What does it mean to "cut-off 10% your weakest exam"?

Question

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.

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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.

Answer 1

My best guess would be

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

Answer 2

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 $(\sum_{t_i\neq t_{low}} t_i w_i) + t'_{low}w_{low}$