With replacement:
Here in the with replacement case the
, where
.
Without replacement:
In the without replacement case
Now we have to calculate the probability
for each distribution and study if binomial
distribution approximates the hypergeometric distribution for in
creasing values of
.
R-codes:
m=2:5
N=10^m
p=0.25
n=0.1*N
a=p*N
b=N-a
x=0.02*N
p1=round(phyper(x,a,b,n),digits = 8)
p2=round(pbinom(x,n,p),digits = 8)
P=matrix(c(n,p1,p2),ncol = 3,byrow = F)
colnames(P)=(c("n","Hypergeometric","Binomial"))
rownames(P)=sprintf("P(X<=%s)",x)
P
Output:
n Hypergeometric Binomial P(X<=2) 10 0.52166241 0.52559280 P(X<=20) 100 0.13579398 0.14883105 P(X<=200) 1000 0.00005045 0.00010898 P(X<=2000) 10000 0.00000000 0.00000000
Conclusion:
From the above table we can see the probabilities from the
binomial distribution is getting closer and closer to the values of
probability from the hypergeometric distribution as the
increases. So here we can say the binomial distribution
approximates the hypergeometric distribution very well as for
sufficiently large
.
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