Question

Four rats are selected at random from a cage of 5 male (M) and 6 female (F) rats. If the random variable Y is concerned with the number of female rats taken out of the cage, construct the probability distribution and answer the questions below. (input up to 4 decimal places)A What is the probability that all selected rats are female?
What is the probability that only one of the selected rats is female?
c. What is the probability that at least two of the selected rats is female?
What is the mean of the probability distribution?
E. What is the standard deviation of the probability distribution?​

Answer 1

Using the hypergeometric distribution, it is found that the distribution is:

P(X = 0) = 0.0152

P(X = 1) = 0.1818

P(X = 2) = 0.4545

P(X = 3) = 0.3030

P(X = 4) = 0.0455

Hence:

A) There is a 0.0455 = 4.55% probability that all selected rats are female.

B) There is a 0.1818 = 18.18% probability that only one of the selected rats is female.

C) There is a 0.803 = 80.3% probability that at least two of the selected rats is female.

D) The mean of the probability distribution is of 2.18.

E) The standard deviation of the probability distribution is of 0.833.

The rats are chosen without replacement, hence the hypergeometric distribution is used to solve this question.

What is the hypergeometric distribution formula?

The formula is:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• N is the size of the population.

• n is the size of the sample.

• k is the total number of desired outcomes.

In this problem:

• There is a total of 11 rats, hence N = 11.

• 6 of the rats are female, hence k = 6.

• 4 rats are taken, hence n = 4.

The distribution is the probability of each outcome, hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,11,4,6) = \frac{C_{6,0}C_{5,4}}{C_{11,4}} = 0.0152$

$P(X = 1) = h(1,11,4,6) = \frac{C_{6,1}C_{5,3}}{C_{11,4}} = 0.1818$

$P(X = 2) = h(2,11,4,6) = \frac{C_{6,2}C_{5,2}}{C_{11,4}} = 0.4545$

$P(X = 3) = h(3,11,4,6) = \frac{C_{6,3}C_{5,1}}{C_{11,4}} = 0.3030$

$P(X = 4) = h(4,11,4,6) = \frac{C_{6,4}C_{5,0}}{C_{11,4}} = 0.0455$

Item a:

P(X = 4) = 0.0455, hence:

There is a 0.0455 = 4.55% probability that all selected rats are female.

Item b:

P(X = 1) = 0.1818, hence:

There is a 0.1818 = 18.18% probability that only one of the selected rats is female.

Item c:

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.4545 + 0.3030 + 0.0455 = 0.803$

There is a 0.803 = 80.3% probability that at least two of the selected rats is female.

Item d:

The mean of the hypergeometric distribution is:

$\mu = \frac{nk}{N}$

Hence:

$\mu = \frac{4(6)}{11} = 2.18$

The mean of the probability distribution is of 2.18.

Item e:

The standard deviation of the hypergeometric distribution is:

$\sigma = \sqrt{\frac{nk(N-k)(N-n)}{N^2(N-1)}}$

Hence:

$\sigma = \sqrt{\frac{4(6)(11-6)(11-4)}{11^2(11-1)}} = 0.833$

The standard deviation of the probability distribution is of 0.833.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394