
Thank You 
 
It really works fine and i do understand it. Now i can sleep, its 0200 hours here.

Why don't you just do this:

H<-rnorm(100, mean=50000, sd=3000)
par(las=1)
hist(H, breaks=seq(40000, 60000, 1000), freq=F)
f<- function(x) exp(-(x-5000)^2/18000000)/sqrt(18000000*pi)
x<- seq(40000, 60000, 100)
lines(x, dnorm(x, 50000, 3000))



On Sat, Sep 12, 2009 at 5:47 PM, KABELI MEFANE<kabelimefane@yahoo.co.uk> wrote:
>  Dear All
>
> let me go one step further by asking you if you could help me show that the distribution of this data in normal.  have a little idea (by trial and error) but i seem to not fully understand how its done.
>
> H<-rnorm(100, mean=50000, sd=3000)
> par(las=1)
> hist(H, breaks=seq(40000, 60000, 1000), freq=F)
> f<- function(x) exp(-(x-5000)^2/18000000)/sqrt(18000000*pi)
> x<- seq(40000, 60000, 1000)
> lines(x,f(x))
>
> The code works when  its like this:
>
> H<-rnorm(100, mean=5000, sd=1000)
> par(las=1)
> hist(H, breaks=seq(2000, 8000, 100), freq=F)
> f<- function(x) exp(-(x-5000)^2/2000000)/sqrt(2000000*pi)
> x<- seq(2000, 8000, 50)
> lines(x,f(x))
>
> Please tell me or advice of an easier or understandable code
>
>
>
>
>
>
>
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>
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?



      
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