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 solve these probability questions, we need to use the properties of the normal distribution. Specifically, we will use the z-score formula, which allows us to find the probability associated with a particular value from a normally distributed population.
The z-score formula is:
z = (X - μ) / σ
Where:
X is the value we're interested in (debt amount in this case)
μ is the mean of the population (given as $15,015)
σ is the standard deviation of the population (given as $3,730)
To find the probability using the z-score, we will use a standard normal distribution table or a calculator. Alternatively, we can use statistical software like Excel or Python.
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 first:
z = (18,000 - 15,015) / 3,730 ≈ 0.7995
Using the standard normal distribution table or calculator, we find that the probability corresponding to a z-score of 0.7995 is approximately 0.7867. However, we need the probability of the debt being greater than $18,000, so we subtract this probability from 1:
P(X > 18,000) ≈ 1 - 0.7867 ≈ 0.2133
Therefore, the probability that the debt for a randomly selected borrower with good credit is more than $18,000 is approximately 0.2133.
b. To find the probability that the debt for a randomly selected borrower with good credit is less than $10,000, we follow the same process:
z = (10,000 - 15,015) / 3,730 ≈ -1.35
Using the standard normal distribution table or calculator, we find that the probability corresponding to a z-score of -1.35 is approximately 0.0885.
P(X < 10,000) ≈ 0.0885
Therefore, the probability that the debt for a randomly selected borrower with good credit is less than $10,000 is approximately 0.0885.
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 individual probabilities for each bound and subtract them:
z1 = (12,000 - 15,015) / 3,730 ≈ -0.81
z2 = (18,000 - 15,015) / 3,730 ≈ 0.7995
Using the standard normal distribution table or calculator, we find that the probability corresponding to a z-score of -0.81 is approximately 0.2083, and the probability corresponding to a z-score of 0.7995 is approximately 0.7867.
P(12,000 < X < 18,000) ≈ 0.7867 - 0.2083 ≈ 0.5784
Therefore, the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 is approximately 0.5784.
d. To find the probability that the debt for a randomly selected borrower with good credit is no more than $14,000, we can use the same approach:
z = (14,000 - 15,015) / 3,730 ≈ -0.273
Using the standard normal distribution table or calculator, we find that the probability corresponding to a z-score of -0.273 is approximately 0.3957.
P(X ≤ 14,000) ≈ 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.