Question

For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015 (BusinessWeek, March 20, 2006). Assume the standard deviation is $3,730 and that debt amounts are normally distributed.

a. What is the probability that the debt for a randomly selected borrower with good credit is more than $18,000 (to 4 decimals)?


b. What is the probability that the debt for a randomly selected borrower with good credit is less than $10,000 (to 4 decimals)?


c.What is the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 (to 4 decimals)?


d. What is the probability that the debt for a randomly selected borrower with good credit is no more than $14,000 (to 4 decimals)?

Answer 1

a. What is the probability that the debt for a randomly selected borrower with good credit is more than $18,000 (to 4 decimals)?

p(x > 18000)
z = ( x - μ ) / σ
z = (18000 -15015)/3730
z =2985/3730 = 0.8
P( z > 0.8) = 1-.7881= 0.2119

b. What is the probability that the debt for a randomly selected borrower with good credit is less than $10,000 (to 4 decimals)?

P( x <10000)
z= ( x - μ ) / σ
z = (10000-15015)/3730
z =- 5015/3730 = -1.3445
P( z < -1.34) = 0.0901

c.What is the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 (to 4 decimals)?


P(12000<x<18000)
z = ( x - μ ) / σ
z = (12000-15015)/3730
z =- 3015/3730 = -.81
z = ( x - μ ) / σ
z = 3015/3730 = 0.80
P(-0.81<z<0.8) = 0.7881 - 0.2090 = 0.5791

There is 0.5791 probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000.

d. What is the probability that the debt for a randomly selected borrower with good credit is no more than $14,000 (to 4 decimals)?

P( x <14000)
z= ( x - μ ) / σ
z = (14000-15015)/3730
z = -1015/3730
z = -0.27
P( z < -0.27) = 0.3936

Answer 2

Kuai is all wrong.

a. 0.2061
b.
c. 0.5906
d. 0.3819

Answer 3

Kuai is all right. He solved questions absolutely in correct way.