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)?

To find the probabilities for each question, we need to use the properties of the standard normal distribution. We can transform the given information about the distribution of debt into the standard normal distribution by using the z-score formula:

z = (X - μ) / σ

where:
- X is the debt amount
- μ is the mean debt amount ($15,015)
- σ is the standard deviation ($3,730)

For all the calculations, we will use the standard normal distribution table or a statistical software that can provide the probabilities associated with specific z-scores.

a. To find the probability that the debt for a randomly selected borrower with good credit is more than $18,000, we need to find the z-score corresponding to $18,000 and then find the area to the right of that z-score in the standard normal distribution table.

z = (18,000 - 15,015) / 3,730
z ≈ 0.799

Using the standard normal distribution table, we can find that the area to the right of z = 0.799 is approximately 0.2119. Therefore, the probability that the debt for a randomly selected borrower with good credit is more than $18,000 is approximately 0.2119.

b. To find the probability that the debt for a randomly selected borrower with good credit is less than $10,000, we need to find the z-score corresponding to $10,000 and then find the area to the left of that z-score in the standard normal distribution table.

z = (10,000 - 15,015) / 3,730
z ≈ -1.344

Using the standard normal distribution table, we can find that the area to the left of z = -1.344 is approximately 0.0909. Therefore, the probability that the debt for a randomly selected borrower with good credit is less than $10,000 is approximately 0.0909.

c. To find the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000, we need to find the z-scores corresponding to $12,000 and $18,000, and then find the area between those two z-scores in the standard normal distribution table.

z1 = (12,000 - 15,015) / 3,730
z1 ≈ -0.809

z2 = (18,000 - 15,015) / 3,730
z2 ≈ 0.799

Using the standard normal distribution table, we can find that the area to the left of z = -0.809 is approximately 0.2114, and the area to the left of z = 0.799 is approximately 0.7881. Therefore, the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 is approximately 0.7881 - 0.2114 = 0.5767.

d. To find the probability that the debt for a randomly selected borrower with good credit is no more than $14,000, we need to find the z-score corresponding to $14,000 and then find the area to the left of that z-score in the standard normal distribution table.

z = (14,000 - 15,015) / 3,730
z ≈ -0.273

Using the standard normal distribution table, we can find that the area to the left of z = -0.273 is approximately 0.3957. Therefore, the probability that the debt for a randomly selected borrower with good credit is no more than $14,000 is approximately 0.3957.

Note: The probabilities given are approximate, as they were rounded to four decimals for simplicity.

To solve these probability questions, we will use the Z-score formula and the standard normal distribution. The formula for the Z-score is given by:

Z = (X - μ) / σ

where:
X = the given debt amount
μ = the mean debt amount
σ = the standard deviation

For part (a), where we want to find the probability that the debt is more than $18,000:

Z = (18,000 - 15,015) / 3,730

Now, we can find the probability using the Z-table or a calculator. From the Z-table, we find that the Z-value of 18,000 corresponds to a probability of 0.7549 (rounded to 4 decimals). However, since we want the probability of the debt being more than $18,000, we need to subtract this probability from 1:

P(X > 18,000) ≈ 1 - 0.7549 = 0.2451 (rounded to 4 decimals)

Therefore, the probability that the debt for a randomly selected borrower with good credit is more than $18,000 is approximately 0.2451.

For part (b), where we want to find the probability that the debt is less than $10,000:

Z = (10,000 - 15,015) / 3,730

Again, we find the Z-value using the Z-table or a calculator. From the Z-table, we find that the Z-value of 10,000 corresponds to a probability of 0.0811 (rounded to 4 decimals). Therefore:

P(X < 10,000) ≈ 0.0811

The probability that the debt for a randomly selected borrower with good credit is less than $10,000 is approximately 0.0811.

For part (c), where we want to find the probability that the debt is between $12,000 and $18,000:

Z1 = (12,000 - 15,015) / 3,730
Z2 = (18,000 - 15,015) / 3,730

Again, using the Z-table or a calculator, we find the probabilities corresponding to Z1 and Z2. We then subtract the probability of Z2 from Z1 to get the probability of being between $12,000 and $18,000:

P(12,000 < X < 18,000) ≈ P(Z1 < Z < Z2)

Finally, for part (d), where we want to find the probability that the debt is no more than $14,000:

Z = (14,000 - 15,015) / 3,730

Again, using the Z-table or a calculator, we find the probability corresponding to the Z-value of 14,000:

P(X ≤ 14,000) ≈ Probability calculated from the Z-table

Note: The actual Z-values and probabilities may vary slightly depending on the level of accuracy used in the Z-table or calculator.