Question

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)?

Answer 1

`E(X)= n(M)/(N)=10 (5)/(25)=2`

`Var(X)=n (M)/(N)(N-M)/(N)(N-n)/(N-1)=10(5)/(25)(25-5)/(25)(25-10)/(25-1)=1`

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:

`P(X=k)= ((MCk)(N-M C n-k))/(NCn)`

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:

`E(X)= n(M)/(N)`

`Var(X)=n (M)/(N)(N-M)/(N)(N-n)/(N-1)`

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:

`E(X)= n(M)/(N)=10 (5)/(25)=2`

`Var(X)=n (M)/(N)(N-M)/(N)(N-n)/(N-1)=10(5)/(25)(25-5)/(25)(25-10)/(25-1)=1`

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

So for this case we want this probability:

`P(X=0)= ((5C0)(25-5 C 10-0))/(25C10)=(1*184756)/(3268760)=0.0565`

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

So for this case we want this probability:

`P(X=5)= ((5C5)(25-5 C 10-5))/(25C10)=(1*15504)/(3268760)=0.00474`