Question

In southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditional student teaching programs. A group of ten candidates for four local teaching positions consisted of six who had enrolled in paid internships and four who enrolled in traditional student teaching programs. All ten candidates appear to be equally qualified, so four are randomly selected to fill the open positions. Let Y be the number of internship trained candidates who are hired.

1. Does Y have a binomial or hypergeometric distribution? Why?
A. A binomial distribution, as the probability of being chosen on a trial is either a success or a failure.
B. A binomial distribution, as the probability of being chosen on a trial is dependent of the outcomes of previous trials.
C. A hypergeometric distribution, as the probability of being chosen on a trial is dependent on the outcome of previous trials.
D. A hypergeometric distribution, as the probability of being chosen on a trial is independent of the outcomes of previous trials.
2. Find the probability that two or more internship trained candidates are hired.
3. What is the mean and standard deviation of Y?

Answer 1

Follows are the solution to this question:

Explanation:

In question 1:

The answer is "option b".

In question 2:

`N= 11 \\\\r=7\\\\n=4\\\\p(y)=\frac{\binom{r}{y} \binom{N-r}{n-y}}{\binom{N}{n}}`

similarly

`p(3)+p(4) = \frac{\binom{7}{3} \binom{11-7}{4-3}}{\binom{11}{4}} +\frac{\binom{7}{4} \binom{11-7}{4-4}}{\binom{11}{4}}`

`=0.4232 +0.1061\\\\=0.5303`

In question 3:

`\mu =n\\\\(r)/(N) = 4 * (7)/(11)= (28)/(11)=2.5454 \\\\\sigma= n \cdot (r)/(N) (N-r)/(N) (N-n)/(N-1)\\\\`

`= 4 * (7)/(11) * (4)/(11) * (7)/(10)\\\\=0.6479`