search for: qnorm

I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1...

dear all, here's a couple of questions that puzzled me in these last hours: ##### issue 1 qnorm(1-10e-100)!=qnorm(10e-100) qnorm(1-1e-10) == -qnorm(1e-10) # turns on to be FALSE. Ok I'm not a computer scientist but, # but I had a look at the R inferno so I write: all.equal(qnorm(1-1e-10) , -qnorm(1e-10)) # which turns TRUE, as one would expect, but all.equal(qnorm(1-1e-100) , -qnorm(...

Hello All I am looking for high precision values for the normal distribution in the tail,(1e-10 and 1 - 1e-10) as the R package that I am using sets any number which is out of this range to these values and then calls the qnorm and qt function. What I have noticed is that the qnorm implementation in R is not symmetric when looking at the tails. This is quite surprising to me, as it is well known that this distribution is symmetric, and I have seen implementations in other languages that are symmetric. I have checked the...

...blems has bothered me for the lase couple of hours. > 1e-100==0 [1] FALSE > (1-1e-100)==1 [1] TRUE How can I tell R that 1-1e-100 does not equal to 1, actually, I found out that > (1-1e-16)==1 [1] FALSE > (1-1e-17)==1 [1] TRUE The reason I care about this is that I was try to use qnorm() in my code, for example, > qnorm(1e-100) [1] -21.27345 and if I want to find qnorm(x) where x is very close to 1, say x=1-1e-100, then you would think using qnorm(1-x, lower.tail=F) would give me something other than INF, but that does not work since R would recognize x==1 in this case and...

...s2,9,1) pars2 points(d,premat2%*%pars2,cex=0.5,col=rgb(0,0.6,0.3,0.2)) #premat2%*%pars2 = cdif #confidence intervals when seWithMean = FALSE devAskNewPage(ask=T) plot(f2) points(d,cdif,cex=0.5,col=rgb(0,1,0,0.2)) Vp2<-f2$Vp[5:13,5:13] se2<-sqrt(diag(premat2%*%Vp2%*%t(premat2))) points(d,cdif+qnorm(0.975)*se2,cex=0.5,col=rgb(1,0,0,0.2)) points(d,cdif-qnorm(0.975)*se2,cex=0.5,col=rgb(0,0,1,0.2)) #numerical output for the confidence bands is given by #cdif+qnorm(0.975)*se2 #cdif-qnorm(0.975)*se2 #confidence intervals when seWithMean = TRUE devAskNewPage(ask=T) plot(f2,seWithMean=T) points(d,cd...

Hi, does the shapiro wilk test in R-1.3.0 work correctly? Maybe it does, but can anybody tell me why the following sample doesn't give "W = 1" and "p-value = 1": R> x<-1:9/10;x [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 R> shapiro.test(qnorm(x)) Shapiro-Wilk normality test data: qnorm(x) W = 0.9925, p-value = 0.9986 I can't imagine a sample being more "normal" than this. Furthermore, the Kolmogorov-Smirnov test gives a p-value of 1. R> ks.test(qnorm(x),"pnorm",mean=0,sd=1) One-sample...

how R implement qnorm() I wonder anyone knows the mathematical process that R calculated the quantile? The reason I asked is soly by curiosity. I know the probability of a normal distribution is calculated through integrate the Gaussian function, which can be implemented easily (see code), while the calculation of qu...

Full_Name: Xiong Guanglei Version: 1.8.1 OS: Linux Submission from: (NULL) (202.38.103.50) qnorm(x) when x>1.0

Hi everyone, I have the following problem. I have some small p-values but when I use qnorm(1-4e-30) I get an error. Is there anyway to get around this? -- Thanks, Jim. [[alternative HTML version deleted]]

Cordial saludo para cada uno. De manera amable les pido ayuda para acceder al código R usado para el algoritmo de la función qnorm. Gracias por su ayuda. César Escalante C. [[alternative HTML version deleted]]

Trenkler, Dietrich said: > > I found the following strange behavior using qnorm() and pnorm(): > > > x<-8.21;x-qnorm(pnorm(x)) > [1] 0.0004638484 > > x<-8.28;x-qnorm(pnorm(x)) > [1] 0.07046385 > > x<-8.29;x-qnorm(pnorm(x)) > [1] 0.08046385 > > x<-8.30;x-qnorm(pnorm(x)) > [1] -Inf > qnorm(1-.Machine$double.eps) [1] 8.12589

You may want to look into using the log option to qnorm e.g., in round figures: > log(1e-300) [1] -690.7755 > qnorm(-691, log=TRUE) [1] -37.05315 > exp(37^2/2) [1] 1.881797e+297 > exp(-37^2/2) [1] 5.314068e-298 Notice that floating point representation cuts out at 1e+/-308 or so. If you want to go outside that range, you may need explicit...

Hello all, I have a question about basic statistics. Given a PDF value of 0.328161, how can I find out the value of -0.625 in R? It is like reversing the dnorm function but I do not know how to do it in R. > pdf.xb <- dnorm(-0.625) > pdf.xb [1] 0.328161 > qnorm(pdf.xb) [1] -0.444997 > pnorm(pdf.xb) [1] 0.628605 Many thanks, Edwin -- View this message in context: http://r.789695.n4.nabble.com/dnorm-and-qnorm-tp3040427p3040427.html Sent from the R help mailing list archive at Nabble.com.

...represent alpha value; e.g.: x <- seq(-3.5,5, length=1000) y <- dnorm(x) # Plot distribution A plot(y~x, type='l',axes=F,xlab="",ylab="",lwd=2) # Plot distribution B y2 <- dnorm(x-1.5) lines(y2~x,lwd=2) # Plot vertical line for alpha value abline(h=0) segments(qnorm(.5)+1.5,0,qnorm(.5)+1.5,dnorm(qnorm(.5))) text(2,0.2,"Power") Now I want to shadow area labeled as "Power". Any suggestion how to do that using 'polygon' function? Thanks in advance for any suggestion. Andrej

A recent message on ansari.test() prompted me to play with the examples. This doesn't work for me in R version 2.4.1 R> ansari.test(rnorm(100), rnorm(100, 0, 2), conf.int = TRUE) Error in uniroot(ab, srange, tol = 1e-04, zq = qnorm(alpha/2, lower = FALSE)) : object "ab" not found It looks like there's a small typo in ccia() inside ansari.test.default() in which 'ab' is passed to uniroot rather than 'ab2'. The definition of ccia below was just copied from https://svn.r-project.org/R/trunk/src/li...

Hi, Am not sure if my code itself is correct. Here's what am trying to do: Minimize integration of a function of gaussian distributed variable 'x' over the interval qnorm(0.999) to Inf by changing value of parameter 'mu'. mu is the shift in mean of 'x'. Code: # x follows gaussian distribution # fx2 to be minimized by changing values of mu # integration to be done over the interval (qnorm(0.999),Inf) p<-0.009 #constant R<-0.25 # constant e<...

dear all, it will just take you a minute to tell me the difference between qnorm and qqnorm. are they same or is there any difference between them?? regards

Dear R users, I was wondering from where I could get the C source code to compute pnbinom() and qnorm() ? (I would use R in batch mode but I find the startup time prohibitive, unless there is a way to speed it up) I searched the Web and it clearly is part of the R distribution, I just don't know how to extract them. Thanking you ! Markus Loecher Princeton, NJ [[alternative HTML version dele...

...value, so that P(X>c)=0.10. using R >>question 2) Graph the function N(2,3) and with this graph, explain what you do in question number 1. I just found question number one but not the second one. So, I'd like to make a plot form this distribution N (2,3) using the functions plot and qnorm. I couldn't find how make it in R. Thanks for any hint. >> Best regards, *>> Talina>>* *>> * [[alternative HTML version deleted]]

Hello, could You please help: I am looking for a way to formulate test accuracy measures such as test sensitivity, specificity, predictive values, and correct classification rate using p/d/qnorm. The tests' primary values follow a bimodal distribution, which is modelled by a mixture of two normal distributions: p * dnorm ((x - u1) / s1) / s1 + (1 - p) * dnorm ((x - u2) / s2) / s2) {written like in Venables & Ripley, 2nd Ed., p. 288}. So, the first and second summand would c...