Answer:
E(X) = 6
Var(X) = 3.394
Step-by-step explanation:
Let X represent the number of carp caught out of the 20 fishes caught. Now, if we are to assume that each
of the (100, 20) ways to catch the 20 fishes will be equally likely.
Thus, we can say that X fulfills a hypergeometric
distribution with parameters as follows;
n = 20, N = 100, k = 30
Formula for expected mean value in hypergeometric distribution is;
E(X) = nk/N
E(X) = (20 × 30)/100
E(X) = 6
Formula for variance is;
Var(X) = (nk/N) × [((n - 1)(k - 1)/(N-1))) + (1 - nk/N)]
Var(X) = ((20 × 30)/100) × [((20 - 1)(30 - 1)/(100 - 1)) + (1 - (20 × 30/100)]
Var(X) = 6 × 0.5657
Var(X) = 3.394
