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A student gets the following marks.

50 out of 100

120 out of 150

30 out of 50

In first method : I calculate the percentage as (sum of obtained marks) / (Total marks) * 100. Hence [(50 + 120 + 30) / (100 + 150 + 50)] * 100 = (200 / 300) * 100 = 66.66%

In second method : I calculate individual percentages and divide by three as:

(50% + 80% + 60%) / 3 = (190 / 3) = 63.33%

Why are the two percentages different and which one is the correct percentage?

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    $\begingroup$ Is one point in the first test equally important as one point in the last test? $\endgroup$ – 5xum Oct 9 '18 at 7:18
  • $\begingroup$ Nothing is mentioned about the importance - only the scores are different. $\endgroup$ – Nijaguna Oct 9 '18 at 7:35
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  1. First notice that the two results are close together, although you can probably create some strange case in which they aren't close.
  2. Second there is no "right" answer. Whoever creates the scoring scheme can decide how to combine the marks.
  3. Third the two answers are the result of two different weighting schemes. There are infinitely many such schemes.
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  • $\begingroup$ Upvote for point 2 alone. Every other answer assumes all points are comparable, and we don't know that. $\endgroup$ – 5xum Oct 9 '18 at 7:20
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Both method are correct, they however have a different meaning.

The first one compute the percentage as if the three test where graded the same way (ie one point equals one point in each of the tests).

The second one assume that all test are equal but graded differently. It thus scale the grading of each test and then take the mean value.

You could also add coefficient to the grades as follow: $(c_1 .50 + c_2.120 + c_3.30) / (c_1.100 + c_2 150 + c_3 50)$

Notice that the first method is taking $c_1=c_2=c_3$ the second one is $c_1=1$ $c_2=100/150$ and $c_3=100/50$.

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This is also called Simpson's paradox. To understand it a bit simply, we can use this example.

Say we have two companies, one with 10 employees and the other with 100; and each has sales of $1000. Here are the two calculations.

Sales/Employees(A) = 1000/10 = $100 Sales/Employees(b) = 1000/100 = $10

Average of Above = 100 + 10/2 = $55

Total sale: 1000 + 1000 = $2000 Total employees : 100 + 10 = 110

Average of Above = 2000/110 = 18.18

You can see the drastic difference between the two, as we have taken an extreme example.

Edit: I'd say that yes, it will depend on the scoring system.

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  • $\begingroup$ What if the two tests are equally difficult and are just scored differently? You can't be sure that one or the other method is correct, it depends on the scoring system. $\endgroup$ – 5xum Oct 9 '18 at 7:20
  • $\begingroup$ Yes @5xum, you're right. I didn't ponder on that, I was more focussed on Simpson's paradox. $\endgroup$ – PradyumanDixit Oct 9 '18 at 7:41
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Your first method would be correct. The cause for the difference between the two methods is that the second one doesn't take into account the differing amounts of total marks between assignments. This leads to an unequal weighting between assignments that are worth more or less than others. For an extreme example, say a student go the following marks:

  • 2 out of 2 (100%)
  • 20 out of 100 (20%)

The first method would give us a correct answer of $\frac{2+20}{2+100} \approx 21.57\%$.

The second method gives us an incorrect answer of $\frac{100+20}{2} = 60\%$

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    $\begingroup$ What if the two tests are equally difficult and are just scored differently? You can't be sure that one or the other method is correct, it depends on the scoring system. $\endgroup$ – 5xum Oct 9 '18 at 7:19
  • $\begingroup$ Pretty clear that the asker wanted an average of all the marks, but pedantry is welcome. $\endgroup$ – coronermclarson Oct 9 '18 at 7:26
  • $\begingroup$ It's nowhere near clear for me. Maybe he wanted an average of all scores obtained in all the tests, maybe he wanted the average mark obtained from the equally important, but differently scored, tests. $\endgroup$ – 5xum Oct 9 '18 at 7:32
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The second method is wrong, as you are calculating the average of the percentage scores, which is not what we want here. We want to calculate the total percentage of right answers of the student, which is basically what the first method depicts.

And adding up the percentage scores and dividing the result by the number of percentage scores (calculating the mean) I think only applies when you have the percentage scores calculated on the same number of [marks].

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    $\begingroup$ What if the two tests are equally difficult and are just scored differently? You can't be sure that one or the other method is correct, it depends on the scoring system. $\endgroup$ – 5xum Oct 9 '18 at 7:19

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