128
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ stats.stackexchange.com/questions/118/… $\endgroup$ – whuber Aug 26 '12 at 22:20
  • 12
    $\begingroup$ 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 '13 at 13:09
  • 2
    $\begingroup$ 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 '17 at 10:29
  • 1
    $\begingroup$ Standard Variation can be unbiased but Standard Deviation can't because Square root function is non linear. $\endgroup$ – Daksh Gargas Sep 20 at 8:03
85
$\begingroup$

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.

$\endgroup$
  • 58
    $\begingroup$ 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 '12 at 12:40
  • 5
    $\begingroup$ You don't really need both. If you report one, you don't need to report the other $\endgroup$ – Peter Flom - Reinstate Monica Aug 26 '12 at 12:47
  • 8
    $\begingroup$ We need both: standard deviation is good for interpretation, reporting. For developing the theory the variance is better. $\endgroup$ – kjetil b halvorsen Jan 27 '16 at 18:13
  • 4
    $\begingroup$ 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 '17 at 10:22
  • 5
    $\begingroup$ @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 - Reinstate Monica Feb 18 at 20:25
50
$\begingroup$

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.

$\endgroup$
24
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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 '13 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 '17 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 at 10:55
15
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This does a great job explaining why in simple terms. $\endgroup$ – gwg May 16 '15 at 23:47
  • 3
    $\begingroup$ 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 '16 at 19:38
6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

protected by whuber Mar 26 '13 at 14:37

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.