# [R] Normal Distribution Quantiles

Rainer Schuermann Rainer.Schuermann at gmx.net
Sat Jan 8 15:25:06 CET 2011

> 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
>
> >
> > and apologies for the level of question...
> > Rainer
> >
> > ______________________________________________
> > R-help at r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help