Question 867170: Can any expert with math help me with this problem?
Count the probbaility of a bionominally distrubited random variable X to get the value of k = 0,1,2,3 for the occassions (n=10,p=0,1), (n=20,p=0,05), (n=100,p=0,01).
Count approximately the same probability, using the approach of Poisson for the bioniminal distribution.
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
NO problem: .01*100. it is.
Re: TY. .01*100. = 1,
Always a good idea though to check the Arithmetic on Probabilities
Consider myself an Expert, only because I am a Doer.
Have set Up a table and have DONE the Poisson Part of the Comparison:
k avg Prob avg Prob avg Prob
0 1 0.3679 1 0.3679 1 0.3679
1 1 0.3679 1 0.3679 1 0.3679
2 1 0.1839 1 0.1839 1 0.1839
3 1 0.0613 1 0.0613 1 0.0613
k Binom Prob Binom Prob Binom Prob
0 n=10,p=.1 .3487 n=20,p=.05 .3585 n=100,p=.001
1 n=10,p=.1 .3874 n=20,p=.05 .3774 n=100,p=.001
2 n=10,p=.1 .1937 n=20,p=.05 .1887 n=100,p=.001
3 n=10,p=.1 .0574 n=20,p=.05 .0596 n=100,p=.001
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