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?
Thanks in advance,
and apologies for the level of question...
Rainer
Normal Distribution Quantiles
9 messages · Rainer Schuermann, David Winsemius, Charles C. Berry +3 more
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
David. > > Thanks in advance, > and apologies for the level of question... > Rainer > > ______________________________________________ > R-help at 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
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 at 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
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
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
______________________________________________ R-help at 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 at 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.
Charles C. Berry Dept of Family/Preventive Medicine cberry at tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
On Sat, Jan 8, 2011 at 6:25 AM, Rainer Schuermann
<Rainer.Schuermann at gmx.net> wrote:
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.
We do ;-) Is this what you are after? It uses uniroot to find a
value of 'x' (days) such that the difference between the given and
obtained ps is 0 (over a sensible interval).
foo <- function(x, p, mean, sd) {
mu <- mean * x
sigma <- sqrt((sd^2) * x)
p - pnorm(4000, mu, sigma, lower.tail = FALSE)
}
uniroot(foo, interval = c(90, 110), p = .8, mean = 40, sd = 10)
Cheers,
Josh
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 at 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 at 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.
Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/
If I understand what you have said below, it looks like you do NOT have the problem solved manually. You CAN use qnorm , and when you do so, your equation yields a simple quadratic which, of course, has an exact solution that you can calculate in R. Of course, one can use uniroot or whatever to solve the quadratic; or simulation or interpolation using pnorm. But other than the R practice, these are unnecessary and, in this case, a bit silly. Cheers, Bert On Sat, Jan 8, 2011 at 6:25 AM, Rainer Schuermann
<Rainer.Schuermann at gmx.net> 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! 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 at 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 at 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.
Bert Gunter Genentech Nonclinical Biostatistics
1 day later
Just to add to the silly solutions, here's how I would have done it... mu <- 40 sdev <- 10 days <- 100:120 # range to explore p <- 0.8 days[ match(TRUE, qnorm(0.2, mu*days, sqrt(sdev * sdev * days)) >= 4000) ] Michael
On 9 January 2011 08:48, Bert Gunter <gunter.berton at gene.com> wrote:
If I understand what you have said below, it looks like you do NOT have the problem solved manually. You CAN use qnorm , and when you do so, your equation yields a simple quadratic which, of course, has an exact solution that you can calculate in R. Of course, one can use uniroot or whatever to solve the quadratic; or simulation or interpolation using pnorm. But other than the R practice, these are unnecessary and, in this case, a bit silly. Cheers, Bert On Sat, Jan 8, 2011 at 6:25 AM, Rainer Schuermann <Rainer.Schuermann at gmx.net> 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! 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 at 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 at 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.
-- Bert Gunter Genentech Nonclinical Biostatistics
______________________________________________ R-help at 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.
Altogether I got five more or less silly solutions (not my judgment!), some of them further discussed in private mail, for a problem where my expectation was to get a simple one-liner back: "Check ?clt" or so... Fortunately, with all of them I seem to arrive at a result that is consistent with what my expressions evaluates to (104.25) which gives me a great opportunity to play around with the various approaches - "brain fodder" for quite a few days. Great experience, Thanks to all, Rainer