Yes. I will help you with anything
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
Go to Math Papa that'll give you the answers with the work shown and explanations.
I am assuming 28/0. The short answer is that it is infinity. (Or they may want undefined)
Here is a bit longer explination:
Let's start by taking 28/1, that gives 28. What about 28/0.5 that gives 56, and as we keep decreasing the denominator closer to zero then the quotient will become larger and larger. We we reach zero, the quotient becomes so large that it is considered infinity.
