On Sat, 8 Jan 2011, 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.
Your question was 'how do I find the smallest integer $n$ such that...',
right?
Using uniroot and pnorm, you could solve for real $n$ and then round up.
Doing this, I find that in greater than 95% of trials, your bushwalker
would be done in 105 days or less.
Or you could use findInterval, sapply, and pnorm to get all of the $n$s in
one expression.
HTH,
Chuck
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
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David Winsemius, MD
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R-help@r-project.org mailing list
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Charles C. Berry Dept of Family/Preventive Medicine
cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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