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What is a standard deviation, how is it calculated and what is its use in statistics?

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    $\begingroup$ Is this really the kind of question we want on this site? A simple google search, opening any stats book, or checking out wikipedia would instantly provide an answer. $\endgroup$ – PeterR Jul 22 '10 at 17:36
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    $\begingroup$ I don't think the purpose of this site is to answer 6th graders questions. And my kid, when faced with such a question, would google for the answer. If there is a specific part of the definition you don't understand, ask away. But such an unfocused question on such a basic topic indicates (to me anyway) that the poster didn't even try to find an answer. What is going to be next "What is a number and how are they used?" $\endgroup$ – PeterR Jul 23 '10 at 12:05
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    $\begingroup$ I think this question is ok. Actually, it was the most upvoted example on topic question on Area 51. Basics are ok here! $\endgroup$ – Peter Smit Jul 26 '10 at 18:52
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    $\begingroup$ Agreed, it's a valid question. It's also well stated as it asks for example usage and calculation. Surely the purpose of the site is to create a repository for ALL questions statistical. $\endgroup$ – Joel Sep 8 '10 at 13:03
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    $\begingroup$ I agree with Joel. Standard deviation is an important concept in statistics. Would it not be absurd if you couldn't ask a question about it on a site about asking statistical questions. $\endgroup$ – Parbury Apr 17 '11 at 23:00
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Standard deviation is a number that represents the "spread" or "dispersion" of a set of data. There are other measures for spread, such as range and variance.

Here are some example sets of data, and their standard deviations:

[1,1,1]     standard deviation = 0   (there's no spread)  
[-1,1,3]    standard deviation = 1.6 (some spread) 
[-99,1,101] standard deviation = 82  (big spead)

The above data sets have the same mean.

Deviation means "distance from the mean".

"Standard" here means "standardized", meaning the standard deviation and mean are in the same units, unlike variance.

For example, if the mean height is 2 meters, the standard deviation might be 0.3 meters, whereas the variance would be 0.09 meters squared.

It is convenient to know that at least 75% of the data points always lie within 2 standard deviations of the mean (or around 95% if the distribution is Normal).

For example, if the mean is 100, and the standard deviation is 15, then at least 75% of the values are between 70 and 130.

If the distribution happens to be Normal, then 95% of the values are between 70 and 130.

Generally speaking, IQ test scores are normally distributed and have an average of 100. Someone who is "very bright" is two standard deviations above the mean, meaning an IQ test score of 130.

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  • $\begingroup$ Neil, thank you for your answer, could you please explain in more details the part "standard" in the term "standard deviation" term. If it is appropriate could you please touch on the same "standard" in "standard error of mean" term. Thank you in advance. $\endgroup$ – stan Feb 18 '12 at 16:35
  • $\begingroup$ Re your recent edits: in what sense is the SD "standardized"? Usually, it becomes the basis for standardization, but is not itself standardized (such as rescaling it by some estimate of its sampling variation). $\endgroup$ – whuber Feb 19 '12 at 17:37
  • $\begingroup$ It is standardized to be in the same unit as the mean $\endgroup$ – Neil McGuigan Apr 9 '12 at 8:37
  • $\begingroup$ The example with a mean height of 2 metres is a good example of needing to take care of the use of decimals. The same example could be done in centimetres where a standard deviation of 30 centimetres would logically derive from a variance of 900 centimetres. $\endgroup$ – Robert Jones Apr 19 '13 at 23:18
  • $\begingroup$ My impression is that they should be avoided in the primary units of measurement. Consider the results say of an SD of 0.133 in metres converted to decimetres, centimetres and millimetres. Would anyone care to elucidate, please? $\endgroup$ – Robert Jones Apr 19 '13 at 23:40
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A quote from Wikipedia.

It shows how much variation there is from the "average" (mean, or expected/budgeted 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 is spread out over a large range of values.

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When describing a variable we typically summarise it using two measures: a measure of centre and a measure of spread. Common measures of centre include the mean, median and mode. Common measure of spread include the variance and interquartile range.

The variance (represented by the Greek lowercase sigma raised to the power two) is commonly used when the mean is reported. The variance is the average squared deviation of variable. The deviation is calculated by subtracting the mean from each observation. This is squared because the sum would otherwise be zero and squaring removes this problem while maintaining the relative size of the deviations. The problem with using the variation as a measure of spread is that it is in squared units. For example if our variable of interest was height measured in inches then the variance would be reported in squared-inches which makes little sense. The standard deviation (represented by the Greek lowercase sigma) is the square-root of the variance and returns the measure of spread to the original units. This is much more intuitive and is therefore more popular than the variance.

When using the standard deviation, one has to be careful of outliers as they will skew the standard deviation (and the mean) as they are not resistant measures of spread. A simple example will illustrate this property. The mean of my terrible cricket batting scores of 13, 14, 16, 23, 26, 28, 33, 39, and 61 is 28.11. If we consider 61 to be an outlier and deleted it, the mean would be 24.

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    $\begingroup$ Graham, I wonder if there are some typos in your answer. Variance is represented by the Greek lowercase sigma raised to the power of 2 (i.e., $\sigma^2$), and the standard deviation is the square-root of that, or just sigma without an exponent (i.e., $\sigma$). You may want to edit your answer. $\endgroup$ – gung Feb 18 '12 at 20:46
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Here's how I would answer this question using a diagram.

Let's say we weigh 30 cats and calculate the mean weight. Then we produce a scatter plot, with weight on the y axis and cat identity on the x axis. The mean weight can be drawn in as a horizontal line. We can then draw in vertical lines which connect each data point to the mean line - these are the deviations of each data point from the mean, and we call them residuals. Now, these residuals can be useful because they can tell us something about the spread of the data: if there are many big residuals, then cats vary a lot in mass. Conversely, if the residuals are mainly small, then cats are fairly closely clustered around the average weight. So if we could have some metric which tells us the average length of a residual in this data set, this would be a handy way of denoting how much spread there is in the data. The standard deviation is, effectively, the length of the average residual.

I would follow on on from this by giving the calculation for s.d., explaining why we square and then square root (I like Vaibhav's short and sweet explanation). Then I would mention the problems of outliers, as Graham does in his last paragraph.

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If the information required is the distribution of data about the mean, standard deviation comes in handy.

The sum of the difference of each value from the mean is zero (obviously, since the value are evenly spread around the mean), hence we square each difference so as to convert negative values to positive, sum them across the population, and take their square root. This value is then divided by the number of samples (or, the size of the population). This gives the standard deviation.

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  • $\begingroup$ ".hence we square each difference...." We could take the absolute value to get rid of negative values too. So why is squaring a better method since we have to take a square root at the end? Why not just sum the absolute values of the deviations? $\endgroup$ – Dilip Sarwate Jan 30 '12 at 17:30
  • $\begingroup$ Seen This one? link $\endgroup$ – Vaibhav Garg Feb 6 '12 at 7:10
  • $\begingroup$ Yes, I had seen that link before. Had you? I fully understand the reasons why squaring is used, ever since I learned them over $45$ years ago. I was questioning your authoritative use of the word hence in your phrase without any indication that you knew the justification for why the sum of squares is used instead of the sum of absolute values. $\endgroup$ – Dilip Sarwate Feb 6 '12 at 10:41
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    $\begingroup$ @DilipSarwate, with all due respect, Proof by authority does not impress me. The supposition that "hence" is "authoritative" is a "Straw-man" that I'd rather ignore. The level of detail in any given statement is commensurate with the inclination and/or the pedagogical significance of the same in a given context. I'd assume that a person who is asking "What is a standard deviation, how is it ....so forth?" may not wish to be burdened with rigorous mathematical definitions of the same. The simplification is deliberate and, let me assure you, not a result of not being aware. $\endgroup$ – Vaibhav Garg Feb 6 '12 at 11:22
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    $\begingroup$ And what, pray tell, is .."hence we square ..." other than a proof by authority that does not impress you? There is no logical reason why squaring is automatically the solution to the problem as your "hence" implies. $\endgroup$ – Dilip Sarwate Feb 6 '12 at 18:53
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I like to think of it as follows: the standard deviation is the average distance from the average. This is more conceptually useful than mathematically useful, but its a nice way to explain it to the uninitiated.

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A standard deviation is the square root of the second central moment of a distribution. A central moment is the expected difference from the expected value of the distribution. A first central moment would usually be 0, so we define a second central moment as the expected value of the squared distance of a random variable from its expected value.

To put it on a scale that is more in line with the original observations, we take the square root of that second central moment and call it the standard deviation.

Standard deviation is a property of a population. It measures how much average "dispersion" there is to that population. Are all the obsrvations clustered around the mean, or are they widely spread out?

To estimate the standard deviation of a population, we often calculate the standard deviation of a "sample" from that population. To do this, you take observations from that population, calculate a mean of those observations, and then calculate the square root of the average squared deviation from that "sample mean".

To get an unbiased estimator of the variance, you don't actually calculate the average squared deviation from the sample mean, but instead, you divide by (N-1) where N is the number of observations in your sample. Note that this "sample standard deviation" is not an unbiased estimator of the standard deviation, but the square of the "sample standard deviation" is an unbiased estimator of the variance of the population.

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    $\begingroup$ this is an incredibly unclear response. Please try to write in English. $\endgroup$ – Neil McGuigan Jul 19 '10 at 19:38
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    $\begingroup$ maybe so. is a person asking this question a person who walked in off the street, or a person who has at least opened a statistics book. Telling someone the standard deviation is just the square root of the variance is completely begging the question. $\endgroup$ – Baltimark Jul 20 '10 at 1:59
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The best way I have understood standard deviation is to think of a hair dresser! (You need to collect data from a hair dresser and averge her hair cutting speed for this example to work.)

It takes an average of 30 minutes for the hair dresser to cut a persons hair.

Suppose you do the calculation (most software packages will do this for you) and you find that the standard deviation is 5 minutes. It means the following:

  • the hair dresser cuts hair of 68% of her clients within 25 minutes and 35 minutes
  • the hair dresser cuts hair of 96% of her clients within 20 and 40 minutes

How do I know this? You need to look at the normal curve, where 68% falls within 1 standard deviation and 96% falls within 2 standard deviations of the mean (in this case 30 minutes). So you add or subtract the standard deviation from the mean.

If consistency is desired, as in this case, then the smaller the standard deviation, the better. In this case, the hair dresser spends a maximum of about 40 minutes with any given client. You need to cut hair fast in order to run a successful saloon!

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  • $\begingroup$ I don't think you proofread your answer, Adhesh. You've got some contradictory information in here. See whether you agree with my edits, ok? $\endgroup$ – rolando2 Jan 31 '12 at 1:04
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    $\begingroup$ You've only described the standard deviation's interpretation in the case of the normal distribution. The '68% rule' and (and 95% rule) only apply for normally distributed data. At least state that the two bullet points are only true if haircutting times follow a normal distribution. $\endgroup$ – Macro Jan 31 '12 at 2:16
  • $\begingroup$ Macro, I did mention the normal curve and it is a given that if you use the normal curve, the data would follow a normal distribution. $\endgroup$ – Adhesh Josh Jan 31 '12 at 8:25
  • $\begingroup$ @rolando2 I dont seem to understand what is wrong with Adhesh's explanation $\endgroup$ – Amarald Apr 17 '12 at 12:19
  • $\begingroup$ @Amarald - have you clicked on "Jan 31 at 1:06" to see the versions before and after editing? I think the answer is stronger after, though Macro makes an important point too. $\endgroup$ – rolando2 Apr 17 '12 at 21:06

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