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If 1 kid can eat 5 candies, 2 kids can eat 11 candies and 3 kids can eat 25 candies, then how much can 2.5 kids eat?

I was thinking you calculate the average then use that but I cant just use the average right? I was thinking the average was equal to 2 kids and then add the .5 but that doesn't seem right.

David G. Stork
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Crazyluv
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    You could do a linear regression, but I would say half a kid can't eat any candy – J. W. Tanner Aug 05 '19 at 18:19
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    I'm not sure I understand the question. Are these six different kids with different candy-eating capabilities, and you're supposed to estimate the amount of candy that a randomly selected "two-and-a-half" of them can eat? – Connor Harris Aug 05 '19 at 18:20
  • Seems silly, but you can fit a quadratic to these points. You get $p(x)=4 x^2 - 6 x + 7$. Then you can evaluate that at $x=2.5$ to get $17$ if you want, though what it means.... – lulu Aug 05 '19 at 18:26
  • Yeah, it is a silly way to put it. I just got some reference material and it doesn't seem right but I cant figure it out. I can use 2.5 as a reference point for consumption but I only have the data for consumption for 1, 2, and 3 units. The example I have uses the average as an equivalent for 2.5 unit but is that doesn't seem right. – Crazyluv Aug 05 '19 at 19:00
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    @J.W.Tanner Depends on whether or not it's the half with the mouth. – Théophile Aug 05 '19 at 20:16
  • @J.W.Tanner Depending on how the kid is cut, not only would he be able up eat the candy, but you may be able to recycle it for further consumption. Authoritative references: https://youtu.be/G1QAt4bgwys and https://youtu.be/ZdctwUTybfE – Deepak Aug 06 '19 at 00:04

2 Answers2

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Depends on what mean you are aiming for, arithmetic, geometric, harmonic, etc. In this case, I'd suggest arithmetic mean. So, we have 6 kids, eating 41 candies. So the arithmetic mean ( average), any one kid ate, is $${41\over 6}=6+{5\over 6}$$ candies per child. This makes: $$(6+{5\over 6})\cdot (2+{1\over 2})= 12+3+(1+{4\over 6})+{5\over 12}= 17 +{2\over 12} $$ candies for two and a half children.

Of course It depends on how you interpret the question. The ambiguous word is can

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I think this approach might be the right one. Let me know. $$n_{candies}=\frac{1}{2.5}\times\biggl(\frac{5}{1}+\frac{11}{2}+\frac{25}{3}\biggr)=\frac{30+33+50}{2.5\times6}=\frac{113}{15}=7.533333333333333$$

poetasis
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