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
b) ![P(X=2) =\frac{(6C2)(15-6 C 5-2)}{15C5}= \frac{6C2 *(9C3)}{15C5}= 0.4196](https://tex.z-dn.net/?f=%20P%28X%3D2%29%20%3D%5Cfrac%7B%286C2%29%2815-6%20C%205-2%29%7D%7B15C5%7D%3D%20%5Cfrac%7B6C2%20%2A%289C3%29%7D%7B15C5%7D%3D%200.4196)
![P(X \leq 2) = P(X=0) +P(X=1)+P(X=2)](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%202%29%20%3D%20P%28X%3D0%29%20%2BP%28X%3D1%29%2BP%28X%3D2%29)
We can find the individual probabilities:
![P(X=0) =\frac{(6C0)(15-6 C 5-0)}{15C5}= \frac{6C0 *(9C5)}{15C5}=0.0420](https://tex.z-dn.net/?f=%20P%28X%3D0%29%20%3D%5Cfrac%7B%286C0%29%2815-6%20C%205-0%29%7D%7B15C5%7D%3D%20%5Cfrac%7B6C0%20%2A%289C5%29%7D%7B15C5%7D%3D0.0420)
![P(X=1) =\frac{(6C1)(15-6 C 5-1)}{15C5}= \frac{6C1 *(9C4)}{15C5}=0.2517](https://tex.z-dn.net/?f=%20P%28X%3D1%29%20%3D%5Cfrac%7B%286C1%29%2815-6%20C%205-1%29%7D%7B15C5%7D%3D%20%5Cfrac%7B6C1%20%2A%289C4%29%7D%7B15C5%7D%3D0.2517)
And we got:
![P(X \leq 2) = 0.0420 +0.2517+0.4196= 0.7133](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%202%29%20%3D%200.0420%20%2B0.2517%2B0.4196%3D%200.7133)
![P(X\geq 2) = 1-P(X](https://tex.z-dn.net/?f=%20P%28X%5Cgeq%202%29%20%3D%201-P%28X%3C2%29%20%3D%201-P%28X%5Cleq%201%29%20%3D%201-%5BP%28X%3D0%29%2B%20P%28X%3D1%29%5D%3D%201-%5B0.0420%20%2B0.2517%5D%3D0.7063)
c)
And the deviation would be the square root of the variance:
![Sd(X) = \sqrt{0.857}= 0.926](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%20%5Csqrt%7B0.857%7D%3D%200.926)
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=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:
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:
![P(X=2) =\frac{(6C2)(15-6 C 5-2)}{15C5}= \frac{6C2 *(9C3)}{15C5}= 0.4196](https://tex.z-dn.net/?f=%20P%28X%3D2%29%20%3D%5Cfrac%7B%286C2%29%2815-6%20C%205-2%29%7D%7B15C5%7D%3D%20%5Cfrac%7B6C2%20%2A%289C3%29%7D%7B15C5%7D%3D%200.4196)
For ![P(X \leq 2) = P(X=0) +P(X=1)+P(X=2)](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%202%29%20%3D%20P%28X%3D0%29%20%2BP%28X%3D1%29%2BP%28X%3D2%29)
We can find the individual probabilities:
![P(X=0) =\frac{(6C0)(15-6 C 5-0)}{15C5}= \frac{6C0 *(9C5)}{15C5}=0.0420](https://tex.z-dn.net/?f=%20P%28X%3D0%29%20%3D%5Cfrac%7B%286C0%29%2815-6%20C%205-0%29%7D%7B15C5%7D%3D%20%5Cfrac%7B6C0%20%2A%289C5%29%7D%7B15C5%7D%3D0.0420)
![P(X=1) =\frac{(6C1)(15-6 C 5-1)}{15C5}= \frac{6C1 *(9C4)}{15C5}=0.2517](https://tex.z-dn.net/?f=%20P%28X%3D1%29%20%3D%5Cfrac%7B%286C1%29%2815-6%20C%205-1%29%7D%7B15C5%7D%3D%20%5Cfrac%7B6C1%20%2A%289C4%29%7D%7B15C5%7D%3D0.2517)
And we got:
![P(X \leq 2) = 0.0420 +0.2517+0.4196= 0.7133](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%202%29%20%3D%200.0420%20%2B0.2517%2B0.4196%3D%200.7133)
For the last case
we can do this using the complement rule
![P(X\geq 2) = 1-P(X](https://tex.z-dn.net/?f=%20P%28X%5Cgeq%202%29%20%3D%201-P%28X%3C2%29%20%3D%201-P%28X%5Cleq%201%29%20%3D%201-%5BP%28X%3D0%29%2B%20P%28X%3D1%29%5D%3D%201-%5B0.0420%20%2B0.2517%5D%3D0.7063)
Part c
The expected value and variance for this distribution are given by:
And the deviation would be the square root of the variance:
![Sd(X) = \sqrt{0.857}= 0.926](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%20%5Csqrt%7B0.857%7D%3D%200.926)