Normal Distribution Quantiles
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.
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> 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