let's imagine that a student got 0 marks in a exam. And in the next one he got 5 marks. to calculate the percentage of the new mark to the last mark, we usually use (new-last)/last*100%. but the denominator is 0. So the expression should be (5-0)/0*100%, right? But 0 as denominator will result in error in calculator. So how to calculate?
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"He scored" or "He got off the mark" might work, while "He did infinitely better" would be misleading. Alternatively, don't divide and just say "he scored 5 marks (more)". – Henry May 14 '12 at 10:36
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sorry can you be more specific? is it possible to calculate a percentage of a number to zero – Alexander Ho May 14 '12 at 10:38
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6@AlexanderHo: No, this is not possible - there is no way to give a useful percentage value here. This is because there is no number $p$ such that $p \cdot 0 = 5$, so there is also no percentage. – Johannes Kloos May 14 '12 at 11:50
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1Instead of measuring the percentage change in marks, an improvement rating out of $100$ could be given by $100(1-e^{\frac{new \ mark}{old \ mark}})$. Then as old mark $\to 0$, the rating $\to 1$, and the ratings of students who score low and improve slightly do not dwarf that of students who score fairly high then improve a lot. – Angela Pretorius May 18 '12 at 11:06
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@JohannesKloos Could you promote your comment to an answer so as to remove this question from the Unanswered queue? You could also incorporate the alternative method $100(1-\exp(-\frac{\text{new mark}}{\text{old mark}}))$ that Angela proposes. – Lord_Farin Jun 06 '13 at 15:01
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No, this is not possible - there is no way to give a useful percentage value here. This is because there is no number $p$ such that $p\cdot0=5$, so there is also no percentage.
Alternatively, as Angela proposes, instead of measuring the percentage change in marks, an improvement rating out of $100$ could be given by $100(1−\exp(-\frac{\text{new mark}}{\text{old mark}}))$. Then as $\text{old mark}\to0$, the rating goes to $1$, and the ratings of students who score low and improve slightly do not dwarf those of students who score fairly high and subsequently improve a lot.
Lord_Farin
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