91Ó°ÊÓ

The catching of fishes can be modeled as a hypergeometric distribution, where we have a population of size \(N=100\), of which \(K=30\) are carp. A sample of size \(n=20\) is chosen, and we want to find the mean and variance of the number of carp among these 20 fish caught.

Mean of a hypergeometric distribution is given by: Mean (\(E[X]\)) = \(n \times \frac{K}{N}\) Where, \(n\) = number of fish caught \(K\) = number of carp in the pond \(N\) = total number of fish in the pond Mean (\(E[X]\)) = \(20 \times \frac{30}{100}\) = \(6\) So, the mean number of carps in the 20 fish caught is 6.

Variance of a hypergeometric distribution is given by: Variance (\(Var[X]\)) = \(n \times \frac{K}{N} \times \frac{(N-K)}{N} \times \frac{(N-n)}{(N-1)}\) Where, \(n\) = number of fish caught \(K\) = number of carp in the pond \(N\) = total number of fish in the pond Variance (\(Var[X]\)) = \(20 \times \frac{30}{100} \times \frac{(100-30)}{100} \times \frac{(100-20)}{(100-1)}\) = \(20 \times \frac{30}{100} \times \frac{70}{100} \times \frac{80}{99}\) Variance (\(Var[X]\)) = \(2.8282\) So, the variance of the number of carps in the 20 fish caught is approximately 2.8282. In conclusion, when catching 20 fish from the pond, the mean number of carps among them is 6, and the variance is approximately 2.8282, under the assumption that each fish has an equal chance of being caught and we catch each fish independently.