> Sounds like homework, which is not an encouraged use of the Rhelp > list. You can either do it in theory...
It is _from_ a homework but I have the solution already (explicitly got that done first!) - this was the pasted Latex code (apologies for that, but in plain text it looks unreadable[1], and I thought everybody here has his / her favorite Latrex editor open all the time anyway...). I'm just looking, for my own advancement and programming training, for a way of doing that in R - which, from your and Dennis' reply, doesn't seem to exist. I would _not_ misuse the list for getting homework done easily, I will not ask "learning statistics" questions here, and I will always try to find the solution myself before posting something here, I promise! Thanks anyway for the simulation advice, Rainer (4000 - (40*n)) -329 [1] --------------- = ---- 1 200 (10*(n^-)) 2 On Saturday 08 January 2011 14:56:20 you wrote: > > On Jan 8, 2011, at 6:56 AM, Rainer Schuermann wrote: > > > This is probably embarrassingly basic, but I have spent quite a few > > hours in Google and RSeek without getting a clue - probably I'm > > asking the wrong questions... > > > > There is this guy who has decided to walk through Australia, a total > > distance of 4000 km. His daily portion (mean) is 40km with an sd of > > 10 km. I want to calculate the number of days it takes to arrive > > with 80, 90, 95, 99% probability. > > I know how to do this manually, eg. for 95% > > $\Phi \left( \frac{4000-40n}{10 \sqrt{n}} \right) \leq 0.05$ > > find the z score... > > > > but how would I do this in R? Not qnorm(), but what is it? > > Sounds like homework, which is not an encouraged use of the Rhelp > list. You can either do it in theory or you can simulate it. Here's a > small step toward a simulation approach. > > > cumsum(rnorm(100, mean=40, sd=10)) > [1] 41.90617 71.09148 120.55569 159.56063 229.73167 > 255.35290 300.74655 > snipped > [92] 3627.25753 3683.24696 3714.11421 3729.41203 3764.54192 > 3809.15159 3881.71016 > [99] 3917.16512 3932.00861 > > cumsum(rnorm(100, mean=40, sd=10)) > [1] 38.59288 53.82815 111.30052 156.58190 188.15454 > 207.90584 240.64078 > snipped > [92] 3776.25476 3821.90626 3876.64512 3921.16797 3958.83472 > 3992.33155 4045.96649 > [99] 4091.66277 4134.45867 > > The first realization did not make it in the expected 100 days so > further efforts should extend the simulation runs to maybe 120 days. > The second realization had him making it on the 98th day. There is an > R replicate() function available once you get a function running that > will return a specific value for an instance. This one might work: > > min(which(cumsum(rnorm(120, mean=40, sd=10)) >= 4000) ) > [1] 97 > > If you wanted a forum that does not explicitly discourage homework and > would be a better place to ask theory and probability questions, there > is CrossValidated: > http://stats.stackexchange.com/faq > > > > > Thanks in advance, > > and apologies for the level of question... > > Rainer > > > > ______________________________________________ > > 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. > > David Winsemius, MD > West Hartford, CT > > ______________________________________________ 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.