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Say that I have two exam grades:

$$e_1$$

and

$$e_2$$

and that exam $e_1$ is worth $p_1$ of your grade and $p_2$ percent of your grade ($p_1 + p_2 = 1 $). If on top of that weighting I promise to cut off 10% (or $\delta$ percent) of your lowest exam, what would the resulting formula for your grade be (lets try to leave all answers between 0 and 1, don't multiply by 100 to get percent). Is the following correct:

$$e_{final} = p_1e_1(1+\delta) + p_2e_2(1-\delta)$$

assuming that $e_1$ is the largest and $e_2$ score is the lowest.

It seems that the issue with this question is that "cut-off 10% your weakest exam" is not well defined, which explains my issue with trying to understand what that meant. I thought I was well defined, but it seems that its not. Then I think the best approach for this question is to try to come up with what that statement means rigorously/precisely and justify it.

However, it seems natural that whatever answer we suggest, it should have some properties. These are some that I would expect.

  1. The final grade should not be more 100%
  2. You should not force 1 "artificially" (i.e. don't cheat to force property 1).

I know that artificially is not well defined. However, I will try to propose what it means here. Say what I proposed:

$$e_{final} = min(1, p_1e_1(1+\delta) + p_2e_2(1-\delta))$$

Personally, that way of calculating things seems like cheating because I am "forcing" the upper bound, instead of the formula just having that upper bound intrinsically. I would prefer a "re-weighting" strategy, or something like that, such that, the upper bound is natural. I apologize if I don't know how to explain what the properties I want are in a super precise way, but I tried to communicate the intuitions of what correct answer would kind of look like.

Charlie Parker
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  • I'm not sure what you mean by 'cut off 10%', but that formula doesn't look right. What would you expect to happen if you got full marks on each exam? – Jessica B Dec 21 '14 at 18:29
  • @JessicaB I don't know what that means, hence the question. Someone just told me that to calculate grades they cut off 10% but I didn't know what it meant...I thought maybe it was a standard way to interpret that, but any different interpretation with justifications should be fine. – Charlie Parker Dec 21 '14 at 18:31
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    Agreed. 'cut off 10%' is not something I have ever heard used and whoever uses it should be explicitly clear what they mean by it. The formula you propose does seem a bit odd to me since it can allow for final grades over 100%. (full credit on first exam worth 95% of the grade and half credit on second exam worth 5% of the grade, cutoff of 10%) – JMoravitz Dec 21 '14 at 18:32
  • I was expecting probably a re-weighting of the two exams such that 10% of one exam was worth nothing, but 10% of another was worth that "missing" 10%? Thats what my intuition tells me to do, hence my formula, but I am not sure if it makes sense...sorry that I can't be more precise! – Charlie Parker Dec 21 '14 at 18:33
  • I think that whatever way we interpret cutting off 10%, should be consistent with always yielding a max score out of 100%. However, I am not sure how to interpret it and I can't ask however said that to me until next year...I think whatever way we interpret it, as long as it re-weights according to the specified percent and it never gives more than 100% should be good enough. However, no cheating by giving a formula that does something like $min(1,f(p))$ where f(p) is the proposed formula for grade calculation, that seems like cheating to me! – Charlie Parker Dec 21 '14 at 18:35
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    In my opinion, it should also be defined in such a way that allows it to be extended to an arbitrary number of tests in a logical way that continues to well reflect how the student performed on those other tests. – JMoravitz Dec 21 '14 at 19:02
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    @JMoravitz I don't see that it is particularly logical to assume that the method must be extendable. It is entirely possible that the method for balancing grades was chosen based on the knowledge that there were only two tests. – Jessica B Dec 21 '14 at 21:30

2 Answers2

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My best suggestion for how I might interpret the statement and a possible solution would be: In the case of $n$ tests with results labeled $t_1, t_2,\dots, t_n$ scored out of 100 and with weights labeled $w_1, w_2,\dots,w_n$ such that $\sum w_i = 1$ with each $w_i\geq 0$, by having a "cutoff of $\delta$" of the lowest appearing test score ($t_{low}$), you would modify it by $t_{low}\mapsto t'_{low} = \frac{100t_{low}}{100-\delta}$. In the case of multiple tests being tied for lowest score, from among the tests with lowest score pick the one with highest (or tied for highest) weight. In other words, a "cutoff" will change the scoring of the lowest test from being scored out of 100 to being scored out of $100 - \delta$ (for example, with $\delta = 15$ the lowest test will have been scored as though it was graded out of 85 instead of 100)

As an additional example, getting a 70 out of 100 for the lowest test, with a "cutoff of 10" the lowest test becomes now graded as a 70 out of 90, or rather a 77.7 out of 100.

This does still have the problem of if the lowest grade is higher than a $100-\delta$, they will have a greater than 100% score for that test and may end with over 100 average overall, but frankly, if someone has a 99 average before and a 101 average after, they get an A+ either way.

By modifying all scores other than the lowest score, that puts unnecessary stress on the calculations and might have incredibly adverse effects in the case of multiple tests. This simply works the same as adding a curve to the worst test.

The original weighted average was $\sum t_i w_i$ and the new weighted average is $(\sum_{t_i\neq t_{low}} t_i w_i) + t'_{low}w_{low}$

JMoravitz
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  • So if I get the same mark on all tests, I get my score boosted on the test that has the greatest weighting, but if I do very slightly worse on a test with minimal weighting then that one will be changed and my mark will end up significantly lower? – Jessica B Dec 21 '14 at 19:18
  • You are correct in noting that I did not adequately define what happens in the case that there are multiple tests which are tied for lowest score. I'll modify it to mean the test specifically with lowest score and in the case of a tie, the test with lowest score but highest (or tied for highest) weight among those tests with lowest score. And yes, I agree, this system is not particularly fair in the case of wildly varying weights. It works well for systems with equal weights for tests. I prefer curving the final grade at the end than a specific grade in the middle of calculations. – JMoravitz Dec 21 '14 at 19:28
  • Unfortunately, There is little to avoid that problem given the problem statement that we focus our attention on the test with lowest score. If it so happens that the test you did worst on (even by a little bit) was worth a weight of 0, then so be it. – JMoravitz Dec 21 '14 at 19:33
  • My comment doesn't actually depend on duplicate test results - having one result slightly higher than another could still produce wild results. – Jessica B Dec 21 '14 at 21:27
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My best guess would be

$e_1(p_1+\delta)+e_2(p_2-\delta)$.

Jessica B
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  • could you justify your answer a little bit please? – Charlie Parker Dec 21 '14 at 18:42
  • Given a weighting of, eg 40% on exam 1 and 60% on exam 2, this would change it to 50% on each. – Jessica B Dec 21 '14 at 18:43
  • @JMoravitz I would expect that if a student did equally well in the two exams then the result would stay the same. The whole things seems strange to me, and this was the only thing that made sense. If someone got, say, 63% in every piece of assessment, why would you not expect them to get 63% overall? – Jessica B Dec 21 '14 at 19:07
  • whoever down voted the answer, could they explain? I initially thought it was a good answer...however, if you think there is something crucially wrong, then i want to know and learn from it. Thanks. – Charlie Parker Dec 21 '14 at 23:16
  • @JMoravitz This formula does not decrease scores - it increases the weighting of the HIGHER score. It will leave scores alone if the two are equal, but otherwise it always increases the total. – Jessica B Dec 22 '14 at 09:03
  • @JMoravitz how does it decrease scores? sorry if its very obvious for you, I just can't see how its a bad formula. Intuitively, it seems to me that it gives more weight to the better score and a lower weight to your lower score. A reasonable thing to do...no? – Charlie Parker Dec 22 '14 at 17:40
  • I admit that I was wrong and I read the formula incorrectly as being the value of the test being changed, not the weight. You are correct that this will not hurt the scores since for $e_1\geq e_2$ you have $e_1p_1 + e_2p_2 \leq e_1p_1 e_2p_2 + (e_1-e_2)\delta = e_1(p_1+\delta) + e_2(p_2-\delta)$. I retract my earlier comments and apologize. His formula matches your intuition from the comments, but I still hold that it ought not modify tests other than the lowest. In the end, I am curious to find out what the teacher actually meant by his phrase when you find out in the future. – JMoravitz Dec 22 '14 at 19:16