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Curving Grades

Curving test scores requires converting the numeric scores to letter grades under the assumption that the test scores have an approximately normal distribution. The "curve" in question is the "bell curve" or normal distribution curve. The main activity in curving grades is finding the cutoff values (that is the lowest possible scores) that will receive an A, a B, a C, and a D.

The main information you need from the test scores to set up the curve are the mean and standard deviation, along with some indication that the test scores are normally distributed. Note that the maximum possible test score never comes into the problem.

What the cutoff values actually are depends on what percentages of A's, B's, C's and D's the instructor would like to have. Once you know the area below a certain cutoff value, then you can find the actual cutoff value by using the invNorm function on the TI 83/84 calculator.

Example 1 Suppose a 140 point test has scores which are normally distributed with mean 110 and standard deviation 7. Find the cutoff scores for an A, a B, a C and a D, if the top 12% receive A's, the next 12% receive B's, the next 12% receive C's and the next 12% receive D's. If Alice scored 116 points on the test, what would her letter grade with the curve be?

Solution: Starting from the bottom is usually easiest. Since 12% receive A's, 12% receive B's, 12% receive C's, and 12% receive D's, 12% + 12% + 12% + 12% = 48% of students will receive D's or better and 100% - 48% = 52% will receive F's. At this point, you would turn on your calculator, press "2ND" and "VARS" and then choose the "invNorm(" function. Enter 0.52 as the area below, enter 110 as the mean, and 7 as the standard deviation. You should have "invNorm(0.52, 110, 10)" on your calculator screen. Pressing "ENTER" results in 110.351075. The number you are looking for depends on the values of the mean and standard deviaion that you use. But the mean and standard deviation were not determined with a high degree of accuracy, so you should not use more than one decimal place in your answer. Hence, the lowest possible D is about 110.4.

Similarly, 12% + 12% +12% = 36% of students will receive C's or better, and 100% - 36% = 64% of students will receive less than a C. You would use the "invNorm(" function again, but the only thing you would change is the area below. When you have "invNorm(0.64, 110, 7)" on your calculator screen, you press "ENTER". The result is 112.5092116, so the lowest possible C is about 112.5.

12% + 12% = 24% of students will receive A's or B's, and so 100% - 24% = 76% will receive less than a B. Using 0.76 as the area below and keeping the mean and standard deviation the same, you would have "invNorm(0.76, 110, 7)" on your calculator screen. Pressing "ENTER" would produce 114.944118. So the lowest possible B is about 114.9.

12% of students will receive A's, and so 100% - 12% = 88% will receive less than an A. Thus the area below is 0.88. You would put this in your calculator with the mean and standard deviation unchanged. So the calculator screen would look like "invNorm(0.88, 110, 7)". Pressing "ENTER" gives 118.2249075. So lowest possible A is about 118.2

Now consider Alice's test score of 116 points. Since 116 is more than 114.9, which is the lowest possible B, Alice has enough points for a B; but since 116 is less than 118.2, which is the lowest possible A, Alice does not have enough points for an A.. Therefore, Alice's letter grade would be a B with the curve.

Example 2 Suppose a 90 point test has scores which are normally distributed with mean 75 and standard deviation 5. Find the cutoff scores for an A, a B, a C and a D, if the top 8% receive A's, the next 8% receive B's, the next 8% receive C's and the next 8% receive D's. If Bill scored 79 points on the test, what would his letter grade with the curve be?

Solution: Again, we'll start from the bottom. Since 8% receive A's, 8% receive B's, 8% receive C's, and 8% receive D's, 8% + 8% + 8% + 8% = 32% of students will receive D's or better and 100% - 32% = 68% will receive F's. The area below is thus 0.68, and we are told the mean is 75 and the standard deviation is 5. Putting these values in the "invNorm(" function gives "invNorm(0.68, 75, 5)". Pressing "ENTER" gives 77.33849401. Again, the mean and standard deviation are not determined with a high degree of accuracy, so we will not use more than one decimal place. Hence, the lowest D is about 77.3.

Similarly, 8% + 8% + 8% = 24% of students will receive C's or better, and 100% - 24% = 76% of students will receive less than a C. The area below is thus 0.76, and the mean and standard deviation remain the same, so the calculator screen will look like "invNorm(0.76, 75, 5)". Pressing "ENTER" gives 78.53151283. Hence, the lowest possible C is about 78.5.

8% + 8% = 16% of students will receive A's or B's, and so 100% - 16% = 84% will receive less than a B. So the area below is 0.84. The calculator screen should look like "invNorm(0.84, 75, 5)" before pressing "ENTER". After pressing "ENTER", the result is 79.97228945. So the the lowest possible B would be about 80.0.

8% of students will receive A's, and so 100% - 8% = 92% will receive below an A. Using 0.92 as the area below, the calculator screen will look like "invNorm(0.92, 75, 5)". Pressing the "ENTER" button gives 82.0253578. So the lowest possible A is about 82.0.

Now Bill's test score of 79 points is more than 78.5, which is the lowest possible C. So he has enough points for a C. But 79 is less than 80.0, the lowest possible B, so Bill does not have enough for a B. Therefore, Bill's letter grade would be a C with the curve.

Example 3 Suppose a 120 point test has scores which are normally distributed with mean 90 and standard deviation 11. Find the cutoff scores for an A, a B, a C and a D, if the top 5% receive A's, the next 10% receive B's, the next 15% receive C's and the next 20% receive D's. If Cindy scored 107 points on the test, what would her letter grade with the curve be?

Solution: Starting from the bottom again, since 5% receive A's, 10% receive B's, 15% receive C's, and 20% receive D's, 5% + 10% + 15% + 20% = 50% of students will receive D's or better and 100% - 50% = 50% will receive F's.

As before, we press 2ND and VARS and then choose "invNorm(" and press "ENTER". Enter 0.50 as the area below, then 90 as the mean and 11 as the standard deviation. The calculator screen should look like "invNorm(0.50, 90, 11)". Pressing "ENTER" gives the result of 90, since 90 is mean of this distribution and 50% of the data is below the mean of a normal distribution. Thus the lowest possible D is the mean, 90.

Similarly, 5% + 10% + 15% = 30% of students will receive C's or better, and 100% - 30% = 70% of students will receive less than a C. Here after "invNorm(", you would enter 0.70, then 90 and then 11. Separate these numbers by commas, add a right parenthesis after 11, and then press ENTER. The result is 95.76840561. So the lowest possible C is about 95.8.

5% + 10% = 15% of students will receive A's or B's, and so 100% - 15% = 85% will receive less than a B. After "invNorm(", you would enter 0.85, then 90 and then 11. Separate these numbers by commas, add a right parenthesis after 11, and then press ENTER. The result is 101.4007672. Thus, the lowest possible B is about 101.4.

5% of students will receive A's, and so 100% - 5% = 95% will receive less than an A. Following "invNorm(" you would enter 0.95, then 90 and then 11. Separate these numbers by commas, add a right parenthesis after 11, and then press ENTER. The result is 108.0933899. Hence, the lowest possible A is about 108.1.

Now Cindy's test score of 107 points is more than 101.4, which is the lowest possible B. So she has at least a B. But 107 is less than 108.1, the lowest possible A. So Cindy does not have enough points for an A. Therefore, Cindy's letter grade would be a B with the curve.

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