To analyze Julie's claim, we need to compare the relative performance of both Derrick and Julie in terms of z-scores on each exam. Z-scores help to understand how well a student performed relative to the average performance of the class.

Z-Score Formula:

The z-score is calculated by: $z = \frac{X - \mu}{\sigma}$ Where:

• $X$ is the score,
• $\mu$ is the mean,
• $\sigma$ is the standard deviation.

---

Derrick's Z-Scores:

For the first test (mean = 65, standard deviation = 15): $z_1(\text{Derrick}) = \frac{80 - 65}{15} = \frac{15}{15} = 1.000$ For the second test (mean = 80, standard deviation = 10): $z_2(\text{Derrick}) = \frac{80 - 80}{10} = \frac{0}{10} = 0.000$

Julie's Z-Scores:

For the first test: $z_1(\text{Julie}) = \frac{70 - 65}{15} = \frac{5}{15} = 0.333$ For the second test: $z_2(\text{Julie}) = \frac{90 - 80}{10} = \frac{10}{10} = 1.000$

---

Total Z-Scores:

To compare their overall performance, we sum the z-scores for both tests.

• Derrick's total z-score: $z_{\text{total}}(\text{Derrick}) = 1.000 + 0.000 = 1.000$
• Julie's total z-score: $z_{\text{total}}(\text{Julie}) = 0.333 + 1.000 = 1.333$

Conclusion:

Since Julie's total z-score is higher than Derrick's, her claim is correct—she performed better relative to the class averages across both tests.

Final Answer:

A. Julie's claim is correct. Derrick's z-scores are 1.000 for the first test and 0.000 for the second test. Julie's z-scores are 0.333 for the first test and 1.000 for the second test. Derrick's total is 1.000, which is less than Julie's total, 1.333.

---

Do you have any questions or need more details?

---

Here are five related questions to expand on this:

1. How would their performance be different if the standard deviation on the second test was 5 instead of 10?
2. What does a z-score of 0 indicate about Derrick's performance on the second test?
3. How can z-scores be used to compare performance across different exams or subjects?
4. How would the z-scores change if Julie scored 85 on both tests?
5. What would be the impact on the total z-scores if the first test had a lower mean?

Tip:

Z-scores are a useful tool to compare performances across different tests, especially when the means and standard deviations vary.