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S2 help please (poisson distributions)

It is claimed that 95% of people in a certain village are right handed. A random sample of 80 villagers are tested to see if they are right or left handed. Use a Poisson approximation to estimate that the number of people that are right handed is:

a) 80
b) 79
c) at least 78

So I know at first this is a binomial question...n=80 n = 80 p=0.95 p = 0.95

Im going to use z instead of lambda because I Dont know how to make a lambda on the computer :P So X Po(z) X~Po(z)

z = mean = np

so np = z

z = 0.95 x 80 = 76

X Po(76) X~Po(76)

P(X=x)=ez×zxx! P(X=x) = e^{-z} \times \frac{z^x}{x!}

So P(X=80)=e76×768080! P(X=80) = e^{-76} \times \frac{76^{80}}{80!} And I got a math error (because the numbers are too small for the calculator i presume)

I know its probably easier to use a binomial distribution as n is large but p is not small...but the question says use poisson =/

Where have I gone wrong? Or is the book stupid? :P

Thanks all
Reply 1
The only thing I can think of is doing 1 - the probabilities but for left handed so n=0.05 n = 0.05
Reply 2
Ok ive jsut tried doing 1-left handed and 80! is a math error...so could you do leflt handed for P(X=80)=1P(X=0) P(X=80) = 1 - P(X=0) ?
Reply 3
Ok just worked it out...work out P(X=0) for left handed people = P(X=80) right handed

*slap on forehead*

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