Respuesta :
Answer:
a) For this case the random variable X follows a hypergometric distribution.
With parameters N=15, M=6 n =5
Where N=15 is the population size, M=6 is the number of success states in the population, n=5 is the number of draws, k is the number of observed successes
[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]
b) [tex] P(X=2) =\frac{(6C2)(15-6 C 5-2)}{15C5}= \frac{6C2 *(9C3)}{15C5}= 0.4196[/tex]
[tex] P(X \leq 2) = P(X=0) +P(X=1)+P(X=2)[/tex]
We can find the individual probabilities:
[tex] P(X=0) =\frac{(6C0)(15-6 C 5-0)}{15C5}= \frac{6C0 *(9C5)}{15C5}=0.0420[/tex]
[tex] P(X=1) =\frac{(6C1)(15-6 C 5-1)}{15C5}= \frac{6C1 *(9C4)}{15C5}=0.2517[/tex]
And we got:
[tex] P(X \leq 2) = 0.0420 +0.2517+0.4196= 0.7133[/tex]
[tex] P(X\geq 2) = 1-P(X<2) = 1-P(X\leq 1) = 1-[P(X=0)+ P(X=1)]= 1-[0.0420 +0.2517]=0.7063[/tex]
c) [tex]E(X)= n\frac{M}{N}= 5 *\frac{6}{15}=2[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}= 5* \frac{6}{15}* \frac{15-6}{15}*\frac{15-5}{15-1}= 5*\frac{6}{15}*\frac{9}{15}*\frac{10}{14}= 0.857[/tex]
And the deviation would be the square root of the variance:
[tex] Sd(X) = \sqrt{0.857}= 0.926[/tex]
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:
[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]
Where N=15 is the population size, M=6 is the number of success states in the population, n=5 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 kind of a distribution does X have? (name and values of parameters)
For this case the random variable X follows a hypergometric distribution.
With parameters N=15, M=6 n =5
b) Computer P(X=2), P( X<=2), P(X>=2)
Using the pmf we have this:
[tex] P(X=2) =\frac{(6C2)(15-6 C 5-2)}{15C5}= \frac{6C2 *(9C3)}{15C5}= 0.4196[/tex]
For [tex] P(X \leq 2) = P(X=0) +P(X=1)+P(X=2)[/tex]
We can find the individual probabilities:
[tex] P(X=0) =\frac{(6C0)(15-6 C 5-0)}{15C5}= \frac{6C0 *(9C5)}{15C5}=0.0420[/tex]
[tex] P(X=1) =\frac{(6C1)(15-6 C 5-1)}{15C5}= \frac{6C1 *(9C4)}{15C5}=0.2517[/tex]
And we got:
[tex] P(X \leq 2) = 0.0420 +0.2517+0.4196= 0.7133[/tex]
For the last case [tex] P(X \geq 2)[/tex] we can do this using the complement rule
[tex] P(X\geq 2) = 1-P(X<2) = 1-P(X\leq 1) = 1-[P(X=0)+ P(X=1)]= 1-[0.0420 +0.2517]=0.7063[/tex]
Part c
The expected value and variance for this distribution are given by:
[tex]E(X)= n\frac{M}{N}= 5 *\frac{6}{15}=2[/tex]
[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}= 5* \frac{6}{15}* \frac{15-6}{15}*\frac{15-5}{15-1}= 5*\frac{6}{15}*\frac{9}{15}*\frac{10}{14}= 0.857[/tex]
And the deviation would be the square root of the variance:
[tex] Sd(X) = \sqrt{0.857}= 0.926[/tex]
