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I was wondering what the difference between the variance and the standard deviation is.

If you calculate the two values, it is clear that you get the standard deviation out of the variance, but what does that mean in terms of the distribution you are observing?

Furthermore, why do you really need a standard deviation?

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    $\begingroup$ stats.stackexchange.com/questions/118/… $\endgroup$
    – whuber ♦
    Aug 26, 2012 at 22:20
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    $\begingroup$ You probably got the answer by now. Still this link has the simplest and best explanation. mathsisfun.com/data/standard-deviation.html $\endgroup$
    – user20726
    Feb 11, 2013 at 13:09
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    $\begingroup$ Standard deviation is useful as the value is in the same scale as the data from which it was computed. If measuring meters, the standard deviation will be meters. Variance, in contrast, will be meters squared. $\endgroup$
    – Vladislavs Dovgalecs
    Nov 8, 2017 at 10:29
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    $\begingroup$ Standard Variation can be unbiased but Standard Deviation can't because Square root function is non linear. $\endgroup$
    – Daksh Gargas
    Sep 20, 2019 at 8:03

7 Answers 7

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The standard deviation is the square root of the variance.

The standard deviation is expressed in the same units as the mean is, whereas the variance is expressed in squared units, but for looking at a distribution, you can use either just so long as you are clear about what you are using. For example, a Normal distribution with mean = 10 and sd = 3 is exactly the same thing as a Normal distribution with mean = 10 and variance = 9.

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    $\begingroup$ yeah thats the mathematical way to explain these two parameters, BUT whats the logical explenation? Why do I really ned two parameters to show the same thing(the deviation around the arithmetical mean)... $\endgroup$
    – Le Max
    Aug 26, 2012 at 12:40
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    $\begingroup$ You don't really need both. If you report one, you don't need to report the other $\endgroup$
    – Peter Flom
    Aug 26, 2012 at 12:47
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    $\begingroup$ We need both: standard deviation is good for interpretation, reporting. For developing the theory the variance is better. $\endgroup$
    – kjetil b halvorsen ♦
    Jan 27, 2016 at 18:13
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    $\begingroup$ The benefit of reporting standard deviation is that it remains in the scale of data. Say, a sample of adult heights is in meters, then standard deviation will also be in meters. $\endgroup$
    – Vladislavs Dovgalecs
    Nov 8, 2017 at 10:22
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    $\begingroup$ @RushatRai When dealing with sums of random variables, variances get added together. For independent random variables, $Var(\sum X_i) = \sum Var(X_i)$. A similar expression exists in the general case without independence (with a correction using covariance terms). In general, the square root transformation complicates things and makes standard deviation more difficult to work with analytically. $\endgroup$
    – knrumsey
    Feb 18, 2019 at 20:25
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You don't need both. They each have different purposes. The SD is usually more useful to describe the variability of the data while the variance is usually much more useful mathematically. For example, the sum of uncorrelated distributions (random variables) also has a variance that is the sum of the variances of those distributions. This wouldn't be true of the SD. On the other hand, the SD has the convenience of being expressed in units of the original variable.

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If John refers to independent random variables when he says "unrelated distributions," then his response is correct. However, to answer your question, there are several points that can be added:

  1. The mean and variance are the two parameters that determine a normal distribution.

  2. The Chebyshev inequality bounds the probability of a observed random variable being within $k$ standard deviations of the mean.

  3. The standard deviation is used to normalize statistics for statistical tests (e.g. the known standard deviation is used to normalize a sample mean for the $z$ test that the mean differs from $0$ or the sample standard deviation is used to normalize the sample mean when the true standard deviation is unknown, resulting in the $t$ test).

  4. For a normal distribution $68\%$ percent of the distribution is within $1$ standard deviation. $95.4\%$ within $2$ standard deviations and over $99\%$ within $3$ standard deviations.

  5. The margin of error is expressed as a multiple of the standard deviation of the estimate.

  6. Variance and bias are measures of uncertainty in a random quantity. The mean square error for an estimate equals the variance + the squared bias.

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    $\begingroup$ You should probably not say "natural parameter", which are mean divided by variance, and 1 divided by variance: en.wikipedia.org/wiki/Natural_parameter $\endgroup$
    – Neil G
    Feb 2, 2013 at 6:31
  • $\begingroup$ According to the wikipedia link the natural parameter(s) for the normal distribution in terms of its exponential family form depends on whether or not $\sigma$ is assumed to be known or unknown. But I get your point and have taken "natural parameters" out of my answer. $\endgroup$
    – Michael R. Chernick
    Jul 21, 2017 at 12:34
  • $\begingroup$ In point 3, shouldn't it be "standard deviation is used to standardise statistics" instead of normalise? $\endgroup$
    – Harry
    Jul 29, 2019 at 10:55
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The variance of a data set measures the mathematical dispersion of the data relative to the mean. However, though this value is theoretically correct, it is difficult to apply in a real-world sense because the values used to calculate it were squared. The standard deviation, as the square root of the variance gives a value that is in the same units as the original values, which makes it much easier to work with and easier to interpret in conjunction with the concept of the normal curve.

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    $\begingroup$ This does a great job explaining why in simple terms. $\endgroup$
    – jds
    May 16, 2015 at 23:47
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    $\begingroup$ Another good point to make would be that each metric sd and var measure the spread of the variable about the mean. Taking the square root of the variance to get the standard deviation could be viewed as a scaling factor applied to get the metric back into units of the variable. $\endgroup$
    – Matt L.
    Jan 27, 2016 at 19:38
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In terms of the distribution they're equivalent (yet obviously not interchangeable), but beware that in terms of estimators they're not: the square root of an estimate of the variance is NOT an (unbiased) estimator of the standard deviation. Only for a moderately large number of samples (and depending on the estimators) the two approach each other. For small sample sizes you need to know the parametric form of the distribution to convert among the two, which can become slightly circular.

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While calculating the variance, we squared the deviations. It mean that if the given data (observations) is in meters, it will become meter square. Hope it's not correct representation about the deviations. So, we square root again (SD) that is nothing but SD.

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In adition to Hassan's response, you need to be careful on interpreting standard deviation. Some people define it as the mean distance between every observation and its mean, but this is the definition of mean absolute deviation (MAD), thus wrong.

For a better understanding of both concepts, variance and SD, I highly recommend Taleb's video series on statistics (first lesson is about SD).

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