Professor Cramer determine a final grade based on attendance, two papers, three

Question

Professor Cramer determine a final grade based on attendance, two papers, three major tests, and a final exam. Each of these activities has a total of 100 possible points. However, the activities carry different weights. Attendance is worth 5%, each paper is worth 9%, each test is worth 18%, and the mil is worth23%. (a) what is the average for a student with 66 on attendance, 88 on the first paper, 86 on the second paper, 83 on test1, middot82 on test 2, 81 on test 2, 81 on test 3, and 74 on the final exam? 82. (b) Compute the average for a student with the above scores on the papers, tests, and final exam, but with a score of only 19 on attendance. 80

Explanation / Answer

a) The average score is obtained by multiplying the respective scores with their weights. Therefore here it is computed as:

Average = 0.05*66 + 0.09*88 + 0.09*86 + 0.18*83 + 0.18*82 + 0.18*81 + 0.23*74 = 80.26

Therefore the answer here is 80.2 ( rounded to one decimal place )

b) Similarly now here the mean score is computed as:

Average = 0.05*19 + 0.09*88 + 0.09*86 + 0.18*83 + 0.18*82 + 0.18*81 + 0.23*74 = 77.91

Therefore the answer here is 77.9 ( rounded to one decimal place )

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