Respuesta :
Answer:
a) For this case the random variable X follows a hypergometric distribution.
b) [tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]
c) [tex]P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565[/tex]
d) [tex]P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474[/tex]
Step-by-step explanation:
The hypergometric 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:
[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]
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:
[tex]E(X)= n\frac{M}{N}[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}[/tex]
a. What is the distribution of X?
For this case the random variable X follows a hypergometric distribution.
b. 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:
[tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]
c. What is the probability that none of the animals in the second sample are tagged?
So for this case we want this probability:
[tex]P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565[/tex]
d. What is the probability that all of the animals in the second sample are tagged?
So for this case we want this probability:
[tex]P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474[/tex]
