Calculate the finite population corrections

(a) For Exercises 3-141 and 3-142, for which exercise should the binomial approximation to the distribution of X be better?

(b) For Exercise 3-141, calculate P(X = 1) and P(X = 4), assuming that X has a binomial distribution, and compare these results to results derived from the hypergeometric distribution.

(c) For Exercise 3-142, calculate P(X = 1) and P(X = 4), assuming that X has a binomial distribution, and compare these results to the results derived from the hypergeometric distribution.

(d) Use the binomial approximation to the hypergeometric distribution to approximate the probabilities in Exercise 3-146. What is the finite population correction in this exercise?

Step 1 of 5:

Here, the values for N, n and K are given.We need to find the finite population correction and also we need to find out the required values.

Step 2 of 5:

(a)

Here for the given two values of N and n, we need to find the finite population correction and also we have to check which of the given distribution has the Binomial distribution.

The finite population correction is given by

For N=100 and n=4,the finite population correction is given by

=

=0.9697

For N=20 and n=4, the finite population correction is given by

=

=0.8421

Here we can observe that the finite population correction for the population with N=100 and n=4 is greater than the population with N=20 and n=4.

So the population with larger finite population correction has the better approximation to the Binomial distribution.

Thus, the finite population correction is 0.9697 for N=100,n=4,0.8421 for N=20,n=4 and the first sample has the better approximation to the Binomial distribution.

Step 3 of 5:

(b)

It is given that X has the Binomial distribution with N=100,n=4,K=20.

We need to find the values of P(X=1) and P(X=4).

Also, we have to assume that X has the Hypergeometric distribution and we have to find the values of P(X=1),P(X=4) to compare it with the values obtained from the Binomial assumption.

We have X~B(n=4,p)

The probability of success p is given by

p=

=

=0.2

So P(X=x) becomes

P(X=x)=

So,P(X=1) becomes

P(X=1)=

=(4)(0.2)(0.512)

=0.4096

P(X=4)=

=(1)(0.0016)(1)

=0.0016

Now using the Hypergeometric distribution P(X=x) is given by

P(X=x)=

Now,P(X=1) becomes

P(X=1)=

=

=

=0.4190

P(X=4)=

=

=0.00124

Thus, using the Binomial approximation P(X=1)=4096 and P(X=4)=0.0016

Using the Hypergeometric approximation,P(X=1)=0.4190 and P(X=4)=0.0012.

Here we can observe that the probabilities obtained using the Binomial approximation agrees well with the probabilities obtained using the Hypergeometric approximation.