Q1 - The score on the final exam had equal weight with the score on Exam 2 in computing the final score (Yes/No)?Just thought I'll add my solution here for the first question (yes/no) - since I didn't see anyone attempt something similar.
Step1: Let's assume x, y and z as the weights for each exam that's used to calculate the final score.
By definition x + y + z, i.e., sum of the weights should be equal to 1.
Eq.1) x + y + z = 1
Step 2: Next let's take 2 samples from the table given. Feel free to take any sample here, but for ease of calculation I'll be taking samples where the final score is a whole-number. I'll be multiplying the intermediate scores (exam1, exam2, etc) by weights to express the final score.
Sample 1: Russel - Intermediate Scores: 51, 69, 72; Final Score: 66
Eq. 2) 51x + 69y + 72z = 66
Sample 2: Nguyen - Intermediate Scores: 70, 74, 72; Final Score: 72
Eq. 3) 70x + 74y + 72z = 72
Step 3: Since, we're only interested in weights for exam 2 and final exam, let's simplify the equations 2 and 3 using equation 1 to remove the scores for Exam 1. To simplify equation 2) - multiply equation 1 by (x's co-efficient; 51) and subtract from equation 2. Perform same for equation 3.
Subtracting 51*Eq. 1 from Eq. 2
Eq. 4) 18y + 21z = 15 (Divide by 3 further)
6y + 7z = 5
Subtracting 70*Eq. 1 from Eq. 3
Eq. 5) 4y + 2z = 2
Step 4: Since, we're asked to determine if the weights
y and
z are same, we can use
proof by contradiction, i.e., assume they are the same and prove by contradiction that they are not. This is the easiest step and can be done much earlier. But, I've included it here to run through the reasoning more clearly. Applying the logic we get the following -
Eq. 4.1) 13y = 5
Eq. 5.1) 6y = 2
We can see that we're getting two values of y, which isn't possible. Hence, we can conclude that the weights for exam 2 and final exam are not the same.