> > The formula I generally use to compute the standard deviation is the
> square root of the variance where the variance is E(X^2) - (E(x))^2.
Even if that is what you want, I hope you don't _implement_ it that way (eg as
mean(x^2)-mean(x)^2. _Serious_ numerical rounding issues...
#Example:
The help page ?sd (I only checked version 3.1.0) has as the 1st
sentence in the details section: "Like ‘var’ this uses denominator n -
1." and on the help page ?var in the details section (though down in
the 3rd paragraph from the end) there is this sentence: "The
denominator n - 1 is used which gi
Hi Bob,
You may check:
library(multicon)
?popsd()
sd(1:3)
#[1] 1
popsd(1:3)
#[1] 0.8164966
sd(1:3)*sqrt(2/3)
#[1] 0.8164966
A.K.
The formula I generally use to compute the standard deviation is the square
root of the variance where the variance is E(X^2) - (E(x))^2. That is, the
divisor is
Which formula for standard deviation are you using?
If you know the population mean then you should divide by n (3 in this
case), but if you don't know the population mean and use the mean
calculated from the sample then it is more usual to use n-1 as the
denominator (this makes the variance an un
Hi,
Please check this link:
http://stats.stackexchange.com/questions/25956/what-formula-is-used-for-standard-deviation-in-r
A.K.
It is my understanding that the R function SD finds the standard deviation of a
random variable or a list. Please consider the following list: { 1, 2, 3 }. I
claim t
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