Question

A lake contains 4 distinct type of fish. Suppose that each fish caught is equally likely
to be any one of these types. Let Y denote the number of fish that need be caught to
obtain at least one of each type.


(a) Give an interval (a, b) such that P(a ≤ Y ≤ b) ≥ 0.90.
(b) Using the one-sided Chebyshev inequality, how many fish need we plan on catching
so as to be at least 90% certain of obtaining at least one of each type?

Answer 1

Step-by-step explanation:

So first for part of it is usually the property of why that in between A and B is nine Is .9. So we use the normal table. You can find at the Z score at the 90 is 1.645. So that means A Is -1.645 and B is 1.645. He is the lower limit and B is the Hyah limited. So that's part A. And for part B We have that poverty debt one. Mhm. The parameter X. He is greater than Parameter X one is 29. So proud of that Z score is greater than Z one is also .9. So You have that .9.5 poverty of Z. We could be zero and Z one is .9. So at this level we have that from the normal table, Z one will be 1.28. That is the number of each that we expected. So he's already answered