Problem 20 - Basic Statistics for Business and Economics

Question

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?

Step-by-Step Solution

Answer

a. 0.3085 b. 0.2902 c. 0.5987

Step 1: Understand the Problem

We are given a normal distribution with a mean (\( \mu \)) of \(70,000 and a standard deviation (\( \sigma \)) of \)20,000. We need to find probabilities for certain ranges of amounts requested.

Step 2: Find Z-score for Part a ($80,000 or more)

To find the probability of a request being $80,000 or more, we first calculate the Z-score: \[Z = \frac{X - \mu}{\sigma} = \frac{80,000 - 70,000}{20,000} = 0.5\]

Step 3: Calculate Probability for Part a

Using the Z-score from Step 2, we find the probability that \( Z \geq 0.5 \) using a standard normal distribution table. This is equal to \( 1 - P(Z \leq 0.5) \). From the table, \( P(Z \leq 0.5) = 0.6915 \). So, \( P(Z > 0.5) = 1 - 0.6915 = 0.3085 \).

Step 4: Find Z-scores for Part b ($65,000 to $80,000)

We need Z-scores for \( X = 65,000 \) and \( X = 80,000 \): For \(65,000: Z = \frac{65,000 - 70,000}{20,000} = -0.25\)For \(80,000: Z = \frac{80,000 - 70,000}{20,000} = 0.5\)

Step 5: Calculate Probability for Part b

We want \( P(-0.25 < Z < 0.5) \). From the table, \( P(Z < 0.5) = 0.6915 \) and \( P(Z < -0.25) = 0.4013 \). Thus, \( P(-0.25 < Z < 0.5) = 0.6915 - 0.4013 = 0.2902 \).

Step 6: Find Z-score for Part c ($65,000 or more)

For \( X = 65,000 \), we use the Z-score from Step 4: \(Z = -0.25\).

Step 7: Calculate Probability for Part c

We need \( P(Z > -0.25) \). From the table, \( P(Z < -0.25) = 0.4013 \). So, \( P(Z > -0.25) = 1 - 0.4013 = 0.5987 \).

Key Concepts

Z-score calculation

The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. Calculating the Z-score is one of the first steps when dealing with normal distributions, as it standardizes different data points, allowing us to focus on one standard normal distribution. This is important because we use the standard normal distribution table to find probabilities. To calculate the Z-score, the formula is:

• \( Z = \frac{X - \mu}{\sigma} \)

In this formula:

• \( Z \) 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.

For example, if you have a home loan application of \( \\(80,000 \) with a mean of \( \\)70,000 \) and a standard deviation of \( \\(20,000 \), the Z-score calculation would be:

• \( Z = \frac{80,000 - 70,000}{20,000} = 0.5 \)

This tells us how far \( \\)80,000 \) is from the average loan amount in terms of standard deviations.

Probability Calculation

After calculating the Z-score, the next step is to determine the probability that corresponds to this Z-score. We use the standard normal distribution table for this, which shows the probability that a standard normal random variable is less than a given value.
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: \( 1 - 0.6915 = 0.3085 \)

This example shows that there's about a 30.85% chance of a loan application being \( \\(80,000 \) or more. For ranges like between \( \\)65,000 \) and \( \\(80,000 \), we find probabilities on both ends and subtract them, such as:

• \( P(-0.25 < Z < 0.5) = P(Z < 0.5) - P(Z < -0.25) \)
• \( 0.6915 - 0.4013 = 0.2902 \)

This means there's a 29.02% chance of the loan requested falling between \( \\)65,000 \) and \( \$80,000 \).

Standard normal distribution table

The standard normal distribution table, often known as the Z-table, is a mathematical table used to calculate the probability of a statistic falling to the left of a given Z-score in the standard normal distribution. It serves as a powerful tool to interpret Z-scores. By using the table, we convert Z-scores to probabilities, making it possible to better understand and anticipate the likelihood of different outcomes.

• 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.

To better use the table:

• 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.

The table is crucial in probability calculations as it allows quick results without sophisticated software, making it essential for manual calculations in statistics.