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) Find the 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?
(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 advanced computer training is found.

Answer:

Step-by-step explanation:

From the given information:

Assume X represents the no. interviewed until 1 has advanced training.

X obeys a Geometric distribution with parameter 0.3.

X  [tex]\sim[/tex] Geom (0.30)

For geometric distribution, the probability density is:

[tex]P(X =x) = p(1-p) ^{x-1} \ \ \ where; x =1,2,3...[/tex]

TO calculate the required probability;

[tex]P(X =5) =0.30 (1-0.30)^{5-1}[/tex]

[tex]P(X =5) =0.30 (0.70)^{4}[/tex]

[tex]P(X=5) = 0.30 \times 0.2401[/tex]

[tex]\mathbf{P(X=5) = 0.07203}[/tex]

(b)

The expected no. of applicants that need to be interviewed are:

[tex]E(X)=\dfrac{1}{p}[/tex]

[tex]E(X)=\dfrac{1}{0.30}[/tex]

E(X) = 3.33

(c)

The mean and the variance can be computed as:

[tex]E(Y) = \dfrac{1}{p}[/tex]

[tex]E(Y) = \dfrac{1}{0.30}[/tex]

E(Y) = 3.33

[tex]V(Y)=\dfrac{1-p}{p^2}[/tex]

[tex]V(Y)=\dfrac{1-0.3}{0.3^2}[/tex]

[tex]V(Y)=\dfrac{0.7}{0.3^2}[/tex]

[tex]V(Y)=7.778[/tex]

Suppose  C represents the no. of the total cost and given that each interview costs $30.

Then C = 30Y

Recall that; C is constant for a random variable X

∴

E(C) = E(30Y)

E(C) = 30E(Y)

E(C) = 30*3.33

E(C) =99.9

E(C) [tex]\simeq[/tex] 100

V(C) = V(30Y)

V(C) = 900 V(Y)

V(C) = 900*7.778

V(C) = 7000.2

V(C) [tex]\simeq[/tex] 7000

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