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### Problem Information:
Two students, Anna and Megan, take two language exams. Their scores are as follows:
- Anna: 83 on the first exam, 85 on the second exam.
- Megan: 70 on the first exam, 85 on the second exam.

Given statistical information:
- The **first exam**: Mean score = 84, Standard deviation = 6.
- The **second exam**: Mean score = 70, Standard deviation = 13.

### Question (a):
**Which of Anna and Megan qualify for language honors?**
- To qualify for language honors, the student must maintain at least an **85 average** across all language courses taken.

**Solution for (a):**
- Anna’s average: \( \frac{83 + 85}{2} = 84 \). This is **less than** 85, so she **does not qualify**.
- Megan’s average: \( \frac{70 + 85}{2} = 77.5 \). This is also **less than** 85, so she **does not qualify** either.

Thus, the answer to **(a)** is:
- **D. Neither Anna nor Megan qualify.**

### Question (b):
**Which student performed better overall?**
To determine overall performance, we can compare how far their scores deviate from the class means using **z-scores**.

For Anna:
- **First exam**: \( z = \frac{83 - 84}{6} = \frac{-1}{6} \approx -0.17 \).
- **Second exam**: \( z = \frac{85 - 70}{13} = \frac{15}{13} \approx 1.15 \).

For Megan:
- **First exam**: \( z = \frac{70 - 84}{6} = \frac{-14}{6} \approx -2.33 \).
- **Second exam**: \( z = \frac{85 - 70}{13} = \frac{15}{13} \approx 1.15 \).

Now, let's average the z-scores for each:
- Anna: \( \frac{-0.17 + 1.15}{2} = 0.49 \).
- Megan: \( \frac{-2.33 + 1.15}{2} = -0.59 \).

Since Anna has the **higher average z-score**, she performed better overall.

Thus, the answer to **(b)** is:
- **Anna performed better overall**.

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Let me know if you'd like further details or have any questions!

Here are 5 related questions for further exploration:
1. What is the interpretation of a z-score?
2. How does standard deviation impact performance comparisons?
3. What would Megan need to score on the second exam to qualify for language honors?
4. Can a student with a negative z-score still be considered a high performer?
5. How does the difference in standard deviations between the two exams affect score comparison?

**Tip:** When comparing performances, z-scores are a useful tool because they normalize scores based on the class mean and standard deviation!

Two language majors, Anna and Megan, took exams in two languages. Anna scored 83 on both exams. Megan scored 70 on the first exam and 85 on the second exam. Overall, student scores on the first exam had a mean of 84 and a standard deviation of 6, and the second exam scores had a mean of 70 and a standard deviation of 13. To qualify for language honors, a major must maintain at least an 85 average across all language courses taken. So far, which of Anna and Megan qualify?

Learn how to calculate averages and z-scores to compare performances between students Anna and Megan based on two language exams. This problem covers key statistical concepts such as averages, standard deviation, and z-scores. Suitable for students in Grades 9-12.

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