# Simple question about Standard Deviation.

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Helen Kirby am 8 Jan. 2017
Kommentiert: Walter Roberson am 9 Jan. 2017
I have a number of data points, lets say in a vector v, and lets say there are "num" of them. If I write sd = std(v) did it assume a sample i.e. it used num-1 (in the denominator) or did I get a population standard dev i.e. it used num? How can I request one or the other?
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### Akzeptierte Antwort

the cyclist am 8 Jan. 2017
Bearbeitet: the cyclist am 8 Jan. 2017
By default, it will give the sample standard deviation. Call it as
std(x,1)
to get the population. That is explained in the documentation for std, in the section describing the input argument weight.
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John BG am 8 Jan. 2017
look for standard deviation or std

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### Weitere Antworten (1)

Helen Kirby am 8 Jan. 2017
Could I ask yet another question on this theme - and yes I have read the documentation and it doesn't answer this question. Say if you have x = [1,2,3,4,5,6] and w = [5,7,10,8,12,3] and we want to find the weighted std for a population, how do I write the command for a POPULATION? I understand for a sample it is:
StdSamp = std(x,w) If you put the 1 as the 3rd parameter, it does not interpret it as pop.
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Walter Roberson am 9 Jan. 2017
You cannot combine the two weighting schemes.
std(x) is normalized by N-1. std(x,1) is normalized by N. std(x,1) works out to be the same as std(x, ones(size(x)) .
std(x,w,1) means to proceed along dimension 1. Your data was row vectors, so that did not work. But you could use
std(x(:), w(:), 1)
if you had particular reason for wanting to specifically process along the first dimension.

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