In southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditional student teaching programs. A group of eight candidates for three local teaching positions consisted of five who had enrolled in paid internships and three who enrolled in traditional student teaching programs. All eight candidates appear to be equally qualified, so three are randomly selected to fill the open positions. Let Y be the number of internship trained candidates who are hired.
a. Does Y have a binomial or hypergeometric distribution? Why?
b. Find the probability that two or more internship trained candidated are hired.
c. Find the standard deviation and mean of Y.

A) Y has Hypergeometric distribution. This is because, the population is not definite and is largely a finite number of people that have one of the two characteristics. Also, a sample of 8 was selected from this population and a random variable of interest that contained the number of people with one of the characteristics.

B) We are told total candidates were 8.

Thus; N = 8

3 were for local teaching positions, thus; n = 3

5 were for paid internship, thus; r = 5

Now, the formula for Hypergeometric distribution is;

P(y) = {(rCy) × [(N-r)C(n-y)]}/(NCn)

Where C represents combination.

Thus, probability that two or more candidates were hired is;

P( X ≥ 2) = P(2) + P(3)

Now, N - r = 8 - 5 = 3

For P(2); n - y = 3 - 2 = 1

For P(3); n - y = 3 - 3 = 0

Thus;

P(2) = {(5C2) × (3C1}/(8C3)

P(2) = 30/56

P(2) = 0.5357

P(3) = {(5C3) × (3C0}/(8C3)

P(3) = 10/56

P(3) = 0.1786

Thus;

P( X ≥ 2) = 0.5357 + 0.1786

P( X ≥ 2) = 0.7143

C) Formula for standard deviation in Hypergeometric distribution is;

σ = √[(nr/N) × ((N - r)/n) × ((N - n)/(N - 1))]

nr/N = (3 × 5/8) = 15/8

((N - r)/N) = (8 - 5)/8 = 3/8

((N - n)/(N - 1)) = (8 - 3)/(8 - 1) = 5/7

σ = √((15/8) × (3/8) × (5/7))

σ =√0.5022

σ = 0.7087

Formula for mean in Hypergeometric distribution is;

E(Y) = nr/N = 3 × 5/8 = 1.875