The problem refers to monthly mortgage payments with the following parameters:

• Mean $\mu = 982$ dollars
• Standard deviation $\sigma = 180$ dollars
• The distribution is normally distributed, and the question asks to find probabilities based on this information.

Part (a) More than $1410

The probability given for $X > 1410$ is already stated as: $P(X > 1410) = 0.0090$

Part (b) Between $690 and $1240

We need to find: $P(690 < X < 1240)$

To solve this:

1. Standardize the values using the Z-score formula: $Z = \frac{X - \mu}{\sigma}$

2. Calculate Z-scores for $690 and $1240: $Z_1 = \frac{690 - 982}{180} = \frac{-292}{180} \approx -1.622$ $Z_2 = \frac{1240 - 982}{180} = \frac{258}{180} \approx 1.433$

3. Using the Standard Normal Distribution Table (or Z-table), we find the probabilities for the Z-scores:

   • For $Z_1 = -1.622$, the probability is approximately $P(Z_1) \approx 0.0525$.
   • For $Z_2 = 1.433$, the probability is approximately $P(Z_2) \approx 0.9236$.

4. Finally, subtract the smaller probability from the larger one: $P(690 < X < 1240) = P(Z_2) - P(Z_1) = 0.9236 - 0.0525 = 0.8711$

Thus, the probability is approximately 0.8711.

Would you like a detailed explanation on any of these steps?

Related Questions:

1. How do you calculate Z-scores for values outside the mean?
2. How does the normal distribution help in mortgage payment predictions?
3. What are the uses of the Standard Normal Distribution Table in probability?
4. How is probability linked to real-world financial decisions like mortgages?
5. What is the significance of calculating probabilities between two payment values?

Tip:

Always round intermediate Z-scores and final probability answers to at least four decimal places to maintain accuracy.