Chapter 7: Problem 20
The amounts of money requested on home loan applications at Down River Federal Savings follow the normal distribution, with a mean of \(\$ 70,000\) and a standard deviation of \(\$ 20,000 .\) A loan application is received this morning. What is the probability: a. The amount requested is \(\$ 80,000\) or more? b. The amount requested is between \(\$ 65,000\) and \(\$ 80,000 ?\) c. The amount requested is \(\$ 65,000\) or more?
Short Answer
Step by step solution
Understand the Problem
Find Z-score for Part a ($80,000 or more)
Calculate Probability for Part a
Find Z-scores for Part b ($65,000 to $80,000)
Calculate Probability for Part b
Find Z-score for Part c ($65,000 or more)
Calculate 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 calculation
- \( Z = \frac{X - \mu}{\sigma} \)
- \( Z \) is the Z-score.
- \( X \) is the value in question, for which you want to find the probability.
- \( \mu \) is the mean of the distribution.
- \( \sigma \) is the standard deviation of the distribution.
- \( Z = \frac{80,000 - 70,000}{20,000} = 0.5 \)
Probability Calculation
For a Z-score of 0.5, the table tells us that \( P(Z \leq 0.5) = 0.6915 \). However, if you want to find the probability of the variable being more than this value, you calculate:
- \( P(Z > 0.5) = 1 - P(Z \leq 0.5) \)
- Which leads us to: \( 1 - 0.6915 = 0.3085 \)
- \( P(-0.25 < Z < 0.5) = P(Z < 0.5) - P(Z < -0.25) \)
- \( 0.6915 - 0.4013 = 0.2902 \)
Standard normal distribution table
- If the Z-score is positive, the table will give the probability a value is less than the given Z-score.
- If the Z-score is negative, it represents the probability of the statistic being lower than that Z-score.
- Always remember the total area under the curve is 1, which represents 100% probability.
- Find the intersection of the row and column corresponding to your Z-score for the cumulative probability.
- For a probability greater than a certain Z-score, subtract the table probability from 1.