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Re: Geiger, not binomial ?



At 11:09 3/23/00 -0600, Glenn Carlson wrote:
There is every reason to expect your data to fit a binomial
distribution since it is the correct distribution. However, as Mr.
Cleyet correctly points out, the Poisson distribution is more useful
here because of the huge number of nuclei in your sample (on the order
of 10^23) and the virtually zero probability that any one nucleus will
decay during the counting interval.

//// As you can see
in the table below, your data with a 0.5 second counting interval fits
a Poisson distribution very well. I leave it to you and your students
to analyze your data with a Gaussian distribution.

#counts #times c*t c! P(k;mu(c)) Expected #times = %diff =
(c) (t) (Poisson) P(k;mu(c))*sum(c*t) |Exp-meas|/Exp*100

0 3321 0 1 0.109586319 3294.712675
/////////
sum(t) = 30065
sum(c*t) = 66475

mean(c) = mu(c) = 2.211042741

A final note. Remember to keep in mind that your measured count rate
is not the same thing as the decay rate of your sample.....
Glenn A. Carlson, P.E.


Hmmm...it seems we can all learn a little something from this
puzzle of Ludwik's.
For example, in Glenn's working above, it would be better to
say that 2.211 is his *estimate* of the mean.

Using my (incomparable?) least squares method, I see that it is more
likely to be 2.212 +- 0.005 in round numbers - speaking of which,
it is less than pukka to be quoting 10 places of precision anywhere
in this table: this is statistics, not quantum chromodynamics! :-)




brian whatcott <inet@intellisys.net>
Altus OK