A statistics professor gives a test and finds that the scores are normally distributed with a mean of 25 (out of 100) and a standard deviation of 5. She plans to curve the scores in one of two ways.
1) She could add 50 points to each grade.
2) She could use the following scheme:
A: Top 10% earn A's
B: Scores above the bottom 70% and below the top 10% earn B's
C: Scores above the bottom 30% and below the top 30% earn C's
D: Scores above the bottom 10% and below the top 70% earn D's
F: Bottom 10% earn F's
If the professor uses the first method, what would the new mean and standard deviation be? Is this a fair method for curving?
If the professor uses the second method, what are the numerical limits for each letter grade?
Which method do you think is fairer? Why?
Just remember, under 1) the distribution is still "tight" around the mean.
On 2) 10 precent get A's, and 10 percent get F's. Did they deserve them?
Fairer? Strickly opinion.
Adding 50 points to each score only changes the mean. These tests are only ordinal scales.
http://www.members.cox.net/dagershaw/lol/TwoPlusTwo.html
The cutting points for grades are arbitrary and can vary from teacher to teacher and subject to subject.
1) If the professor adds 50 points to each grade, the new mean would be 25 + 50 = 75. The standard deviation would remain the same at 5.
To check if this method is fair, we need to consider the effect on the scores. Adding 50 points to each grade would greatly inflate the scores. It would not accurately reflect the students' actual performance or differences among them. As a result, this method is not fair in terms of accurately representing students' abilities.
2) To determine the numerical limits for each letter grade using the second method, we need to calculate the z-scores corresponding to the given percentiles:
Top 10%: z-score = invNorm(0.90) ≈ 1.28
Bottom 10%: z-score = invNorm(0.10) ≈ -1.28
Bottom 30%: z-score = invNorm(0.30) ≈ -0.52
Top 30%: z-score = invNorm(0.70) ≈ 0.52
Using these z-scores, we can find the corresponding raw scores:
A: scores ≥ μ + (1.28 × σ) = 25 + (1.28 × 5) = 31.4 (rounded to 31)
B: 31 < scores < 37
C: 19 < scores < 25
D: 6 < scores < 19
F: scores ≤ 6
In this method, the numerical limits for each letter grade are:
A: scores ≥ 31
B: 31 < scores < 37
C: 19 < scores < 25
D: 6 < scores < 19
F: scores ≤ 6
To determine which method is fairer, we need to consider the goals and purpose of grading.
If the goal is to accurately reflect the students' performance and distinguish between different levels of achievement, the second method is likely fairer. It takes into account the distribution of the scores and assigns different letter grades based on percentiles, which gives a more nuanced representation of students' abilities.
On the other hand, if the goal is to boost all students' grades without regard to their actual performance, the first method of adding 50 points would be perceived as "fair" by those who benefit from the score inflation. However, it may not accurately reflect the students' abilities and could disadvantage those who performed well initially.
Overall, fairness in grading depends on the purpose, goals, and philosophies of the educational institution and the intended outcomes of grading.
To find the new mean and standard deviation if the professor uses the first method (adding 50 points to each grade), we can simply add 50 to the mean, while keeping the standard deviation the same.
New mean = old mean + 50 = 25 + 50 = 75
New standard deviation = old standard deviation = 5
Therefore, with the first method, the new mean would be 75 and the standard deviation would remain 5.
Now let's move on to the second method. The professor wants to assign letter grades based on specific percentage ranges. Here are the numerical limits for each letter grade:
A: Top 10% earn A's
B: Scores above the bottom 70% and below the top 10% earn B's
C: Scores above the bottom 30% and below the top 30% earn C's
D: Scores above the bottom 10% and below the top 70% earn D's
F: Bottom 10% earn F's
To calculate the limits for each grade, we need to determine the corresponding scores at each percentile. From the information given, we know that the mean is 25 and the standard deviation is 5.
Using a normal distribution table or statistical software, we can find the z-scores that correspond to the desired percentiles.
For example, to find the score that corresponds to the top 10% (A grade), we find the z-score at the 90th percentile. The z-score at the 90th percentile is approximately 1.28.
Using the formula z-score = (x - mean) / standard deviation, we can rearrange it to find the desired score:
x = mean + (z-score * standard deviation)
x = 25 + (1.28 * 5)
x = 25 + 6.4
x = 31.4
Therefore, the limit for an A grade is a score above 31.4.
Similarly, we can calculate the limits for the other grades:
B grade:
Bottom 70% = 11th percentile = z-score of approximately -1.28
Top 10% = 90th percentile = z-score of approximately 1.28
Using the formula, we can calculate the limits:
Bottom limit for B grade = 25 + (-1.28 * 5) = 18.6
Top limit for B grade = 25 + (1.28 * 5) = 31.4
C grade:
Bottom 30% = 35th percentile = z-score of approximately -0.52
Top 30% = 70th percentile = z-score of approximately 0.52
Using the formula, we can calculate the limits:
Bottom limit for C grade = 25 + (-0.52 * 5) = 22.4
Top limit for C grade = 25 + (0.52 * 5) = 27.6
D grade:
Bottom 10% = 10th percentile = z-score of approximately -1.28
Top 70% = 70th percentile = z-score of approximately 0.52
Using the formula, we can calculate the limits:
Bottom limit for D grade = 25 + (-1.28 * 5) = 18.6
Top limit for D grade = 25 + (0.52 * 5) = 27.6
F grade:
Bottom 10% = 10th percentile = z-score of approximately -1.28
Using the formula, we can calculate the limit:
Bottom limit for F grade = 25 + (-1.28 * 5) = 18.6
Now, let's discuss which method is fairer. This is a subjective question and can depend on various factors and perspectives.
The first method (adding 50 points to each grade) simply shifts the distribution upwards by a fixed amount. This can provide an equal increase to all students' scores but doesn't necessarily reflect their relative performance compared to their peers. For example, if a student originally scored 80 (out of 100), their score would increase to 130. However, if another student originally scored 95, their score would increase to 145. This method may not accurately represent the comparative performance.
The second method takes into account the distribution of scores and assigns grades based on relative performance. It differentiates between top, middle, and bottom performers. By following the specified percentage ranges, it may provide a more meaningful representation of students' performance compared to their peers.
Overall, the fairness of a grading method depends on the specific context, objectives, and values of the evaluator.