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Determine the area under the standard normal curve that lies to the right of the z-score −1.12 and to the left of the z-score −0.79.

To solve this problem, we need to determine two specific areas from the standard normal distribution table based on the given z-scores.

1. **Area to the right of \( z = -1.12 \):**
   The table provides the area to the left of any given z-score. To find the area to the right of \( z = -1.12 \), we need to subtract the area to the left of \( z = -1.12 \) from 1.
   - From the table, the area to the left of \( z = -1.12 \) is \( 0.1314 \).
   - Therefore, the area to the right of \( z = -1.12 \) is:
     \[
     1 - 0.1314 = 0.8686
     \]

2. **Area to the left of \( z = -0.79 \):**
   We directly use the table to find the area to the left of \( z = -0.79 \). However, since the table does not provide a value for \( z = -0.79 \) exactly, we can use the closest z-scores:
   - For \( z = -0.80 \), the area is \( 0.2119 \).
   - For \( z = -0.78 \), the area is \( 0.2177 \).

   Using linear interpolation between these two values:
   \[
   \text{Area for } z = -0.79 \approx \frac{(0.2177 - 0.2119)}{(0.78 - 0.80)} \times (-0.79 + 0.80) + 0.2119
   \]
   Simplifying:
   \[
   \text{Area for } z = -0.79 \approx 0.2148
   \]

3. **Combined Area:**
   The combined area is the region to the left of \( z = -0.79 \) and to the right of \( z = -1.12 \). To get this, we subtract the area to the left of \( z = -1.12 \) from the area to the left of \( z = -0.79 \):
   \[
   0.2148 - 0.1314 = 0.0834
   \]

Thus, the area under the standard normal curve between \( z = -1.12 \) and \( z = -0.79 \) is **0.0834**.

Would you like further details or clarification on any step? 

Here are 5 related questions to expand this topic:
1. How do you use a z-table for positive z-scores?
2. What is the significance of z-scores in standard normal distribution?
3. How can you calculate areas when z-scores fall between table values?
4. What is the relationship between z-scores and probability in real-world applications?
5. Can you explain how interpolation is used in statistics for approximating values?

**Tip:** When using the standard normal table, remember it always provides cumulative probabilities from the far left to the given z-score.

Determine the area under the standard normal curve that lies to the right of the z-score −1.12 and to the left of the z-score −0.79.

Learn how to calculate the area under the standard normal curve between z-scores -1.12 and -0.79. This problem involves using the standard normal distribution table and linear interpolation for approximation.

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