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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

Expert verified
a. 0.9800; b. 0.2520; c. 0.3428

Step by step solution

01

Understanding the Problem

The problem provides that the average credit card debt for college seniors is $3262 with a standard deviation of $1100, following a normal distribution. We need to calculate the probabilities for different debt scenarios.
02

Identifying the Parameters for Z-Score

To find probabilities for specific debt levels, we will convert the debt amounts into Z-scores. The formula for a Z-score is given by \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the debt amount, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
03

Calculating Z-Score for Part a

For part a, \( X = 1000 \). The Z-score is calculated as \( Z = \frac{1000 - 3262}{1100} \approx -2.057 \). We will look for \( P(Z \, > \, -2.057) \), which involves finding the complementary area of the corresponding value in the Z-table.
04

Finding Probability for Part a

Using the Z-table, the probability \( P(Z \, < \, -2.057) \approx 0.0200 \). Therefore, \( P(Z \, > \, -2.057) = 1 - 0.0200 = 0.9800 \). This means the probability that a senior owes at least $1000 is 0.9800.
05

Calculating Z-Score for Part b

For part b, \( X = 4000 \). The Z-score is \( Z = \frac{4000 - 3262}{1100} \approx 0.671 \). So, we need to find \( P(Z \, > \, 0.671) \).
06

Finding Probability for Part b

The Z-table gives \( P(Z \, < \, 0.671) \approx 0.7480 \). So, \( P(Z \, > \, 0.671) = 1 - 0.7480 = 0.2520 \). Hence, the probability that a senior owes more than $4000 is 0.2520.
07

Calculating Z-Scores for Part c

For part c, calculate Z-scores for both \( X = 3000 \) and \( X = 4000 \). For \( X = 3000 \), \( Z = \frac{3000 - 3262}{1100} \approx -0.238 \). For \( X = 4000 \), as calculated, \( Z \approx 0.671 \).
08

Finding Probability for Part c

Find \( P(-0.238 < Z < 0.671) \). Use the Z-table to get \( P(Z < -0.238) \approx 0.4052 \) and \( P(Z < 0.671) \approx 0.7480 \). Therefore, the probability is \( 0.7480 - 0.4052 = 0.3428 \). This is the probability that a senior owes between \(3000 and \)4000.

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Key Concepts

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

Z-score
When working with a normal distribution, the Z-score is essential to understand how far away a particular value is from the mean. In simple terms, it tells us how many standard deviations a specific value is away from the average. This is incredibly useful in determining probabilities for different values in a normal distribution.
To calculate a Z-score, use the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
    • Here,
      • \( X \) is the value we are interested in,
      • \( \mu \) is the mean of the distribution, and
      • \( \sigma \) is the standard deviation.
    Using this formula helps in transforming any data point into a more standardized value, making it easier to compare with a standard normal distribution.
    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
Standard deviation (\( \sigma \) is a crucial concept in statistics that measures how spread out the values are in a dataset. It indicates the extent of variation or dispersion from the mean (\( \mu \).
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.
If the debt amounts have a higher standard deviation, it means students' debts vary more widely from the average debt amount.
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
Probability calculation in a normal distribution often involves using a Z-score to find out how likely it is for a random variable to fall within a certain range. When given a normally distributed dataset, determining probabilities for specific events can be easily tackled using the Z-table and Z-scores.
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.
With the Z-scores:
  • 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.
This involves subtracting the cumulative probabilities from the Z-table to find either the probability greater than or less than a certain Z-score. This process helps make informed predictions based on the normal distribution of data.

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