Question

Susan is a sales representative who has a history of making a successful sale from about 90% of her sales contacts. If she makes 12 successful sales this week, Susan will get a bonus. Let n be a random variable representing the number of contacts needed for Susan to get the 12th sale.

(a) Explain why a negative binomial distribution is appropriate for the random variable n.

We have binomial trials for which the probability of success is p = 0.90 and failure is q = 0.10; k is a fixed whole number > 1.We have poisson trials for which the probability of success is p = 0.90 and failure is q = 0.10; k is a fixed whole number ≥1.     We have binomial trials for which the probability of success is p = 0.90 and failure is q = 0.10; k is a fixed whole number ≥1.We have binomial trials for which the probability of success is p = 0.10 and failure is q = 0.90; k is a fixed whole number ≥1.

Write out the formula for P(n) in the context of this application. (Use C(a,b) as the notation for "a choose b".)
P(n) =  

(b) Compute P(n = 12), P(n = 13), and  P(n = 14). (Use 4 decimal places.)

P(12)
P(13)
P(14)

(c) What is the probability that Susan will need from 12 to 14 contacts to get the bonus? (Use 4 decimal places.)


(d) What is the probability that Susan will need more than 14 contacts to get the bonus? (Use 4 decimal places.)


(e) What are the expected value μ and standard deviation σ of the random variable n? (Use 2 decimal places.)

μ
σ

Interpret these values in the context of this application.

The expected contact in which the eleventh sale occurs is μ, with a standard deviation of σ.The expected contact in which the twelfth sale occurs is μ, with a standard deviation of σ.     The expected contact in which the twelfth sale occurs is σ, with a standard deviation of μ.The expected contact in which the thirteenth sale occurs is μ, with a standard deviation of σ.

Answer 1

a)

We have binomial trials for which the probability of success is p = 0.90 and failure is q = 0.10; k is a fixed whole number > 1

P(n) = C(n-1,11)*(0.9)¹²(0.1)^n-12

b)

P(12)=0.2824

P(13)=0.3389

P(14)=0.2203

c)

probability that Susan will need from 12 to 14 contacts to get the bonus :

probability =

P(12<=X<=14)=

0.8416

d )

probability that Susan will need more than 14 contacts to get the bonus =1-0.8416 =0.1584

e)

here mean of distribution=μ=r/p=

13.33

standard deviation σ=√(r(1-p))/p=

1.22

The expected contact in which the twelfth sale occurs is μ, with a standard deviation of σ.