Hi,
Just learned another way to calculate sd for a frequency distribution
matrix:
> p <- matrix(c(10,3,20,4,30,5),ncol=2,byrow=TRUE)
> p
[,1] [,2]
[1,] 103
[2,] 204
[3,] 305
> rep(p[,1],p[,2])
[1] 10 10 10 20 20 20 20 30 30 30 30 30
> sd(rep(p[,1],p[,2]))
[1] 8.348471
Hi,
For the second one,
sqrt((sum(p[,1]^2*p[,2])-(sum(p[,1]*p[,2]))^2/sum(p[,2]))/(sum(p[,2])-1)),
please refer to the following link for an example to explain how it works.
http://www.lboro.ac.uk/media/wwwlboroacuk/content/mlsc/downloads/var_stand_deviat_group.pdf
For the first one:
sd(unli
Thank you, I'll try as soon as possible!
Il 13/Feb/2015 22:28 "JS Huang [via R]" <
ml-node+s789695n470323...@n4.nabble.com> ha scritto:
> Or if you want to perform the calculation without using sd:
>
> sqrt((sum(p[,1]^2*p[,2])-(sum(p[,1]*p[,2]))^2/sum(p[,2]))/(sum(p[,2])-1))
>
>
>
Or if you want to perform the calculation without using sd:
sqrt((sum(p[,1]^2*p[,2])-(sum(p[,1]*p[,2]))^2/sum(p[,2]))/(sum(p[,2])-1))
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Hi,
Try this.
> sd(unlist(sapply(1:dim(p)[1],function(i)rep(p[i,1],p[i,2]
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