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)?

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

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

a. Well, let's calculate that probability. We know that the mean debt is $15,015, and the standard deviation is $3,730. So, to find the probability that the debt is more than $18,000, we need to calculate the z-score and then find the corresponding probability using a standard normal distribution table.

The z-score formula is: z = (x - μ) / σ, where x is the debt amount, μ is the mean, and σ is the standard deviation.

Plugging in the values, we get: z = (18,000 - 15,015) / 3,730

Calculating this, we find: z ≈ 0.799

Now, we need to find the probability associated with this z-score. From the standard normal distribution table, we find the value 0.799 corresponds to a probability of approximately 0.7881.

So, the probability that the debt for a randomly selected borrower with good credit is more than $18,000 is approximately 0.7881 (to 4 decimals).

b. To find the probability that the debt is less than $10,000, we can use the same z-score formula and find the corresponding probability from the standard normal distribution table.

Again, plugging in the values: z = (10,000 - 15,015) / 3,730

Calculating this, we get: z ≈ -1.346

From the standard normal distribution table, we find that a z-score of -1.346 corresponds to a probability of approximately 0.0884.

So, the probability that the debt for a randomly selected borrower with good credit is less than $10,000 is approximately 0.0884 (to 4 decimals).

c. To find the probability that the debt is between $12,000 and $18,000, we need to calculate the z-scores for both values and find the difference between their probabilities.

For $12,000: z1 = (12,000 - 15,015) / 3,730
For $18,000: z2 = (18,000 - 15,015) / 3,730

Calculating these, we get: z1 ≈ -0.806 and z2 ≈ 0.799

From the standard normal distribution table, we find that a z-score of -0.806 corresponds to a probability of approximately 0.2092, and a z-score of 0.799 corresponds to a probability of approximately 0.7881.

So, the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 is approximately (0.7881 - 0.2092) = 0.5789 (to 4 decimals).

d. Finally, to find the probability that the debt is no more than $14,000, we can calculate the z-score and find the corresponding probability from the standard normal distribution table.

Using the z-score formula: z = (14,000 - 15,015) / 3,730

Calculating this, we get: z ≈ -0.274

From the standard normal distribution table, we find that a z-score of -0.274 corresponds to a probability of approximately 0.3957.

So, the probability that the debt for a randomly selected borrower with good credit is no more than $14,000 is approximately 0.3957 (to 4 decimals).

To answer these questions, we will use the concept of the standard normal distribution and the z-score.

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 area under the standard normal curve to the right of $18,000. To do this, we first need to calculate the z-score associated with $18,000.

The z-score formula is given by (X - μ) / σ, where X is the value you want to convert to a z-score, μ is the mean, and σ is the standard deviation.

In this case, X = $18,000, μ = $15,015, and σ = $3,730.

Z = (18,000 - 15,015) / 3,730 ≈ 0.799

Next, we need to find the area to the right of the z-score of 0.799 using either a table or a calculator. Using a standard normal distribution table, we find that the probability associated with a z-score of 0.799 is approximately 0.7881.

Since we want the probability more than $18,000, we subtract this value from 1:

Probability = 1 - 0.7881 ≈ 0.2119 (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 0.2119.

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 area under the standard normal curve to the left of $10,000.

Using the same process as in part a, we calculate the z-score:

Z = (10,000 - 15,015) / 3,730 ≈ -1.35

Using the standard normal distribution table, we find that the probability associated with a z-score of -1.35 is approximately 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 area under the standard normal curve between these two values.

First, we calculate the z-scores for the boundaries:

Z1 = (12,000 - 15,015) / 3,730 ≈ -0.809
Z2 = (18,000 - 15,015) / 3,730 ≈ 0.799

Using the standard normal distribution table, we find that the probability associated with a z-score of -0.809 is approximately 0.2112, and the probability associated with a z-score of 0.799 is approximately 0.7881.

To find the probability between these two z-scores, we subtract the smaller probability from the larger probability:

Probability = 0.7881 - 0.2112 ≈ 0.5769

Therefore, the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 is approximately 0.5769.

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 area under the standard normal curve to the left of $14,000.

Using the same process as in part b, we calculate the z-score:

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

Using the standard normal distribution table, we find that the probability associated with a z-score of -0.273 is approximately 0.3931.

Therefore, the probability that the debt for a randomly selected borrower with good credit is no more than $14,000 is approximately 0.3931.

Kuai is all wrong.

a. 0.2061
b.
c. 0.5906
d. 0.3819