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Topic: Combinations with variable probability?
Replies: 10   Last Post: Jan 19, 2009 6:09 AM

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petertwocakes@googlemail.com

Posts: 27
Registered: 1/8/09
Re: Combinations with variable probability?
Posted: Jan 18, 2009 8:10 AM
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On 16 Jan, 23:32, Ray Vickson <RGVick...@shaw.ca> wrote:
> On Jan 16, 11:47 am, petertwocakes <petertwoca...@googlemail.com>
> wrote:
>
>
>

> > On 16 Jan, 19:00, Matt <matt271829-n...@yahoo.co.uk> wrote:
>
> > > On Jan 16, 5:51?pm, petertwocakes <petertwoca...@googlemail.com>
> > > wrote:

>
> > > > A population of N=1000 listen to a particular radio show for one
> > > > month. Each person listens for a different amount of time, t,
> > > > expressed as a proportion of the total time the show occupied over the
> > > > month from, and the times are normally distributed between 0.0 to 1.0.

>
> > > This is not possible. The normal distribution always ranges from -oo
> > > to +oo. You may have in mind some sort of "truncated" normal
> > > distribution, or you may need to come up with a different model for
> > > the listening times.

>
> > > > Each person can tell us exactly how much time, t, ?they listened for.
>
> > > > The show's output comprises a playlist of 100 songs, all played a
> > > > different amount of times, but we know the proportion of air-time, a,
> > > > occupied by each song. "Song A" ?accounted for a = 0.15 total air-
> > > > time.

>
> > > > Given ?a sample of n= 20 people at random, ?can we estimate the
> > > > probability that exactly k of them heard the song?

>
> > > You need more information. For starters, what does "heard" mean? Heard
> > > the whole song? Heard any part of it?

>
> > > > Is this even possible without exhaustively calculating p for each
> > > > possible combination?

>
> > > > If instead we ask a random sample of 20 people if they heard the song,
> > > > and 5 of them have, what is the probability that that particular
> > > > outcome occured, given that we know a =0.15, and the value of t for
> > > > each person.

>
> > > > Although at first it looks like a variation on the hypergeometric
> > > > distribution, I'm guessing that's no use because of the variable
> > > > probabilities?

>
> > > > What sort of things should I be studying to figure this one out?
>
> > Yes, approximately normal, truncated at zero and 1.0, with a mean of
> > 0.5, SD ?0.34

>
> Why use a truncated normal? I would seriously doubt that truncating a
> normal has anything to do with the actual distribution. It is much
> more likely that a Beta distribution is the applicable distribution in
> such a case.
>
> R.G. Vickson
>

> > "Heard" means heard any part of it. There is no unmentioned bias in
> > any of the conditions
> > If there's any other missing information, please infer ideal default
> > values that would make either part solveable.


OK, exploring the Beta distribution now.




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