# Why is the variance squared, and does it mean the same thing as the standard deviation? [duplicate]

I am still confused, despite similar questions being asked, about the difference between the variance and standard deviation in statistics. Why is the variance squared?

• I am not clear what you are asking here... The variance is squared because that's it's definition. It is not the same as the standard deviation, it is the sd squared (they are only the same when they = 0 or 1). Or do you want to know how each is used? – Peter Flom Sep 17 '13 at 10:54
• Variance and standard deviation are both useful. For example, with independent variables, the variance of their sum is the sum of their variances. Simple properties of variances allow us to partition variances in ANOVA for example; things are more complex if you try to work from standard deviations (which don't partition the same way; their squares do). But on the other hand, variances aren't in the original units, the way standard deviations are, so standard deviations, too, are important in their own right. – Glen_b -Reinstate Monica Sep 17 '13 at 10:56
• Lots of useful explanations: stats.stackexchange.com/questions/118/… stats.stackexchange.com/questions/35123/… At one level the reason is simply that the variance is squared because it's convenient that way. Perhaps it's best to follow the links and to ask a more focused question if it's still unclear. – Gala Sep 17 '13 at 12:04
• Perhaps it is worth pointing out that the variance is itself is rarely_squared;_ the variance (let's denote it by $V$) is a quantity that happens to be the square of the standard deviation $\sigma$, $V=\sigma^2$, but $(\text{variance})^2 = \sigma^4$ is not encountered as frequently as $V =\sigma^2$. – Dilip Sarwate Sep 17 '13 at 12:54

It is not easy to get an intuition behind standard deviation $\sigma$, as one gets easily about mean as soon as one sees them.
Now why variance is squared of standard deviation is perhaps an outcome of the definition of the variance ($Var[X] = \Sigma_i (x-u)^2 f(x) dx)$ which requires that the difference term $(x - u)$ should be squared. Now one can protest that one need not use $(x - u)^2$ in this definition and live with some other metric such as $|(x-u)|$.