Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and released to mix into the population. after they have had an opportunity to mix, a random sample of 10 of these animals is selected. let x = the number of tagged animals in the second sample. assuming there is a total of 25 animals of this type in the region, what are e(x) and var(x)?

Step-by-step explanation:

Previous concepts

The hypergeometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes  

The expected value and variance for this distribution are given by:

What is the distribution of X?  

For this case the random variable X follows a hypergometric distribution.

Compute the values for E(X) and Var(X)  

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

What is the probability that none of the animals in the second sample are tagged?  

So for this case we want this probability:

What is the probability that all of the animals in the second sample are tagged?  

So for this case we want this probability:

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

it has the same slope and a different y-intercept

hope this : )