4 – 6,5 = –2,5

X5 – Xср = 12 – 6,5 = 5,5

X6 – Xср = 7 – 6,5 = 0,5

3. Каждую полученную разницу возвести в квадрат:

(X1 – Xср)² = (2,5)² = 6,25

(X2 – Xср)² = (–4,5)² = 20,25

(X3 – Xср)² = (–1,5)² = 2,25

(X4 – Xср)² = (–2,5)² = 6,25

(X5 – Xср)² = 5,5² = 30,25

(X6 – Xср)² = 0,5² = 0,25

4. Сделать сумму полученных значений:

Σ (Xi – Xср)² = 6,25 + 20,25+ 2,25+ 6,25 + 30,25 + 0,25 = 65,5

5. Поделить на размер выборки, вычитав перед этим 1 (т.е. на n–1):

(Σ (Xi – Xср)²)/(n-1) = 65,5 / (6 – 1) = 13,1

6. Найти квадратный корень:

S = √((Σ (Xi – Xср)²)/(n–1)) = √ 13,1 ≈ 3,6193

Дисперсия и стандартное отклонение

Стандартное отклонение равно квадратному корню из дисперсии (S = √D). То есть, если у вас уже есть стандартное отклонение и нужно рассчитать дисперсию, нужно лишь возвести стандартное отклонение в квадрат (S² = D).

Дисперсия — в статистике это «среднее квадратов отклонений от среднего». Чтобы её вычислить нужно:

Вычесть среднее значение из каждого числа
Возвести каждый результат в квадрат (так получатся квадраты разностей)
Найти среднее значение квадратов разностей.

Ещё расчёт дисперсии можно сделать по этой формуле:

Где:
S² — выборочная дисперсия,
Xi — величина отдельного значения выборки,
Xср (может появляться как X̅) — среднее арифметическое выборки,
n — размер выборки.

Правило трёх сигм

Это правило гласит: вероятность того, что случайная величина отклонится от своего математического ожидания более чем на три стандартных отклонения (на три сигмы), почти равна нулю.

Правило трёх сигм

Глядя на рисунок нормального распределения случайной величины, можно понять, что в пределах:

одного среднеквадратического отклонения заключаются 68,26% значений (Xср ± 1σ или μ ± 1σ),
двух стандартных отклонений — 95,44% (Xср ± 2σ или μ ± 2σ),
трёх стандартных отклонений — 99,72% (Xср ± 3σ или μ ± 3σ).

Это означает, что за пределами остаются лишь 0,28% — это вероятность того, что случайная величина примет значение, которое отклоняется от среднего более чем на 3 сигмы.

Стандартное отклонение в excel

Вычисление стандартного отклонения с «n – 1» в знаменателе (случай выборки из генеральной совокупности):

1. Занесите все данные в документ Excel.

2. Выберите поле, в котором вы хотите отобразить результат.

3. Введите в этом поле «=СТАНДОТКЛОНА(«

4. Выделите поля, где находятся данные, потом закройте скобки.

2. Выберите поле, в котором вы хотите отобразить результат.

3. Введите в этом поле «=СТАНДОТКЛОНА(«

4. Выделите поля, где находятся данные, потом закройте скобки.

Формулы вычисления стандартного отклонения sigma сигма стандартное отклонение в эксель excel

5. Нажмите Ввод (Enter).

В случае если данные представляют всю генеральную совокупность (n в знаменателе), то нужно использовать функцию СТАНДОТКЛОНПА.

Формулы вычисления стандартного отклонения sigma сигма стандартное отклонение в эксель excel

Коэффициент вариации

Коэффициент вариации — отношение стандартного отклонения к среднему значению, т.е. Cv = (S/μ) × 100% или V = (σ/X̅) × 100%.

Стандартное отклонение делится на среднее и умножается на 100%.

Можно классифицировать вариабельность выборки по коэффициенту вариации:

при <10% выборка слабо вариабельна,
при 10% – 20 % — средне вариабельна,
при >20 % — выборка сильно вариабельна.

Узнайте также про:

Корреляции,
Метод Крамера,
Метод наименьших квадратов,
Теорию вероятностей
Интегралы.

Стандартное отклонение и стандартная ошибка: в чем разница?

17 авг. 2022 г.
читать 2 мин

В статистике студенты часто путают два термина: стандартное отклонение и стандартная ошибка .

Стандартное отклонение измеряет, насколько разбросаны значения в наборе данных.

Стандартная ошибка — это стандартное отклонение среднего значения в повторных выборках из совокупности.

Давайте рассмотрим пример, чтобы ясно проиллюстрировать эту идею.

Пример: стандартное отклонение против стандартной ошибки

Предположим, мы измеряем вес 10 разных черепах.

Для этой выборки из 10 черепах мы можем вычислить среднее значение выборки и стандартное отклонение выборки:

Предположим, что стандартное отклонение оказалось равным 8,68. Это дает нам представление о том, насколько распределен вес этих черепах.

Но предположим, что мы собираем еще одну простую случайную выборку из 10 черепах и также проводим их измерения. Более чем вероятно, что эта выборка из 10 черепах будет иметь немного другое среднее значение и стандартное отклонение, даже если они взяты из одной и той же популяции:

Теперь, если мы представим, что мы берем повторные выборки из одной и той же совокупности и записываем выборочное среднее и выборочное стандартное отклонение для каждой выборки:

Теперь представьте, что мы наносим каждое среднее значение выборки на одну и ту же строку:

Стандартное отклонение этих средних значений известно как стандартная ошибка.

Формула для фактического расчета стандартной ошибки:

Стандартная ошибка = s/ √n

куда:

s: стандартное отклонение выборки
n: размер выборки

Какой смысл использовать стандартную ошибку?

Когда мы вычисляем среднее значение данной выборки, нас на самом деле интересует не среднее значение этой конкретной выборки, а скорее среднее значение большей совокупности, из которой взята выборка.

Однако мы используем выборки, потому что для них гораздо проще собирать данные, чем для всего населения. И, конечно же, среднее значение выборки будет варьироваться от выборки к выборке, поэтому мы используем стандартную ошибку среднего значения как способ измерить, насколько точна наша оценка среднего значения.

Вы заметите из формулы для расчета стандартной ошибки, что по мере увеличения размера выборки (n) стандартная ошибка уменьшается:

Стандартная ошибка = s/ √n

Это должно иметь смысл, поскольку большие размеры выборки уменьшают изменчивость и увеличивают вероятность того, что среднее значение нашей выборки ближе к фактическому среднему значению генеральной совокупности.

Когда использовать стандартное отклонение против стандартной ошибки

Если мы просто заинтересованы в измерении того, насколько разбросаны значения в наборе данных, мы можем использовать стандартное отклонение .

Однако, если мы заинтересованы в количественной оценке неопределенности оценки среднего значения, мы можем использовать стандартную ошибку среднего значения .

В зависимости от вашего конкретного сценария и того, чего вы пытаетесь достичь, вы можете использовать либо стандартное отклонение, либо стандартную ошибку.

Коэффициент вариации

Стандартное отклонение делится на среднее и умножается на 100%.

Можно классифицировать вариабельность выборки по коэффициенту вариации:

при <10% выборка слабо вариабельна,
при 10% – 20 % — средне вариабельна,
при >20 % — выборка сильно вариабельна.

Узнайте также про:

Корреляции,
Метод Крамера,
Метод наименьших квадратов,
Теорию вероятностей
Интегралы.

Стандартное отклонение и стандартная ошибка: в чем разница?

17 авг. 2022 г.
читать 2 мин

В статистике студенты часто путают два термина: стандартное отклонение и стандартная ошибка .

Стандартное отклонение измеряет, насколько разбросаны значения в наборе данных.

Стандартная ошибка — это стандартное отклонение среднего значения в повторных выборках из совокупности.

Давайте рассмотрим пример, чтобы ясно проиллюстрировать эту идею.

Пример: стандартное отклонение против стандартной ошибки

Предположим, мы измеряем вес 10 разных черепах.

Для этой выборки из 10 черепах мы можем вычислить среднее значение выборки и стандартное отклонение выборки:

Теперь представьте, что мы наносим каждое среднее значение выборки на одну и ту же строку:

Стандартное отклонение этих средних значений известно как стандартная ошибка.

Формула для фактического расчета стандартной ошибки:

Стандартная ошибка = s/ √n

куда:

s: стандартное отклонение выборки
n: размер выборки

Какой смысл использовать стандартную ошибку?

Стандартная ошибка = s/ √n

Когда использовать стандартное отклонение против стандартной ошибки

Cumulative probability of a normal distribution with expected value 0 and standard deviation 1

In statistics, the standard deviation is a measure of the amount of variation or dispersion of a set of values.^[1] A low standard deviation indicates that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range.

Standard deviation may be abbreviated SD, and is most commonly represented in mathematical texts and equations by the lower case Greek letter σ (sigma), for the population standard deviation, or the Latin letter s, for the sample standard deviation.

The standard deviation of a random variable, sample, statistical population, data set, or probability distribution is the square root of its variance. It is algebraically simpler, though in practice less robust, than the average absolute deviation.^[2]^[3] A useful property of the standard deviation is that, unlike the variance, it is expressed in the same unit as the data.

The standard deviation of a population or sample and the standard error of a statistic (e.g., of the sample mean) are quite different, but related. The sample mean’s standard error is the standard deviation of the set of means that would be found by drawing an infinite number of repeated samples from the population and computing a mean for each sample. The mean’s standard error turns out to equal the population standard deviation divided by the square root of the sample size, and is estimated by using the sample standard deviation divided by the square root of the sample size. For example, a poll’s standard error (what is reported as the margin of error of the poll), is the expected standard deviation of the estimated mean if the same poll were to be conducted multiple times. Thus, the standard error estimates the standard deviation of an estimate, which itself measures how much the estimate depends on the particular sample that was taken from the population.

In science, it is common to report both the standard deviation of the data (as a summary statistic) and the standard error of the estimate (as a measure of potential error in the findings). By convention, only effects more than two standard errors away from a null expectation are considered «statistically significant», a safeguard against spurious conclusion that is really due to random sampling error.

When only a sample of data from a population is available, the term standard deviation of the sample or sample standard deviation can refer to either the above-mentioned quantity as applied to those data, or to a modified quantity that is an unbiased estimate of the population standard deviation (the standard deviation of the entire population).

Basic examples[edit]

Population standard deviation of grades of eight students[edit]

Suppose that the entire population of interest is eight students in a particular class. For a finite set of numbers, the population standard deviation is found by taking the square root of the average of the squared deviations of the values subtracted from their average value. The marks of a class of eight students (that is, a statistical population) are the following eight values:

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

${displaystyle mu ={frac {2+4+4+4+5+5+7+9}{8}}={frac {40}{8}}=5.}$

First, calculate the deviations of each data point from the mean, and square the result of each:

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

The variance is the mean of these values:

${displaystyle sigma ^{2}={frac {9+1+1+1+0+0+4+16}{8}}={frac {32}{8}}=4.}$

and the population standard deviation is equal to the square root of the variance:

This formula is valid only if the eight values with which we began form the complete population. If the values instead were a random sample drawn from some large parent population (for example, they were 8 students randomly and independently chosen from a class of 2 million), then one divides by 7 (which is n − 1) instead of 8 (which is n) in the denominator of the last formula, and the result is In that case, the result of the original formula would be called the sample standard deviation and denoted by s instead of sigma . Dividing by n − 1 rather than by n gives an unbiased estimate of the variance of the larger parent population. This is known as Bessel’s correction.^[4]^[5] Roughly, the reason for it is that the formula for the sample variance relies on computing differences of observations from the sample mean, and the sample mean itself was constructed to be as close as possible to the observations, so just dividing by n would underestimate the variability.

Standard deviation of average height for adult men[edit]

If the population of interest is approximately normally distributed, the standard deviation provides information on the proportion of observations above or below certain values. For example, the average height for adult men in the United States is about 70 inches, with a standard deviation of around 3 inches. This means that most men (about 68%, assuming a normal distribution) have a height within 3 inches of the mean (67–73 inches) – one standard deviation – and almost all men (about 95%) have a height within 6 inches of the mean (64–76 inches) – two standard deviations. If the standard deviation were zero, then all men would be exactly 70 inches tall. If the standard deviation were 20 inches, then men would have much more variable heights, with a typical range of about 50–90 inches. Three standard deviations account for 99.73% of the sample population being studied, assuming the distribution is normal or bell-shaped (see the 68–95–99.7 rule, or the empirical rule, for more information).

Definition of population values[edit]

Let μ be the expected value (the average) of random variable X with density f(x):

${displaystyle mu equiv operatorname {E} [X]=int _{-infty }^{+infty }xf(x),mathrm {d} x}$

The standard deviation σ of X is defined as

${displaystyle sigma equiv {sqrt {operatorname {E} left[(X-mu )^{2}right]}}={sqrt {int _{-infty }^{+infty }(x-mu )^{2}f(x),mathrm {d} x}},}$

which can be shown to equal ${textstyle {sqrt {operatorname {E} left[X^{2}right]-(operatorname {E} [X])^{2}}}.}$

Using words, the standard deviation is the square root of the variance of X.

The standard deviation of a probability distribution is the same as that of a random variable having that distribution.

Not all random variables have a standard deviation. If the distribution has fat tails going out to infinity, the standard deviation might not exist, because the integral might not converge. The normal distribution has tails going out to infinity, but its mean and standard deviation do exist, because the tails diminish quickly enough. The Pareto distribution with parameter has a mean, but not a standard deviation (loosely speaking, the standard deviation is infinite). The Cauchy distribution has neither a mean nor a standard deviation.

Discrete random variable[edit]

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

${displaystyle sigma ={sqrt {{frac {1}{N}}left[(x_{1}-mu )^{2}+(x_{2}-mu )^{2}+cdots +(x_{N}-mu )^{2}right]}},{text{ where }}mu ={frac {1}{N}}(x_{1}+cdots +x_{N}),}$

or, by using summation notation,

${displaystyle sigma ={sqrt {{frac {1}{N}}sum _{i=1}^{N}(x_{i}-mu )^{2}}},{text{ where }}mu ={frac {1}{N}}sum _{i=1}^{N}x_{i}.}$

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

${displaystyle sigma ={sqrt {sum _{i=1}^{N}p_{i}(x_{i}-mu )^{2}}},{text{ where }}mu =sum _{i=1}^{N}p_{i}x_{i}.}$

Continuous random variable[edit]

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

${displaystyle sigma ={sqrt {int _{mathbf {X} }(x-mu )^{2},p(x),mathrm {d} x}},{text{ where }}mu =int _{mathbf {X} }x,p(x),mathrm {d} x,}$

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

${displaystyle {sqrt {left(e^{sigma ^{2}}-1right)e^{2mu +sigma ^{2}}}}.}$

Estimation[edit]

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).

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 technically 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[edit]

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 s_N, 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:^[6]

${displaystyle s_{N}={sqrt {{frac {1}{N}}sum _{i=1}^{N}left(x_{i}-{bar {x}}right)^{2}}},}$

where ${displaystyle {x_{1},,x_{2},,ldots ,,x_{N}}}$ are the observed values of the sample items, and is the mean value of these observations, while the denominator N stands for the size of the sample: this is the square root of the sample variance, which is the average of the squared deviations about the sample mean.

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.^[7] 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 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[edit]

If the biased sample variance (the second central moment of the sample, which is a downward-biased estimate of the population variance) is used to compute an estimate of the population’s standard deviation, the result is

${displaystyle s_{N}={sqrt {{frac {1}{N}}sum _{i=1}^{N}left(x_{i}-{bar {x}}right)^{2}}}.}$

Here taking the square root introduces further downward bias, by Jensen’s inequality, due to the square root’s 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 applying Bessel’s correction, using N − 1 instead of N to yield the unbiased sample variance, denoted s²:

${displaystyle s^{2}={frac {1}{N-1}}sum _{i=1}^{N}left(x_{i}-{bar {x}}right)^{2}.}$

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 deviations from the mean, ${displaystyle textstyle (x_{1}-{bar {x}},;dots ,;x_{n}-{bar {x}}).}$

Taking square roots reintroduces bias (because the square root is a nonlinear function which does not commute with the expectation, i.e. often ), yielding the corrected sample standard deviation, denoted by s:

${displaystyle s={sqrt {{frac {1}{N-1}}sum _{i=1}^{N}left(x_{i}-{bar {x}}right)^{2}}}.}$

As explained above, while s² is an unbiased estimator for the population variance, s is still a biased estimator for the population standard deviation, though markedly less biased than the uncorrected sample standard deviation. This estimator is commonly used and generally known simply as the «sample standard deviation». The bias may still be large for small samples (N less than 10). As sample size increases, the amount of bias decreases. We obtain more information and the difference between ${frac {1}{N}}$ and ${displaystyle {frac {1}{N-1}}}$ becomes smaller.

Unbiased sample standard deviation[edit]

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 can be given by replacing N − 1 with N − 1.5, yielding:

${displaystyle {hat {sigma }}={sqrt {{frac {1}{N-1.5}}sum _{i=1}^{N}left(x_{i}-{bar {x}}right)^{2}}},}$

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%.

A more accurate approximation is to replace above with .^[8]

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

${displaystyle {hat {sigma }}={sqrt {{frac {1}{N-1.5-{frac {1}{4}}gamma _{2}}}sum _{i=1}^{N}left(x_{i}-{bar {x}}right)^{2}}},}$

where γ₂ denotes the population excess kurtosis. The excess kurtosis may be either known beforehand for certain distributions, or estimated from the data.^[9]

Confidence interval of a sampled standard deviation[edit]

The standard deviation we obtain by sampling a distribution is itself not absolutely accurate, both for mathematical reasons (explained here by the confidence interval) and for practical reasons of measurement (measurement error). The mathematical effect can be described by the confidence interval or CI.

To show how a larger sample will make the confidence interval narrower, consider the following examples:
A small population of N = 2 has only 1 degree of freedom for estimating the standard deviation. The result is that a 95% CI of the SD runs from 0.45 × SD to 31.9 × SD; the factors here are as follows:

${displaystyle Pr left(q_{frac {alpha }{2}}<k{frac {s^{2}}{sigma ^{2}}}<q_{1-{frac {alpha }{2}}}right)=1-alpha ,}$

where ${displaystyle q_{p}}$ is the p-th quantile of the chi-square distribution with k degrees of freedom, and 1-alpha is the confidence level. This is equivalent to the following:

${displaystyle Pr left(k{frac {s^{2}}{q_{1-{frac {alpha }{2}}}}}<sigma ^{2}<k{frac {s^{2}}{q_{frac {alpha }{2}}}}right)=1-alpha .}$

With k = 1, ${displaystyle q_{0.025}=0.000982}$ and ${displaystyle q_{0.975}=5.024}$ . The reciprocals of the square roots of these two numbers give us the factors 0.45 and 31.9 given above.

A larger population of N = 10 has 9 degrees of freedom for estimating the standard deviation. The same computations as above give us in this case a 95% CI running from 0.69 × SD to 1.83 × SD. So even with a sample population of 10, the actual SD can still be almost a factor 2 higher than the sampled SD. For a sample population N=100, this is down to 0.88 × SD to 1.16 × SD. To be more certain that the sampled SD is close to the actual SD we need to sample a large number of points.

These same formulae can be used to obtain confidence intervals on the variance of residuals from a least squares fit under standard normal theory, where k is now the number of degrees of freedom for error.

Bounds on standard deviation[edit]

For a set of N > 4 data spanning a range of values R, an upper bound on the standard deviation s is given by s = 0.6R.^[10]
An estimate of the standard deviation for N > 100 data taken to be approximately normal follows from the heuristic that 95% of the area under the normal curve lies roughly two standard deviations to either side of the mean, so that, with 95% probability the total range of values R represents four standard deviations so that s ≈ R/4. This so-called range rule is useful in sample size estimation, as the range of possible values is easier to estimate than the standard deviation. Other divisors K(N) of the range such that s ≈ R/K(N) are available for other values of N and for non-normal distributions.^[11]

Identities and mathematical properties[edit]

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:

{displaystyle {begin{aligned}sigma (c)&=0sigma (X+c)&=sigma (X),sigma (cX)&=|c|sigma (X).end{aligned}}}

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

sigma (X+Y)={sqrt {operatorname {var} (X)+operatorname {var} (Y)+2,operatorname {cov} (X,Y)}}.,

where ${displaystyle textstyle operatorname {var} ,=,sigma ^{2}}$ and stand for variance and covariance, respectively.

The calculation of the sum of squared deviations can be related to moments calculated directly from the data. In the following formula, the letter E is interpreted to mean expected value, i.e., mean.

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

The sample standard deviation can be computed as:

${displaystyle s(X)={sqrt {frac {N}{N-1}}}{sqrt {operatorname {E} left[(X-operatorname {E} [X])^{2}right]}}.}$

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

${displaystyle {sqrt {{frac {1}{N}}sum _{i=1}^{N}left(x_{i}-{bar {x}}right)^{2}}}={sqrt {{frac {1}{N}}left(sum _{i=1}^{N}x_{i}^{2}right)-{bar {x}}^{2}}}={sqrt {left({frac {1}{N}}sum _{i=1}^{N}x_{i}^{2}right)-left({frac {1}{N}}sum _{i=1}^{N}x_{i}right)^{2}}},}$

which means that the standard deviation is equal to the square root of the difference between the average of the squares of the values and the square of the average value.

See computational formula for the variance for proof, and for an analogous result for the sample standard deviation.

Interpretation and application[edit]

Example of samples from two populations with the same mean but different standard deviations. Red population has mean 100 and SD 10; blue population has mean 100 and SD 50.

A large standard deviation indicates that the data points can spread 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. These standard deviations 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[edit]

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).

Experiment, industrial and hypothesis testing[edit]

Standard deviation is often used to compare real-world data against a model to test the model.
For example, in industrial applications the weight of products coming off a production line may need to comply with a legally required value. By weighing some fraction of the products an average weight can be found, which will always be slightly different from the long-term average. By using standard deviations, a minimum and maximum value can be calculated that the averaged weight will be within some very high percentage of the time (99.9% or more). If it falls outside the range then the production process may need to be corrected. Statistical tests such as these are particularly important when the testing is relatively expensive. For example, if the product needs to be opened and drained and weighed, or if the product was otherwise used up by the test.

In experimental science, a theoretical model of reality is used. Particle physics conventionally uses a standard of «5 sigma» for the declaration of a discovery. A five-sigma level translates to one chance in 3.5 million that a random fluctuation would yield the result. This level of certainty was required in order to assert that a particle consistent with the Higgs boson had been discovered in two independent experiments at CERN,^[12] also leading to the declaration of the first observation of gravitational waves.^[13]

Weather[edit]

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.

Finance[edit]

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^[14] (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, 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 technical analysis tool. For example, the upper Bollinger Band is given as ${displaystyle textstyle {bar {x}}+nsigma _{x}.}$ 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.

Financial time series are known to be non-stationary series, whereas the statistical calculations above, such as standard deviation, apply only to stationary series. 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[edit]

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. That is indeed the case. To move orthogonally from L to the point P, one begins at the point:

{displaystyle M=left({bar {x}},{bar {x}},{bar {x}}right)}

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) ${textstyle {sqrt {sum _{i}left(x_{i}-{bar {x}}right)^{2}}}}$ 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[edit]

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.

Distance from mean	Minimum population
	50%
2σ	75%
3σ	89%
4σ	94%
5σ	96%
6σ	97%
	${displaystyle 1-{frac {1}{k^{2}}}}$ ^[15]
${displaystyle {frac {1}{sqrt {1-ell }}},sigma }$

Rules for normally distributed data[edit]

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 states 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

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

where μ is the expected value of the random variables, σ equals their distribution’s standard deviation divided by 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:

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

where is the error function. The proportion that is less than or equal to a number, x, is given by the cumulative distribution function:

${displaystyle {text{Proportion}}leq x={frac {1}{2}}left[1+operatorname {erf} left({frac {x-mu }{sigma {sqrt {2}}}}right)right]={frac {1}{2}}left[1+operatorname {erf} left({frac {z}{sqrt {2}}}right)right]}$

.^[16]

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:

Confidence interval	Proportion within	Proportion without
Percentage	Percentage	Fraction
0.318639σ	25%	75%	3 / 4
0.674490σ	50%	50%	1 / 2
0.977925σ	66.6667%	33.3333%	1 / 3
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 / 1000
3.890592σ	99.99%	0.01%	1 / 10000
4σ	99.993666%	0.006334%	1 / 15787
4.417173σ	99.999%	0.001%	1 / 100000
4.5σ	99.9993204653751%	0.0006795346249%	1 / 147159.5358 6.8 / 1000000
4.891638σ	99.9999%	0.0001%	1 / 1000000
5σ	99.9999426697%	0.0000573303%	1 / 1744278
5.326724σ	99.99999%	0.00001%	1 / 10000000
5.730729σ	99.999999%	0.000001%	1 / 100000000
6σ	99.9999998027%	0.0000001973%	1 / 506797346
6.109410σ	99.9999999%	0.0000001%	1 / 1000000000
6.466951σ	99.99999999%	0.00000001%	1 / 10000000000
6.806502σ	99.999999999%	0.000000001%	1 / 100000000000
7σ	99.9999999997440%	0.000000000256%	1 / 390682215445

Relationship between standard deviation and mean[edit]

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, …, x_n are real numbers and define the function:

${displaystyle sigma (r)={sqrt {{frac {1}{N-1}}sum _{i=1}^{N}left(x_{i}-rright)^{2}}}.}$

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

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[edit]

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. Assuming statistical independence of the values in the sample, the standard deviation of the mean is related to the standard deviation of the distribution by:

${displaystyle sigma _{text{mean}}={frac {1}{sqrt {N}}}sigma }$

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):

${displaystyle {begin{aligned}operatorname {var} (X)&equiv sigma _{X}^{2}operatorname {var} (X_{1}+X_{2})&equiv operatorname {var} (X_{1})+operatorname {var} (X_{2})end{aligned}}}$

(Statistical independence is assumed.)

${displaystyle operatorname {var} (cX_{1})equiv c^{2},operatorname {var} (X_{1})}$

hence

${displaystyle {begin{aligned}operatorname {var} ({text{mean}})&=operatorname {var} left({frac {1}{N}}sum _{i=1}^{N}X_{i}right)={frac {1}{N^{2}}}operatorname {var} left(sum _{i=1}^{N}X_{i}right)&={frac {1}{N^{2}}}sum _{i=1}^{N}operatorname {var} (X_{i})={frac {N}{N^{2}}}operatorname {var} (X)={frac {1}{N}}operatorname {var} (X).end{aligned}}}$

Resulting in:

$sigma _{text{mean}}={frac {sigma }{sqrt {N}}}.$

In order to estimate the standard deviation of the mean $sigma _{text{mean}}$ it is necessary to know the standard deviation of the entire population sigma beforehand. However, in most applications this parameter is unknown. For example, if a series of 10 measurements of a previously unknown quantity is performed in a laboratory, it is possible to calculate the resulting sample mean and sample standard deviation, but it is impossible to calculate the standard deviation of the mean. However, one can estimate the standard deviation of the entire population from the sample, and thus obtain an estimate for the standard error of the mean.

Rapid calculation methods[edit]

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

${displaystyle s_{j}=sum _{k=1}^{N}{x_{k}^{j}}.}$

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

${displaystyle sigma ={frac {sqrt {Ns_{2}-s_{1}^{2}}}{N}}}$

Where N, as mentioned above, is the size of the set of values (or can also be regarded as s₀).

Similarly for sample standard deviation,

${displaystyle s={sqrt {frac {Ns_{2}-s_{1}^{2}}{N(N-1)}}}.}$

In a computer implementation, as the two 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.^[17] 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:

${displaystyle {begin{aligned}A_{0}&=0A_{k}&=A_{k-1}+{frac {x_{k}-A_{k-1}}{k}}end{aligned}}}$

where A is the mean value.

${displaystyle {begin{aligned}Q_{0}&=0Q_{k}&=Q_{k-1}+{frac {k-1}{k}}left(x_{k}-A_{k-1}right)^{2}=Q_{k-1}+left(x_{k}-A_{k-1}right)left(x_{k}-A_{k}right)end{aligned}}}$

Note: $Q_{1}=0$ since k-1=0 or $x_{1}=A_{1}$

Sample variance:

${displaystyle s_{n}^{2}={frac {Q_{n}}{n-1}}}$

Population variance:

${displaystyle sigma _{n}^{2}={frac {Q_{n}}{n}}}$

Weighted calculation[edit]

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

${displaystyle s_{j}=sum _{k=1}^{N}w_{k}x_{k}^{j}.,}$

And the standard deviation equations remain unchanged. 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:

${displaystyle {begin{aligned}W_{0}&=0W_{k}&=W_{k-1}+w_{k}end{aligned}}}$

and places where 1/n is used above must be replaced by w_i/W_n:

${displaystyle {begin{aligned}A_{0}&=0A_{k}&=A_{k-1}+{frac {w_{k}}{W_{k}}}left(x_{k}-A_{k-1}right)Q_{0}&=0Q_{k}&=Q_{k-1}+{frac {w_{k}W_{k-1}}{W_{k}}}left(x_{k}-A_{k-1}right)^{2}=Q_{k-1}+w_{k}left(x_{k}-A_{k-1}right)left(x_{k}-A_{k}right)end{aligned}}}$

In the final division,

${displaystyle sigma _{n}^{2}={frac {Q_{n}}{W_{n}}},}$

and

${displaystyle s_{n}^{2}={frac {Q_{n}}{W_{n}-1}},}$

${displaystyle s_{n}^{2}={frac {n'}{n'-1}}sigma _{n}^{2},}$

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.

History[edit]

The term standard deviation was first used in writing by Karl Pearson in 1894, following his use of it in lectures.^[18]^[19] This was as a replacement for earlier alternative names for the same idea: for example, Gauss used mean error.^[20]

Higher dimensions[edit]

The standard deviation ellipse (green) of a two-dimensional normal distribution

In two dimensions, the standard deviation can be illustrated with the standard deviation ellipse (see Multivariate normal distribution § Geometric interpretation).

References[edit]

^ Bland, J.M.; Altman, D.G. (1996). «Statistics notes: measurement error». BMJ. 312 (7047): 1654. doi:10.1136/bmj.312.7047.1654. PMC 2351401. PMID 8664723.
^ Gauss, Carl Friedrich (1816). «Bestimmung der Genauigkeit der Beobachtungen». Zeitschrift für Astronomie und Verwandte Wissenschaften. 1: 187–197.
^ Walker, Helen (1931). Studies in the History of the Statistical Method. Baltimore, MD: Williams & Wilkins Co. pp. 24–25.
^ Weisstein, Eric W. «Bessel’s Correction». MathWorld.
^ «Standard Deviation Formulas». www.mathsisfun.com. Retrieved 21 August 2020.
^ Weisstein, Eric W. «Standard Deviation». mathworld.wolfram.com. Retrieved 21 August 2020.
^ «Consistent estimator». www.statlect.com. Retrieved 10 October 2022.
^ Gurland, John; Tripathi, Ram C. (1971), «A Simple Approximation for Unbiased Estimation of the Standard Deviation», The American Statistician, 25 (4): 30–32, doi:10.2307/2682923, JSTOR 2682923
^ «Standard Deviation Calculator». PureCalculators. 11 July 2021. Retrieved 14 September 2021.
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External links[edit]

«Quadratic deviation», Encyclopedia of Mathematics, EMS Press, 2001 [1994]
«Standard Deviation Calculator»