Susan is a sales representative who has a history of making a successful sale fr
ID: 3132852 • Letter: S
Question
Susan is a sales representative who has a history of making a successful sale from about 74% 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 = 0.74 and failure is q = 0.26; k is a fixed whole number > 1.We have poisson trials for which the probability of success is p = 0.74 and failure is q = 0.26; k is a fixed whole number 1. We have binomial trials for which the probability of success is p = 0.26 and failure is q = 0.74; k is a fixed whole number 1.We have binomial trials for which the probability of success is p = 0.74 and failure is q = 0.26; 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.)
(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 and standard deviation of the random variable n? (Use 2 decimal places.)
Interpret these values in the context of this application.
The expected contact in which the eleventh sale occurs is , with a standard deviation of .
The expected contact in which the thirteenth sale occurs is , with a standard deviation of .
The expected contact in which the twelfth sale occurs is , with a standard deviation of .
The expected contact in which the twelfth sale occurs is , with a standard deviation of .
P(12) P(13) P(14)Explanation / Answer
a. Negative binomial distribution is appropriate for the random variable n : We have binomial trials for which the probability of success is p = 0.74 and failure is q = 0.26; k is a fixed whole number 1.
b. P(n) = (x1C r1)(1p)xrpr
P(12) = 11C11 ( 0.74)12 (0.26)0 = 0.02697
P(13) = 12C11 ( 0.74)13 (0.26)1 = 0.06225
P(14) = 13C11 ( 0.74)14 (0.26)2 = 0.07786
c. P (X = 12, 13,14) = P(12) + P(13)+P(14) = 0.02697+0.06225+0.07786 = 0.1671
d. P ( X > 14) = 1 - P (X = 12, 13,14) = 1 - 0.1671 = 0.8329.
e. Mean = r/p = 12/0.74 = 16.22 ~ 17
Standard deviation = r(1p)/ p2 = 12(0.26) / 0.742 = 5.6976
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