AlphaCentauri4ly wrote:
I think the best way to quickly validate Q#1 is to consider an evenly spaced set of E1, E2 & F like that of Ardanin:
The Arithmetic mean of 83, 84 and 85(E2,F,E1) is clearly 84 which is equal to the final score. The only way it's possible is if either they are equally weighed or weighed more toward the average. For example: Average of the set 83,84,84,84,84,85 would still be 84 as adding average to a set would not change the average.
On the other hand, any number other than the value of mean is added to the set then the new mean will certainly deviate from the average. For example in case of E2&F weighing the same(Barring E1):
83, 83,84,84,85-> The Mean shifts from 84 to a number between 83 and 84.
Hence, this set alone proves that E2&F can't have equal weights without E1 carrying the same weight. Either way, we got both Yes and No, so we cannot determine the weights to compare.
AndrewN Was my thought process correct?
Watch those superlatives—here,
best in
best way. But yes, I agree that your thought process is sound. In keeping with my simple-is-often-better approach, I looked to the first few rows of data only to deduce that no, the exams in question cannot be given equal weight in computing the final score.
An easy way to work with averages is to set the average and then calculate the value above or below that average for each data point. In this manner, you can quickly work through a set to figure out the average, without working through lengthy calculations. For instance, say that I had the following set: {4, 9, 22, 28, 37}. Of course, I could add them up and divide by 5, or I could set the average to something that looked reasonable—say, 20—and then simply keep track of how far above or below my target average each number in the set was:
20 | {4, 9, 22, 28, 37}
20 | {-16, -11, +2, +8, +17}
Since the sum of the numbers in brackets is 0, I know the average is 20, right on the nose. What if I had selected 15?
15 | {4, 9, 22, 28, 37}
15 | {-11, -6, +7, +13, +22}
Note that the sum of the numbers in brackets is now +25. When I average this value by dividing by the five data points, I get +5. Thus, after correcting the estimate by adding 5, I get the true average: 15 + 5 = 20.
How does this relate to the question at hand? Well, we are
given the
final score, so we can set this value as our average. We can see quite clearly that the second row values will average to 84: {85, 83, 84} is {+1, -1, 0}. But if we check the top row, the picture comes out a bit different, relative to the final score of 86.5: {89, 87, 85}. That should be 87 (from {+2, 0, -2}) if everything carries equal weight. So, now the question becomes, how can the score be 86.5? If the latter two tests carry equal weight, they would average to 86, and then they would have to carry a lot more weight than Exam 1 to pull the final score down to 86.5. It seems more reasonable that that 86.5 comes from averaging the first two exam scores and then balancing that out with the final exam score: 89 and 87 average to 88; 88 and 85 average to 86.5 (1.5 from each value). This looked promising. Now, I just wanted proof by checking the third row:
67.75 | {65, 70, 68}
The average of 65 and 70 is 67.5 (2.5 from each value).
67.75 | {67.5, 68}
Yep, that one checks out. The average would be 67.75.
I know the process looks lengthy when I type everything out, but trust me, once the first domino fell, the answer followed soon after. Thank you for thinking to ask me about this one. I have to run to my next lesson!
- Andrew
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