Poisson Approximation to Binomial

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Using a binomial distribution with n=10 and p=0.5 does not meet the criteria for a Poisson approximation, as n is not large enough and np exceeds 10. Despite this, calculations show that the Poisson approximation yields a probability of 3.4%, while the binomial formula results in 1%. The range calculated using the mean and standard deviation is between 1.8% and 8.4%. The confusion arises from the expectation that the binomial result should fall within this range, highlighting the limitations of using Poisson in this scenario. Overall, the approximation is not reliable due to the failure to meet the necessary conditions.
Victor Frankenstein
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For a binomial distribution with n=10 and p=0.5 ,we should not use the poisson approximation because both of the conditions n>=100 and np<=10 are not satisfied. SUppose we go way out on a limb and use the Poisson aproximation anyway. Are the resulting probabilities unacceptable approximations? Why or why not ?
 
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Have you worked out the actual value and the approximation?
 
n=10, p=1/2, x=1, mean=n*p= 5

using the poisson formula i got 3.4%
using binomial formula I got 1%
using (mean)+/- 2(standard deviation) i got 8.4 - 1.8

I don't know If its right, confused because should'nt the binomial furmula give the answer within the maximum and minumum usual values ?
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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