Complete the following CNP Bank Card Case and answer the accompanying questions
ID: 3309205 • Letter: C
Question
Complete the following CNP Bank Card Case and answer the accompanying questions
Before banks issue a credit card, they usually rate or score the customer in terms of his or her projected probability of being a profitable customer. A typical scoring table appears below.
Age
Under 25 25–29 30–34 35 +
(12 pts.) (5 pts.) (0 pts.) (18 pts.)
Time at Same Address
<1 yr. 1–2 yrs. 3–4 yrs. 5+ yrs.
(9 pts.) (0 pts.) (13 pts.) (20 pts.)
Auto age
None 0–1yr. 2–4 yrs. 5+ yrs.
(18 pts.) (12 pts.) (13 pts.) (3 pts.)
Monthly Car Payment
None $1–$99 $100–$299 $300+
(15 pts.) (6 pts.) (4 pts.) (0 pts.)
Housing Cost
$1–$199 $200–$399 Owns Lives with relatives
(0 pts.) (10 pts.) (12 pts.) (24 pts.)
Checking/Savings Accounts
Both Checking Only Savings Only Neither
(15 pts.) (3 pts.) (2 pts.) (0 pts.)
The score is the sum of the points on the six items. For example, Sushi Brown is under 25 years old (12 pts.), has lived at the same address for 2 years (0 pts.), owns a 4-year-old car (13 pts.), with car payments of $75 (6 pts.), housing cost of $200 (10 pts.), and a checking account (3 pts.). She would score 44. A second chart is then used to convert scores into the probability of being a profitable customer. A sample chart of this type appears below.
Score 30 40 50 60 70 80 90
Probability .70 .78 .85 .90 .94 .95 .96
Sushi’s score of 44 would translate into a probability of being profitable of approximately .81. In other words, 81 percent of customers like Sushi will make money for the bank card operations. Here are the interview results for three potential customers.
David Edward Ann
Name Born Brendan McLaughlin
Age 42 23 33
Time at same address 9 2 5
Auto age 2 3 7
Monthly car payment $140 $99 $175
Housing cost $300 $200 Owns clear
Checking/savings accounts Both Checking only Neither
1. Score each of these customers and estimate their probability of being profitable.
2. What is the probability that all three are profitable?
3. What is the probability that none of them are profitable?
4. Find the entire probability distribution for the number of profitable customers among this group of three.
5. Write a brief summary of your findings.
Explanation / Answer
1. David
Time at address
Hence, Probability of David being profitable is 0.95.
Edward
Time at address
Edward has a probability of 0.81 by interpolating the value of probability between scores 40 and 50.
P(44) = P(40)+(44-40)*(P(50)-P(40))/(50-40)
Ann
Time at address
The probability of Ann being profitable is 0.77 by interpolating between scores 30 and 40.
P(39) = P(30)+(39-30)*(P(39)-P(30))/(40-30)
Hence the probabilities are:
David: 0.95
Edward: 0.81
Ann: 0.77
2.
The probabiity of all 3 being profitable = P(D)*P(E)*P(A) {Assuming Independence}
= 0.95*0.81*0.77
The probabiity of all 3 being profitable = 0.593
3.
The probability of none of them being profitable = (1-P(D))*(1-P(E))*(1-P(A)) {Assuming Independence}
= 0.05*0.19*0.23
The probability of none of them being profitable = 0.0022
4.
From the previous parts, we know that,
P(Number of profitable = 3) = 0.593
P(Number of profitable = 0) = 0.002
P(Number of profitable = 1) = P(D)*(1-P(E))*(1-P(A)) + (1-P(D))*P(E)*(1-P(A)) + (1-P(D))*(1-P(E))*P(A) = 0.95*0.19*0.23 + 0.05*0.81*0.23 + 0.05*0.19*0.77 = 0.041+0.009+0.007 = 0.057
P(Number of profitable = 2) = P(D)*P(E)*(1-P(A)) + (1-P(D))*P(E)*P(A) + P(D)*(1-P(E))*P(A) = 0.95*0.81*0.23 + 0.05*0.81*0.77 + 0.95*0.19*0.77 = 0.177+0.031+0.140 = 0.348
Hence the probability distribution is:
5. The scoring criteria for credit rating gives the weightages to different factors denoting financial capabilities of an individual. The combined score can be directly converted to probabilities using the chart. The probabilities show that there is very little chance of all 3 being unprofitable and a 94% chance of at least 2 of them being profitable.
Category Value Points Age 42 18Time at address
9 20 Auto age 2 13 Monthly Car Payment 140 4 Housing Cost 300 10 Checking/Savings Both 15 Total 80 points