
Re: Combinations with variable probability?
Posted:
Jan 18, 2009 8:10 AM


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 <matt271829n...@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 airtime, 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.

