We claimed that the binomial distribution is a good model for the hyper geometri

Question

We claimed that the binomial distribution is a good model for the hyper geometric distribution, in the right circumstances. Suppose samples of size 50 are being tested. Consider units defective with probability p = 0.06, so that if the size of the population is N, M = Np units are defective Use R to find the minimum population size N for which no corresponding values in the binomial and hypergeometric PMFs differ by more than 0.01 (In other words, find N for which |P(X_binom = x) - P(X_hyper = x)| lessthanorequalto 0.01. for all values of x in the sample space.) Does your answer for #5

Explanation / Answer

5) Suppose the population size be N. A sample of size n=50 is taken from the population. The probability of defective, p=0.06.

Now pmf of Binomial distribution, f_binom=choose(n,x)*p^x*(1-p)^(n-x); x=0,1,2,...,n.

pmf of Hypergeometric distribution, f_hyper=choose(N*p,x)*(N(1-p),(n-x))/choose(N,n); x=0,1,2,...,n.

We need to find N such that |f_binom(x)-f_hyper(x)| is less than or equal to 0.01 for all x=0,1,2,...,n=50.

The answer is N=613.

The R-code is given below:

p=0.06; n=50
fbinom<-function(x){
return(choose(n,x)*p^x*(1-p)^(n-x))}

fhyper<-function(x,N){
return(choose(N*p,x)*choose(N*(1-p),(n-x))/choose(N,n))}

error<-function(N){
diff<-NULL
for(x in 0:n){
diff[x]=abs(fbinom(x)-fhyper(x,N))
}
return(max(diff))
}

N=51
while(error(N)>0.01){
N=N+1}
N

6) The answer for above question changes if the probability of defective "p" is changed.

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