https://file.aiquickdraw.com/mathbot/file/1728707607914x68c64my.jpg

The image shows a normal distribution curve with two shaded regions under the curve: one between \( z = -0.26 \) and \( z = 0 \), and another between \( z = 0 \) and \( z = 1.98 \). The question asks for the area of these shaded regions, which corresponds to the cumulative probabilities between these z-scores.

To find the total area of the shaded region:

1. **Find the cumulative probability at \( z = 1.98 \)**:
   - From a standard normal distribution table or using a calculator, the cumulative probability for \( z = 1.98 \) is approximately 0.9761.

2. **Find the cumulative probability at \( z = -0.26 \)**:
   - The cumulative probability for \( z = -0.26 \) is approximately 0.3974.

3. **Subtract the cumulative probabilities**:
   \[
   \text{Shaded Area} = P(0 \leq z \leq 1.98) + P(-0.26 \leq z \leq 0)
   \]
   \[
   = (0.9761 - 0.5) + (0.5 - 0.3974)
   \]
   \[
   = 0.4761 + 0.1026 = 0.5787
   \]

Thus, the area of the shaded region is approximately **0.5787**.

Would you like more details or have any questions on this? 

Here are five related questions:
1. What is a standard normal distribution, and how is it used in probability?
2. How do you interpret z-scores in a normal distribution?
3. How can you calculate probabilities for non-standard normal distributions?
4. What is the significance of the area under a normal distribution curve?
5. How do you use a standard normal distribution table?

**Tip**: When calculating areas for standard normal distributions, it's helpful to break the problem into segments around \( z = 0 \) because symmetry simplifies calculations.

Find the area of the shaded region in a normal distribution between z = -0.26 and z = 1.98.

Learn how to calculate the area of the shaded region between z = -0.26 and z = 1.98 under a normal distribution curve. This detailed solution covers cumulative probability and z-scores, essential for understanding normal distributions. Suitable for high school students in Grades 10-12.

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