The first statistics exam had a mean of 66 and a standard deviation of 13 ​points; the second had a mean of 85 and a standard deviation of 4 points. Anna scored a 85 on both tests. Megan scored 92 on the first exam and 78 on the second. They both totaled 170 points on the two​ exams, but Anna claims that her total is better. Explain.
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Part 1
The total of Anna​'s ​z-scores is 
  
enter your response here​, which is greater than Megan​'s total of 
  
enter your response here.
​(Round to one decimal place as​ needed.)

To explain why Anna's total is better than Megan's, we will calculate the **z-scores** for both individuals. A z-score represents how many standard deviations a score is from the mean. By comparing z-scores, we can assess who performed better in relative terms on each exam.

### Formula for z-score:
\[
z = \frac{x - \mu}{\sigma}
\]
where:
- \( x \) is the individual's score,
- \( \mu \) is the mean of the distribution,
- \( \sigma \) is the standard deviation of the distribution.

### Anna's z-scores:
1. **First Exam**:  
   Mean \( \mu_1 = 66 \), Standard Deviation \( \sigma_1 = 13 \), Anna's Score \( x_1 = 85 \)
   \[
   z_1 = \frac{85 - 66}{13} = \frac{19}{13} \approx 1.5
   \]
   
2. **Second Exam**:  
   Mean \( \mu_2 = 85 \), Standard Deviation \( \sigma_2 = 4 \), Anna's Score \( x_2 = 85 \)
   \[
   z_2 = \frac{85 - 85}{4} = \frac{0}{4} = 0
   \]

Thus, **Anna's total z-score** is:
\[
z_{\text{total, Anna}} = 1.5 + 0 = 1.5
\]

### Megan's z-scores:
1. **First Exam**:  
   Mean \( \mu_1 = 66 \), Standard Deviation \( \sigma_1 = 13 \), Megan's Score \( x_1 = 92 \)
   \[
   z_1 = \frac{92 - 66}{13} = \frac{26}{13} = 2
   \]

2. **Second Exam**:  
   Mean \( \mu_2 = 85 \), Standard Deviation \( \sigma_2 = 4 \), Megan's Score \( x_2 = 78 \)
   \[
   z_2 = \frac{78 - 85}{4} = \frac{-7}{4} = -1.75
   \]

Thus, **Megan's total z-score** is:
\[
z_{\text{total, Megan}} = 2 + (-1.75) = 0.25
\]

### Conclusion:
- **Anna's total z-score** is **1.5**.
- **Megan's total z-score** is **0.25**.

Since Anna's z-score is higher, it means she performed better relative to the mean in both exams, even though their total scores were the same. Anna's performance was more consistently above the average across both exams.

### Final Answers:
- The total of Anna's z-scores is **1.5**, which is greater than Megan's total of **0.25**.

---

Would you like a more detailed explanation or clarification on any part of this calculation?

Here are five related questions to expand on the topic:
1. How does a z-score help in comparing scores across different distributions?
2. What are the implications of a negative z-score in terms of performance?
3. Can z-scores be applied in other contexts besides exam scores? If so, how?
4. How do z-scores relate to percentiles in statistics?
5. What are the potential limitations of using z-scores for comparisons in real-world situations?

### Tip:
When comparing performances across different tests or groups, always consider z-scores for a more meaningful comparison, as they standardize scores to a common scale based on mean and standard deviation.

The first statistics exam had a mean of 66 and a standard deviation of 13 points; the second had a mean of 85 and a standard deviation of 4 points. Anna scored 85 on both tests. Megan scored 92 on the first exam and 78 on the second. They both totaled 170 points on the two exams, but Anna claims that her total is better. Explain. The total of Anna's z-scores is ___, which is greater than Megan's total of ___.

Learn how to compare exam performances using z-scores. Anna and Megan scored 170 points on two tests with different means and standard deviations. We break down the calculations for their z-scores, showing why Anna's performance is considered better despite their equal totals. Suitable for high school students in Grades 10-12.

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