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Suppose average score of 46 scores selected from 1,2,3,4,5 is 1.65.Is there any way to find out which are the scores that are selected.

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By the way, unique reconstruction is possible only if: all scores are 1; or one score is 2 and the rest is 1; or all scores are 5; or one score is 4 and the rest is 5. In all other cases, you can change two scores two obtain a different score set with the same average. –  Hagen von Eitzen Oct 3 '12 at 9:50
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Since $46\cdot 1.65$ is not an integer, apparently the average score $1.65$ was obtained from a precise calculation for $\bar x =\frac1{46}\sum x_i$ by rounding to two decimal places. That is, we may assume that the true value is $1.645\le \bar x< 1.655$. This leads to $75.67\le \sum x_i <76.13$ and since $\sum x_i$ must be an integer, we conclude $\sum x_i=76$.

If we assume that among the 46 numbers there are exactly $n_1$ occurance of 1, $n_2$ occurance of 2 etc., then we find that $$\tag176 = \sum x_i = n_1+2n_2+3n_3+4n_4+5n_5$$ and of course $$\tag246 = n_1+n_2+n_3+n_4+n_5.$$ Subtracting $(1)-(2)$ gives $$\tag3n_2+2n_3+3n_4+4n_5=30.$$ There are mqany possible solutions in nonnegative integers for $(3)$ and they all happen to lead to solutions of the original problem. For example, there might be 30 times 2 and 16 times 1. Or there might be 7 times 5 and once 3 and 38 times 1. Or three each of 2, 3, 4, 5 and 34 times 1. Or, or, or, ...

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You’ve probably interpreted the problem correctly, but I’d be more impressed by a student who pointed out that the average can’t be $1.65$. –  Brian M. Scott Oct 3 '12 at 9:28
    
@BrianM.Scott: Maybe. A clever student might even say that an average of 1.65 implies that 666 of the 46 scores taken from $\{1,2,3,4,5\}$ are 42. –  Hagen von Eitzen Oct 3 '12 at 9:35
    
@HagenvonEitzen so to be precise many possible solutions,thnks anywaz for answering –  iJay Oct 3 '12 at 9:40
    
@Hagen: No maybe about it: I would be more impressed! I might be amused by your hypothetical Besserwisser. I’m definitely not impressed by the person who composed the problem. –  Brian M. Scott Oct 3 '12 at 9:40
    
@BrianM.Scott: I agree that the problem should be better formulated. If the problem statement had 1.7 instead of 1.65, we couldn't even find the sum (could be 76 up to 80). If it had the number 2 without knowing it is actually a rounded number, we'd think we can conclude the sum is 92. Then again I am fond of problems involving un-rounding given numbers, e.g. if an opinion poll states that 17% like A and 33% like B and the rest likes C, I start laughing because it is quite likely that they asked only 6 (or maybe 12) people. –  Hagen von Eitzen Oct 3 '12 at 9:48
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If the average of $46$ numbers is $1.65$, then the sum of those numbers must be $46\cdot1.65=75.9$. Is it possible, then, that all of the numbers come from the set $\{1,2,3,4,5\}$?

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