To determine the score required to earn an A+, we need to find the score that corresponds to the top 4% of the exam scores. This involves finding the z-score that corresponds to the 96th percentile of a standard normal distribution (because 100% - 4% = 96%).

Step 1: Find the z-score for the 96th percentile

Using a z-table or statistical software, the z-score corresponding to the 96th percentile is approximately:

$z = 1.75$

Step 2: Convert the z-score to the actual exam score

The z-score formula is:

$z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the score we need to find.
• $\mu = 75$ (mean).
• $\sigma = 4$ (standard deviation).

Rearranging to solve for $X$:

$X = z \cdot \sigma + \mu$

Substitute the values:

$X = 1.75 \cdot 4 + 75$

$X = 7 + 75 = 82$

Final Answer:

The minimum score required to earn an A+ is 82.

Correct option: D) 82

Would you like a detailed explanation of the z-score calculation or further clarification?

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Related Questions:

1. What percentage of students score below 71 on this exam?
2. What is the z-score for a score of 68 on this exam?
3. What score corresponds to the bottom 10% of this exam distribution?
4. If 200 students took the exam, how many scored 82 or above?
5. How would the cutoff for an A+ change if the standard deviation increased to 6 points?

Tip: Understanding percentiles and z-scores is crucial for interpreting normal distributions in exams, grading, and other real-life applications!