# Standard Deviation. Why do we take the square root of the entire equation?

Please forgive my lack of maths knowledge,

It is my understanding that:

Standard Deviation is the average distance from the mean in a data set of numbers.

Therefore it stands to reason that working out the standard deviation of the data set $x_i = \{1,2,3,4,5\}$ would involve the following.

First working out the mean $\mu(x_i) = 3$ and Then working out the sum of the distance from the average $\sum{|x_i-\mu|} = 6$ then we do $\frac{\sum{|x_i-\mu|}}{N} = \frac{6}{5} = 1.2$

This means that, according to my method/thinking, 1.2 is the standard deviation.

However when using the formula $\sqrt{\frac{\sum{(x_i-\mu)}^2}{N}}$ I get $1.414$

Can someone explain why I'm wrong in layman terms. Thankyou

• This is the definition: $\sqrt{\frac{\sum|x_i-\mu|^2}{N} }$ and not $\sqrt{\frac{\sum|x_i-\mu|}{N} }$ as you thought. – zoli Mar 21 '15 at 22:25
• Standard deviation is not the average distance from the mean, as your example shows. ${}\qquad{}$ – Michael Hardy Mar 21 '15 at 22:36
• To get $2.68$ you must have made a mistake in the calculation. I get $\sum(x_i-\mu)^2=10$; with $N=5$ I get $\sqrt(10/5)=1.414$. – bof Mar 21 '15 at 22:41
• There are many nice answers to this over at stats.SE here and also in this MO answer. – aes Mar 21 '15 at 22:44
• @aes Very very true. That link is legit. – Daniel W. Farlow Mar 21 '15 at 22:48

The reason for using standard deviation rather than mean absolute deviation is that the variance of $\{x_i\}_{i=1}^m$ plus the variance of $\{y_j\}_{j=1}^m$ is the variance of $\{x_i+y_j\}_{i=1,\,j=1}^{n,\,m}$ (but only if you define variance in the way that puts $n$ and $m$ rather than the Bessel-corrected $n-1$ and $m-1$ in the denominators). This makes it possible, for example, to apply the central limit theorem to find the probability that when you toss $1800$ coins, the number of heads is between $890$ and $920$. You can find the standard deviation of the number of heads because of the additivity of variances.