On Jan 8, 2011, at 9:25 AM, Rainer Schuermann wrote:
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!
It would seem to be a straightforward application of the central limit
theorem. Sum of normally distributed IID variables with common
mean ... that sort of thing. When I simulated it (n=10,000), I got:
50% 80% 90% 95% 99%
100 103 104 105 106
Thanks anyway for the simulation advice,
Rainer
(4000 - (40*n)) -329
[1] --------------- = ----
1 200
(10*(n^-))
2
I'm not sure how that expression relates to the problem at hand. But
even at my advanced age, I'm open to education.
--
David.
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
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.
