On 20/10/2008, at 9:51 AM, Boks, M.P.M. wrote:
Dear Experts,
Probably trivial, but I am struggling to get what I want:
I need to know how the number of required trials to get a certain
number of successes.
By example:
How many trials do I need to have 98% probability of 50 successes,
when the a priory probability is 0.1 per trial.
The Negative binomial function may do the job (not sure):
NegBinomial {stats}
The Negative Binomial Distribution
Description
Density, distribution function, quantile function and random
generation for the negative binomial distribution with parameters
size and prob.
Usage
dnbinom(x, size, prob, mu, log = FALSE)
pnbinom(q, size, prob, mu, lower.tail = TRUE, log.p = FALSE)
qnbinom(p, size, prob, mu, lower.tail = TRUE, log.p = FALSE)
rnbinom(n, size, prob, mu)
I tried finding out how to do this by using examples, but I am at a
loss. Any help would be much appreciated!
As far as I can see (which is often not very far) the negative
binomial distribution has nothing
to do with it.
You want Pr(X >= 50) = 0.98 where X is binomially distributed with n
= ?, p = 0.1.
Equivalently Pr(X <= 49) = 0.02.
After some trial-and-error I found:
> pbinom(49,645:655,0.1,lower=FALSE)
[1] 0.9786144 0.9792460 0.9798610 0.9804599 0.9810430 0.9816106
0.9821632
[8] 0.9827009 0.9832242 0.9837334 0.9842288
I.e. the *smallest* n that makes Pr(X>=50) >= 0.98 is n = 648.
Note that you have to be careful with the ``at leasts'' here; it's easy
to make parity errors in respect of looking at upper and lower tails
when
dealing the cumulative distribution of a discrete random variable.
cheers,
Rolf Turner
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