The mean and standard deviation in midterm tests of a probability course are 72 and 12, respectively. These quantities for final tests are 68 and 15. What final grade is comparable to Velma's 82 in the midterm.

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BlackTom AI · Accepted Answer

This item is an essay-type prompt, and no multiple-choice options are provided to analyze. I will evaluate the student's solution path and demonstrate the calculation step by step.

First, extract the key data from the prompt: midterm mean = 72, midterm standard deviation = 12, final mean = 68, final standard deviation = 15. Velma’s midterm score is 82.

Next, compute Velma’s z-score on the midterm to quantify how far her score is from the midterm mean in standard deviation units. Using z = (score − mean)/SD, we get:
z_midterm = (82 − 72)/12 = 10/12 ≈ 0.83.

To find the final exam score that has the same relative position (same z-score) on the final distribution, set the final z-score equal to Velma’s midterm z-score:
z_final = (x − 68)/15 = 0.83.
Solve for x: x − 68 = 0.83 × 15 ≈ 12.45, so x ≈ 68 + 12.45 = 80.45.

Thus, a final exam score of about 80.45 would be comparable to Velma’s 82 on the midterm, in the sense of matching their z-scores relative to their respective distributions.

Now, analyze the student’s solution steps: They correctly identified the need to convert Velma’s midterm score to a z-score using the midterm mean and standard deviation. Their calculation z = (82 − 72)/12 = 0.83 is accurate. They then correctly used the final exam distribution parameters and set 0.83 = (x − 68)/15 to solve for x, obtaining x = 68 + 0.83 × 15 ≈ 80.45, which is consistent with the z-score preservation idea. The final statement about the final grade matching Velma’s midterm performance aligns with the computed value.

Consider potential clarifications or edge notes:
- The result 80.45 assumes exact arithmetic; in practice, you might round to an appropriate grade precision (e.g., 80.5 or 80).
- If the final grade scale is discrete (e.g., whole numbers), you might report the nearest whole number, such as 80 or 81, depending on rounding rules.
- If one wanted to discuss the concept more broadly, this approach demonstrates equating percentiles across distributions with different means and variances via z-scores.

If there were multiple answer choices, this reasoning would map to selecting the option that mirrors x ≈ 80.45 as the score on the final exam corresponding to Velma’s midterm performance.

Home Module 08 Assignment

The mean and standard deviation in midterm tests of a probability course are 72 and 12, respectively. These quantities for final tests are 68 and 15. What final grade is comparable to Velma's 82 in the midterm.

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