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
