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For borrowers with good credit scores, the mean debt for revolving and installment accounts is \(\$ 15,015\) (Business Week, March 20,2006 ). Assume the standard deviation is \(\$ 3540\) and that debt amounts are normally distributed. a. What is the probability that the debt for a borrower with good credit is more than \(\$ 18,000 ?\) b. What is the probability that the debt for a borrower with good credit is less than \(\$ 10,000 ?\) c. What is the probability that the debt for a borrower with good credit is between \(\$ 12,000\) and \(\$ 18,000 ?\) d. What is the probability that the debt for a borrower with good credit is no more than \(\$ 14,000 ?\)

Short Answer

Expert verified
a) 0.201, b) 0.078, c) 0.602, d) 0.387

Step by step solution

01

Understanding the Problem

We are given the mean debt amount \(\mu = 15,015\) and standard deviation \(\sigma = 3,540\). Since the distribution is normal, we'll use the Z-score formula \(Z = \frac{X - \mu}{\sigma}\) to find probabilities of different debt amounts. We need to solve four different parts, each focusing on a probability concerning a specific debt amount threshold.
02

Probability for Debt > $18,000

First, we find the Z-score for \(X = 18,000\):\[ Z = \frac{18,000 - 15,015}{3,540} = \frac{2,985}{3,540} \approx 0.843 \]Using the standard normal table, we find the probability \(P(Z < 0.843)\). Subtract this from 1 to find \(P(Z > 0.843)\). Look up in the Z-table:\[ P(Z < 0.843) \approx 0.799 \]Thus:\[ P(Z > 0.843) = 1 - 0.799 = 0.201 \]
03

Probability for Debt < $10,000

Find the Z-score for \(X = 10,000\):\[ Z = \frac{10,000 - 15,015}{3,540} = \frac{-5,015}{3,540} \approx -1.416 \]Look up the Z-table:\[ P(Z < -1.416) \approx 0.078 \]Hence, the probability is approximately 0.078.
04

Probability for $12,000 < Debt < $18,000

Calculate the Z-scores for \(X_1 = 12,000\) and \(X_2 = 18,000\):\[ Z_1 = \frac{12,000 - 15,015}{3,540} = \frac{-3,015}{3,540} \approx -0.851 \]\[ Z_2 = \frac{18,000 - 15,015}{3,540} = \frac{2,985}{3,540} \approx 0.843 \]Using the Z-table, find:\[ P(Z < 0.843) \approx 0.799 \]\[ P(Z < -0.851) \approx 0.197 \]Then:\[ P(-0.851 < Z < 0.843) = 0.799 - 0.197 = 0.602 \]
05

Probability for Debt ≤ $14,000

Calculate the Z-score for \(X = 14,000\):\[ Z = \frac{14,000 - 15,015}{3,540} = \frac{-1,015}{3,540} \approx -0.287 \]Refer to the Z-table:\[ P(Z < -0.287) \approx 0.387 \]Thus, the probability is approximately 0.387.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculations
Probability calculations are a key concept in statistics, allowing us to estimate the chance that a certain event will occur. When dealing with normally distributed data, we use specific tools and formulas, like the Z-score, to help us in these calculations. For instance, let's consider the problem of calculating the probability that a borrower's debt with a good credit score is above or below a specific amount.
The normal distribution is key here because it describes how the data, or in this case, the debts, are spread across the possible amounts. When we know the mean and the standard deviation of this distribution, we can calculate the exact probabilities we are interested in.
For example, if we want to know the probability that debt exceeds $18,000, we first calculate its Z-score, then refer to a standard normal distribution table to find the probability. This is how probability calculations are combined with the Z-score for precise estimations.
Z-score
The Z-score is a statistical measure that tells us how many standard deviations an element is from the mean. It is essential for working with normal distributions. The formula for a Z-score is:
  • \(Z = \frac{X - \mu}{\sigma}\)
where \(X\) is the value we are examining, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
This tool transforms our normal data point, such as a debt amount, into a standard normal variable, allowing us to use standard normal distribution tables to find probabilities.
In the exercise, Z-scores were computed for different debt amounts such as \(18,000, \)10,000, and so on. These Z-scores then provided the necessary information for identifying probabilities related to those debt amounts in a standardized manner. The Z-score is particularly useful because once calculated, it doesn't matter what the units are for \(X\); the Z-score can always be looked up in the standard normal table.
Statistics
Statistics is about gathering, analyzing, interpreting, presenting, and organizing data. It plays a crucial role in virtually every scientific field and helps us make decisions based on data.
In the context of the given problem, statistics provides us the foundational understanding to approach probability and normal distribution issues. By using statistical analysis, we can uncover trends and patterns that would otherwise be invisible in raw data.
In our exercise, statistics help us understand the distribution of debt amounts for borrowers with good credit scores. The calculated mean and standard deviation give us insights into the data set's center and spread, enabling us to predict probabilities for debt levels using the normal distribution model. Statistics empower us to transform empirical data into logical decisions and predictions.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simpler terms, it tells us how spread out the numbers are in a data set. A low standard deviation means that most of the numbers are close to the mean. A high standard deviation indicates that the numbers are spread out over a wider range.
When dealing with a normal distribution, the standard deviation allows us to characterize the spread of the distribution. In our problem, the standard deviation is \(\\( 3540\), indicating how much the individual debts deviate from the mean debt of \(\\) 15,015\).
Understanding the standard deviation enables us to calculate the Z-score for any given debt amount, and hence determine the probability of that amount occurring. The standard deviation is a fundamental concept in statistics because it provides a link between a normal distribution and probability calculations, making it a vital tool for data analysis.

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Most popular questions from this chapter

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