Answer to Question #316318 in Statistics and Probability for vas

Question

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 candidates are hired

(c) What are the mean and standard deviation of Y ?.

Answer 1

$a:\\hypergeometric\,\,because\,\,the\,\,candidates\,\,are\,\,chosen\,\,without\,\,replacement\\parameters\,\,N=8,K=5,n=3\\b:\\P\left( Y\geqslant 2 \right) =P\left( Y=2 \right) +P\left( Y=3 \right) =\frac{C_{5}^{2}C_{3}^{1}}{C_{8}^{3}}+\frac{C_{5}^{3}C_{3}^{0}}{C_{8}^{3}}=0.714286\\c:\\EY=n\frac{K}{N}=3\cdot \frac{5}{8}=1.875\\ \sigma Y=\sqrt{\frac{nK\left( N-K \right) \left( N-n \right)}{N^2\left( N-1 \right)}}=\sqrt{\frac{3\cdot 5\left( 8-5 \right) \left( 8-3 \right)}{8^2\left( 8-1 \right)}}=0.708683$