Question about geom dist
- From: nomail1983@xxxxxxxxxxx
- Date: 3 Jan 2006 13:44:24 -0800
I took an intro stat class recently. The text, written for the class
by the teacher, explains the geometric distribution, but we did not
cover it in the class.
According to the text, if we want to know "the number of
components that we would expect to test before finding a
defective one", the answer is E(X) = 1/p. So if the probability of
a defective component is 1/50, E(X) = 50.
But we also know that the probability that the n-th test
component is defective, P(X = n), is p*(1-p)^(n-1). Thus,
P(X <= n) is SUM(P(X = k), k=1,...,n).
So, wouldn't" the expected number of components tested before
finding a defective one" be the smallest n such that P(X <= n) >=
50%?
In the case of p=1/50, n=35 because P(X <= 34) = 49.7% and
P(X <= 35) = 50.7%. So, isn't 35, not 50, "the number of
components that we would expect to test before finding a
defective one"?
I presume (guess!) that E(X)=1/p is determined by
LIM(SUM(k*P(X = k)), k=1,...,inf). It is true that for 1/50, that
limit is 50. So, based on formulation, isn't that number "the
average number of components in each of many batches that
we would expect to test before finding a defective component
in each batch"?
On the other (third?) hand, I have to admit that I have trouble
reconciling the different numbers for the two different
statements myself. Can someone explain?
.
- Prev by Date: Re: How to generate random X given only min, max, mean?
- Next by Date: Re: the Kruskal-Wallis test?
- Previous by thread: Re: How to generate random X given only min, max, mean?
- Next by thread: Re: the Kruskal-Wallis test?
- Index(es):
Relevant Pages
|
