Hi /sci/, Can you tell me the easiest way to calculate the

>>8110215
calculate the sum of the squares of every number and then subtract the square of the mean from that as many times as you have numbers.

>>8110219
Thank you!

>>8110245
forgot to add: when you've done all that you need to then take the square root

>>8110219
That's not stdev

>>8110219
>>8110275
you're wrong

ffs. google.com

en.wikipedia.org/wiki/Standard_deviation

>>8110279
>>8110285
not that guy but you can prove that this is true yourself from the definition of variance, it's very easy.

Just apply this

[math]\mu = \frac {\sum (x)}{n}\\
\sigma = \sqrt {\frac {\sum \left (x^{2} \right )}{n} - \mu^{2}}[/math]

where x represents your numbers.

>>8110593
variance = (1/n)sum((x-a)^2)

(1/n)sum((x-a)^2) =? sum(x^2) - na^2

(1/n)sum(x^2 + a^2 - 2ax) =? sum(x^2) - na^2

(1/n)sum(x^2) + a^2 - (2a/n)sum(x) =? sum(x^2) - na^2

(1 - 1/n)sum(x^2) - (2a/n)sum(x) + (n+1)a^2 =? 0

So if we have two numbers A and B

(1 - 1/2)(A^2+B^2) - (2(A+B)/4)(A+B) + (2+1)(A+B)^2/4 =? 0

(1/2)(A^2+B^2) + (1/4)(A+B)^2 =? 0

3A^2 + 2AB + 3B^2 =? 0

This only has imaginary solutions, so it's wrong.

>>8110613
That's wrong also. Jesus /sci/.

[eqn]\sigma =\sqrt{ \frac{\sum(x-\mu)^2}{n}}[/eqn]

>>8110907
what an idiot

>>8110925
this

alternatively, 1/n*sum(x^2f(x))-µ^2