Final answer:
To find the minimum scores for an A, B, C, or D in the statistics class, we use the normal distribution to find the corresponding z-scores for each percentile rank and then convert these z-scores to actual scores using the mean of 68 and a standard deviation of 9.
Step-by-step explanation:
To find the lowest score on the assignment that would qualify a student for an A, B, C, or D in a statistics class where the assignment points are normally distributed with a mean of 68 and a standard deviation of 9, we can use the concept of z-scores and the normal distribution.
Steps to Determine the Qualifying Scores
- Identify the percentile rank for each grade category (top 10% for A, next 20% for B, middle 40% for C, next 20% for D).
- Use standard normal distribution tables, computers, or calculators to find the z-score that corresponds to each percentile rank.
- Convert the z-scores to actual scores using the given mean (68) and standard deviation (9).
The formula to convert a z-score to an actual score is: X = μ + (σ × z), where X is the actual score, μ is the mean, σ is the standard deviation, and z is the z-score.
Calculating the Lowest Qualifying Scores
- For an A (top 10%): a z-score approximately 1.28 would be used.
- For a B (next 20%): a z-score approximately 0.84 would be used.
- For a C (middle 40%): two z-scores approximately -0.25 and 0.25 bracket the middle 40%.
- For a D (next 20%): a z-score approximately -0.84 would be used.
After calculating, we will get the respective scores for A, B, C, and D which are the minimum scores needed to qualify for these grades.