Chapter 6: Problem 11
The average credit card debt for college seniors is \(\$ 3262\). If the debt is normally distributed with a standard deviation of \(\$ 1100,\) find these probabilities. a. The senior owes at least \(\$ 1000\). b. The senior owes more than \(\$ 4000\). c. The senior owes between \(\$ 3000\) and \(\$ 4000\).
Short Answer
Step by step solution
Understanding the Problem
Identifying the Parameters for Z-Score
Calculating Z-Score for Part a
Finding Probability for Part a
Calculating Z-Score for Part b
Finding Probability for Part b
Calculating Z-Scores for Part c
Finding Probability for Part c
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Z-score
To calculate a Z-score, use the formula:
- \( Z = \frac{X - \mu}{\sigma} \)
- \( X \) is the value we are interested in,
- \( \mu \) is the mean of the distribution, and
- \( \sigma \) is the standard deviation.
- Here,
For example, if we have average credit card debt \( \mu = 3262 \) and standard deviation \( \sigma = 1100 \), and you want to find the probability for owing \\(1000, you calculate the Z-score. With \( X = 1000 \), the Z-score is \( Z = \frac{1000 - 3262}{1100} \approx -2.057 \).
This means \\)1000 is approximately 2.057 standard deviations below the mean. From here, you would use the Z-score in the Z-table to find probabilities associated with this position.
Standard deviation
A lower standard deviation means the data points are generally close to the mean, while a higher standard deviation indicates more spread or variability in the data points.
In our context of college seniors' credit card debt:
- The mean debt is \\(3262, and
- The standard deviation is \\)1100.
Standard deviation is vital when calculating a Z-score and further understanding the normal distribution, as it scales the deviation of a data point from the mean. This makes comparisons across different datasets possible by standardizing values.
Probability calculation
For example:
- To determine the probability of a senior owing at least \\(1000, calculate the Z-score of \\)1000 and find its position on the Z-table.
- Similarly, for a debt of more than \\(4000, find \( Z \) for \\)4000 and identify the probability.
- For a range (e.g., between \\(3000 and \\)4000), find the Z-scores for both amounts and subtract the probabilities.
- For \( X = 1000 \), the Z-score translates to a probability, showing how much debt is smaller or larger than what would usually be expected.
- The area under the standard normal curve given by Z-table helps in finding probabilities for both less-than and more-than scenarios easily.