Suppose Canadian home-owners owe an average of $185,000 on their mortgages. Assume that mortgage debt is normally distributed in Canada with a standard deviation of $85,000.

Standard Normal Distribution Table

**a.** Albertans are reported to owe $243,600 in mortgage debt, much higher than the Canadian average. What is the probability of randomly selecting a Canadian with mortgage debt that exceeds $243,600?

Round to four decimal places if necessary

**b.** What is the probability of randomly selecting a Canadian with mortgage debt below $97,000?

Round to four decimal places if necessary

**c.** Determine the minimum mortgage debt owing by the 24% of Canadians with the largest mortgages.

Round to the nearest dollar

Let's solve these problems step by step.

We are given the following information:
- The mean mortgage debt, $ \mu = 185,000 $
- The standard deviation of mortgage debt, $ \sigma = 85,000 $

We will use the standard normal distribution (Z-scores) to solve the problems.

---

### **(a)** What is the probability of randomly selecting a Canadian with mortgage debt that exceeds $243,600$?

To find this, we will:
1. Convert the debt amount into a Z-score using the formula:
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   where $X$ is the given debt amount, $ \mu $ is the mean, and $ \sigma $ is the standard deviation.
   
2. Use the standard normal distribution table to find the corresponding probability.

For $ X = 243,600 $:
\[
Z = \frac{243,600 - 185,000}{85,000} = \frac{58,600}{85,000} \approx 0.6894
\]

Using the Z-table (or a calculator), the cumulative probability for $ Z = 0.6894 $ is approximately $ P(Z \leq 0.6894) = 0.7547 $.

Therefore, the probability of exceeding this debt amount is:
\[
P(Z > 0.6894) = 1 - 0.7547 = 0.2453
\]

Thus, the probability is approximately **0.2453**.

---

### **(b)** What is the probability of randomly selecting a Canadian with mortgage debt below $97,000$?

Again, we will:
1. Calculate the Z-score for $ X = 97,000 $:
\[
Z = \frac{97,000 - 185,000}{85,000} = \frac{-88,000}{85,000} \approx -1.0353
\]

2. Use the Z-table to find the cumulative probability for $ Z = -1.0353 $.

The cumulative probability for $ Z = -1.0353 $ is approximately $ P(Z \leq -1.0353) = 0.1506 $.

Thus, the probability of selecting a Canadian with a mortgage debt below $97,000$ is approximately **0.1506**.

---

### **(c)** Determine the minimum mortgage debt owing by the 24% of Canadians with the largest mortgages.

To find this:
1. We need to find the Z-score corresponding to the top 24% (i.e., the 76th percentile). From the Z-table, the Z-score for $ P(Z \leq 0.76) = 0.76 $ is approximately **0.7063**.

2. Convert this Z-score back into the mortgage debt amount using the formula:
   \[
   X = \mu + Z \cdot \sigma
   \]
   where $ \mu = 185,000 $ and $ \sigma = 85,000 $.

\[
X = 185,000 + 0.7063 \cdot 85,000 = 185,000 + 60,035.5 \approx 245,036
\]

Thus, the minimum mortgage debt owing by the top 24% of Canadians is approximately **$245,036**.

---

### Summary of answers:
- **(a)** Probability of mortgage debt exceeding $243,600: **0.2453**
- **(b)** Probability of mortgage debt below $97,000: **0.1506**
- **(c)** Minimum mortgage debt for the top 24%: **$245,036**

---

Would you like further explanation or clarification on any of these steps?

Here are 5 related questions to further explore:
1. How does increasing the standard deviation affect the probabilities in parts (a) and (b)?
2. What is the probability of selecting a Canadian with mortgage debt between $97,000 and $243,600?
3. What mortgage debt amount represents the 90th percentile of Canadians?
4. If Albertans had a standard deviation of $90,000 instead, how would that affect the probability in part (a)?
5. How would the minimum mortgage debt for the top 24% change if the average mortgage debt were $200,000?

**Tip**: When working with normal distributions, always convert the values into Z-scores to use the standard normal distribution table.

Suppose Canadian home-owners owe an average of $185,000 on their mortgages. Assume that mortgage debt is normally distributed in Canada with a standard deviation of $85,000.

**a.** Albertans are reported to owe $243,600 in mortgage debt, much higher than the Canadian average. What is the probability of randomly selecting a Canadian with mortgage debt that exceeds $243,600?

**b.** What is the probability of randomly selecting a Canadian with mortgage debt below $97,000?

**c.** Determine the minimum mortgage debt owing by the 24% of Canadians with the largest mortgages.

Explore probability problems related to Canadian mortgage debt distribution. Calculate the chances of exceeding $243,600 in debt and find the minimum debt for the top 24% of Canadians. Learn Z-score applications with a mean of $185,000 and standard deviation of $85,000.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation