Probability Question
- From: Math1723 <anonym1723@xxxxxxx>
- Date: Fri, 27 Jun 2008 13:05:54 -0700 (PDT)
Suppose you have a (possibly biased) coin which has a probability p of
coming up heads and 1-p coming up tails. Further suppose that Player
A flips that coin n times, yielding x heads and n-x tails. How many
standard deviations away from the norm would A's result be?
I can compute that the probability of A's result P(A) is C(n,x)(p^x)(1-
p)^(n-x) and that the population standard deviation sigma is sqrt(np(1-
p)), and that these two equations will be important for the result.
However, it's not obvious where I go from here. I am essentially
looking for an equation of the form SD(p,n,x).
In a simple case where p=0.5 and n=16, P(A,x)=C(16,x)/2^16 and sigma =
2. Since x=8 is the expected value, SD(0.5,16,8) should be 0; if x=0
or x=16, I would expect SD(0.5,16,x) to be somewhere around 3 range,
since this is what I would guess how many sigma's away such a result
might be. Yet I do not see any obvious (or justifiable) way to yield
this result.
Any suggestions?
.
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