To solve this, we need to compute the numerical limits for each letter grade based on the provided normal distribution, with a mean of 31 and a standard deviation of 8.2.

Steps:

1. Determine the Z-scores for the given percentiles (15%, 30%, 70%, 85%, etc.).

   • The Z-score tells us how many standard deviations a particular value is from the mean.

2. Convert Z-scores to actual test scores using the formula: $X = \mu + Z \cdot \sigma$ Where:

   • $X$ = test score
   • $\mu = 31$ (mean score)
   • $\sigma = 8.2$ (standard deviation)

Percentiles:

• For A (Top 15%): We need the Z-score for the 85th percentile.
• For B (Between bottom 70% and top 15%): Find Z-scores for the 30th and 85th percentiles.
• For C (Between bottom 30% and top 30%): Z-scores for 30th and 70th percentiles.
• For D (Between bottom 15% and top 70%): Z-scores for 15th and 30th percentiles.
• For F (Bottom 15%): Z-score for the 15th percentile.

I'll now calculate the Z-scores and the corresponding intervals.Here are the computed numerical limits for each grade based on the given normal distribution:

• A (Top 15%): Scores above 39.50
• B (Above bottom 70% and below top 15%): Scores between 35.30 and 39.50
• C (Above bottom 30% and below top 30%): Scores between 26.70 and 35.30
• D (Above bottom 15% and below top 70%): Scores between 22.50 and 26.70
• F (Bottom 15%): Scores below 22.50

Would you like more details or further clarification on any part?

Related Questions:

1. How do we interpret Z-scores in a normal distribution?
2. How can we compute percentile ranks for other distributions?
3. What is the impact of increasing the standard deviation on grade cutoffs?
4. How do we calculate Z-scores for other percentiles?
5. How is a curved grading system different from a fixed grading system?

Tip: Z-scores are a powerful tool to standardize values and compare them across different normal distributions.