BONUS QUESTION When looking at the Poisson distribution we s

BONUS QUESTION: When looking at the Poisson distribution, we said that if we have a binomial experiment with a large n and a small p, the binomial pmf can be approximated by a Poisson pmf with mu=np. Let\'s verify this proposition: Suppose that 2% of all copies of a particular book are bound incorrectly. Let X be the number of copies that are not bound correctly in a sample of 100. Use both the binomial and Poisson distributions to calculate P(X

Solution

Binomial distribution with n=100 and p=0.02

P(X=x)=100Cx*(0.02^x)*(0.98^(100-x)) for x=0,1,2,...,100

So P(X<=3) = P(X=0)+P(X=1)+P(X=2)+P(X=3)

=100C0*(0.02^0)*(0.98^(100-0))+...+100C3*(0.02^3)*(0.98^(100-3))

=0.8589616

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Poisson distribution with mean=n*p=100*0.02 =2

P(X=x)=(2^x)*exp(-2)/x!

So P(X<=3)= P(X=0)+P(X=1)+P(X=2)+P(X=3)

=(2^0)*exp(-2)/1+....+(2^3)*exp(-2)/3!

=0.8571235

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Yes, because both probability are closed.