A bank with a home-mortgage loan default rate of 22%, is planning to extend 12 n

Question

A bank with a home-mortgage loan default rate of 22%, is planning to extend 12 new home loans this week. If we let X represent the number of home-loans paid back on time, the probabilities for the values of X can be calculated using the binomial probability distribution. Please answer the following questions based on this information.

a. What is the probability of exactly 7 of the 12 loans defaulting?

b. What is the probability that at least 10 out of the 12 loans will default?

c. What is the probability of more than 8 of the 12 loans defaulting? .

d. What is the probability of at most 6 out of the 12 loans being paid back on time?

e. What is the probability of less than 4 of the 12 loans defaulting? .

f. What is the expected number of defaults out of the next 12 loans?

g. Calculate and interpret the standard deviation for the number of loans paid back on time out of the next 12 loans.

Explanation / Answer

n = 12, p = 0.22, q = 1 - p = 0.78

P(x) = C(n, x) p^x q^(n - x)

(a) P(7) = C(12, 7) 0.22^7 0.78^(12 - 7) = 0.0057

(b) P(x 10) = P(10) + P(11) + P(12)

= C(12, 10) 0.22^10 0.78^(12 - 10) + C(12, 11) 0.22^11 0.78^(12 - 11) + C(12, 12) 0.22^12 0.78^(12 - 12)

= 0

(c) P(x > 8) = P(9) + P(10) + P(11) + P(12)

= C(12, 9) 0.22^9 0.78^(12 - 9) + C(12, 10) 0.22^10 0.78^(12 - 10) + C(12, 11) 0.22^11 0.78^(12 - 11) +
C(12, 12) 0.22^12 0.78^(12 - 12)

= 0.0001

Now take p = 0.78, q = 0.22, n = 12

(d) P(x 6) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6)

= C(12, 0) 0.78^0 0.22^(12 - 0) + C(12, 1) 0.78^1 0.22^(12 - 1) + C(12, 2) 0.78^2 0.22^(12 - 2)

+ C(12, 3) 0.78^3 0.22^(12 - 3) + C(12, 4) 0.78^4 0.22^(12 - 4) + C(12, 5) 0.78^5 0.22^(12 - 5)

+ C(12, 6) 0.78^6 0.22^(12 - 6)

= 0.0304

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.