I don't have any knowledge of statistics beyond high school common sense. Why is the standard deviation usually seen in combinatorics textbooks, and why is the standard deviation defined intentionally? What is its purpose?

Thanks in advance. I have checked in Wikipedia and many other websites to see what this is, but they are not very concise and clear.

  • $\begingroup$ It is not entirely clear what is the question! Please let us know, by combinatics, what do you have in mind? $\endgroup$ – user21436 Mar 24 '12 at 18:35
  • $\begingroup$ Simply put: the variance is a "nice" measure of dispersion (which the answers below have elaborated); unfortunately, since it does not have the same "dimensions" as the original measurements, it is convenient to define a new quantity with compatible dimensions, and taking the square root of the variance (which is always doable since the variance is always nonnegative) is an obvious first step... $\endgroup$ – J. M. is a poor mathematician Mar 25 '12 at 1:31
  • $\begingroup$ There is also the central limit theorem, which shows that the expectation and variance (or standard deviation) are the most salient features of a distribution. $\endgroup$ – Yuval Filmus Mar 25 '12 at 4:01
  • $\begingroup$ Related: Motivation behind standard deviation? $\endgroup$ – Rahul Mar 25 '12 at 4:16

I never felt that I understood the fuss about the standard deviation, until I encountered Chebyshev's inequality.

Suppose you have a bunch of data. Clearly, the mean of the data is somewhere in the middle, and the data are spread out more or less far from the mean. The standard deviation says how far away they are, in the following sense: no matter how the data are distributed, and no matter how spread out they are, at most $1/n^2$ of them are more than $n$ standard deviations from the mean.

For example, if the standard deviation is 2 units, then all but at most 1/9 of the data are between $m - 6$ and $m + 6$, where $m$ is the mean. Often the data will cluster more closely around the mean than this, but even if you know nothing else, you can guarantee that no more than 1/9 of the data are more than 6 units from $m$.

  • $\begingroup$ This is in fact the reasoning behind buzzwords like "six sigma" being thrown around... $\endgroup$ – J. M. is a poor mathematician Mar 25 '12 at 1:29

The original poster asked a further question in a comment:

"Why on the sole purpose for what its being use you are saying, there are no other formula could replace it and share the same popularity as the standard deviation?"

The mean absolute deviation can also be used as a measure of dispersion, but it lacks one kind of well-behavedness that makes the standard deviation superior to it: If you have independent random variables $X_1,\ldots,X_n$, then the standard deviation of the sum $X_1+\cdots+X_n$ is the square root of the sum of the squares of the standard deviations of the individual terms. No such neat formula is known for the mean absolute deviation.

  • $\begingroup$ And of course the square of the differences is continuously differentiable, whereas the sum of the absolute differences is not. $\endgroup$ – user21725 Mar 25 '12 at 1:47
  • $\begingroup$ Usually you'd say "the variance of the sum of the random variables is the sum of the variances of each random variable", though. :P $\endgroup$ – Joe Z. Jan 15 '13 at 1:58

Standard deviation is a measure of how the observations are located in relation to the mean. For example if you have a set of 2 observations : $$ 8 \ ,\ 2\ $$ Their mean is $\bar{x}=\frac{8+2}{2}=5$. But as you can see they are not very "close" to the mean. On the other hand a set of another 2 observations like:$$ 4 \ , \ 6$$ has a mean $\bar{x}=\frac{4+6}{2}=5$ too,but those values are closer to the mean. Therefore the first set has a large standard deviation (SD=3), while the second set has a smaller one(SD=1). These two numbers say that every observation is about 3 away from the mean in the first case, while the other observations are 1 away from the second mean. It's very important in statistics to know the diversity and variability of a set of data in order to analyze them and come to useful conclusions about the population or the sample you're observing.

  • $\begingroup$ Why on the sole purpose for what its being use you are saying, there are no other formula could replace it and share the same popularity as the standard deviation? $\endgroup$ – Victor Mar 24 '12 at 18:29
  • $\begingroup$ I'm not quite sure what you're asking..You wanna know the reason we use SD as a measure of diversity and not something else? $\endgroup$ – chemeng Mar 24 '12 at 18:37
  • $\begingroup$ @Victor there are others quantities we might compute. The popularity of the standard deviation has more to do with the fact that the variance is more tractable than other measures of dispersion and that it just seems to pop up more frequently than other measures of spread, e.g. the standard deviation and the mean are the relevant quantities in the central limit theorem. $\endgroup$ – guy Mar 24 '12 at 21:24

Given a sample $x = \{x_1,x_2,\ldots,x_n\}$ the sample standard deviation is defined as $$ \operatorname{sd}(x) = \sqrt{\frac{1}{n} \sum_{k=1}^n(x_k - \bar{x})^2 } $$ where $\bar{x} = \frac{1}{n} \sum_{k=1}^n$ denotes the sample mean.

Several properties of the standard deviation are worth noting.

  • It does not change if all the values in the sample are shifted by the same amount, $x_k \mapsto x_k + a$.
  • If all the values in the sample are multiplied by a common factor, i.e. $x_k \mapsto \lambda x_k$, the standard variation also gets multiplied by its absolute value, i.e. $\operatorname{sd}(\lambda \cdot x) \mapsto | \lambda | \cdot \operatorname{sd}( x)$.
  • Standard deviation is bounded from above by the maximal deviation from the mean: $\operatorname{sd}(x) \leqslant \sup_k | x_k - \bar{x} |$. In particular standard deviation of the sample with identical elements is zero.

Because of these properties, standard deviation is a measure of scale of the data, which indicates extent the sample elements deviate from the mean, hence the name.

  • 1
    $\begingroup$ It seems you are implicitly (and unnecessarily) assuming that $\lambda \geq 0$. $\endgroup$ – cardinal Mar 25 '12 at 2:42
  • $\begingroup$ @cardinal Thanks for the comment, I have updated the post. $\endgroup$ – Sasha Mar 25 '12 at 3:04

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