To find the minimum score required for an A grade, we need to determine the score that corresponds to the top 12% of the distribution.

Since the scores are normally distributed, we can use the Z-score formula:

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

where:

• $X$ is the score we need to find,
• $\mu = 65.7$ is the mean of the distribution,
• $\sigma = 8.9$ is the standard deviation.

Step 1: Determine the Z-score for the Top 12%

The top 12% corresponds to the 88th percentile of the normal distribution. We look up the Z-score that corresponds to the 88th percentile, which is approximately:

$Z \approx 1.175$

Step 2: Solve for $X$ Using the Z-score Formula

Rearrange the Z-score formula to solve for $X$:

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

Substitute the values:

$X = 1.175 \cdot 8.9 + 65.7$

Step 3: Calculate $X$

$X \approx 1.175 \times 8.9 + 65.7 \approx 10.4575 + 65.7 \approx 76.2$

Rounding to the nearest whole number, the minimum score required for an A grade is:

$\boxed{76}$

Would you like further explanation or have questions?

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Here are some related questions you might consider:

1. What score would be required to secure a B grade?
2. How do you find the score that corresponds to the bottom 8%?
3. How would you find the score threshold for a C grade?
4. What percentile corresponds to the score of 65.7?
5. How would changes in mean or standard deviation affect grade cutoffs?

Tip: Remember, percentiles indicate the percentage of scores below a specific value in a normal distribution.