Mathematics, 20.11.2019 02:31 rogersdeloris1ovgm3b

Suppose that 30% of the applicants for a certain industrial job possess advanced training in computer programming. applicants are interviewed sequentially and are selected at random from the pool. (a) (4 marks) find the probability that the first applicant with advanced training in programming is found on the fifth interview. (b) (2 marks) what is the expected number of applicants who need to be interviewed in order to find the first one with advanced training? (c) (6 marks) let y denote the number of the trial on which the first applicant with computer training was found. if each interview costs $30, find the expected value and variance of the total cost incurred interviewing candidates until an applicant with advanced computer training is found.

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a) There is a 7.20% probability that the first applicant with advanced training in programming is found on the fifth interview.

b) The expected number of interviews to find the first one with advanced training is 3.33.

c) The expected cost is $99.9.

The variance of the cost is $22.053.

Step-by-step explanation:

The negative binomial distribution is the number X of repeated trials to produce r successes with p probability in a binomial experiment.

The probability that it takes n trials for x sucesses is given by:

$P(X = x) = C_{n-1,x-1}*p^{x}*(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

In this problem, we have that:

Suppose that 30% of the applicants for a certain industrial job possess advanced training in computer programming. This means that p = 0.30 .

(a) Find the probability that the first applicant with advanced training in programming is found on the fifth interview.

This is the probability that it takes 5 trials for 1 sucess. So, n = 5, x = 1 .

$P(X = x) = C_{n-1,x-1}*p^{x}*(1-p)^{n-x}$

$P(X = 5) = C_{4,0}*(0.30)^{1}*(0.70)^{4} = 0.0720$

There is a 7.20% probability that the first applicant with advanced training in programming is found on the fifth interview.

(b) What is the expected number of applicants who need to be interviewed in order to find the first one with advanced training?

The expected number of trials to get r sucesses is given by

$E = \frac{r}{p}$

So, with r = 1 .

$E = \frac{r}{p} = \frac{1}{0.3} = 3.33$

The expected number of interviews to find the first one with advanced training is 3.33.

(c) Let Y denote the number of the trial on which the first applicant with computer training was found. If each interview costs $30, find the expected value and variance of the total cost incurred interviewing candidates until an applicant with

There are 3.33 expected interviews before the first candidate with computer training is found. Each interview costs $30. Soo the expected cost is

E = 30*3.33 = 99.9

The expected cost is $99.9.

The variance of the number of expected trials to find r sucesses is given by:

$V = \frac{pr}{(1-pr)^{2}}$

$V = \frac{0.33}{(0.67)^2} = 0.7351$

The variance of the cost is 0.7351*30 = $22.053.

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6(2x + 5y)6 (2x) = 12x6 (5y) = 30y12x + 30y

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