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.
given average score,find the selected numbers
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0By 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. – 2012-10-03
2 Answers
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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0You’ve probably interpreted the problem correctly, but I’d be more impressed by a student who pointed out that the average can’t be $1.65$. – 2012-10-03
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0@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. – 2012-10-03
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0@HagenvonEitzen so to be precise many possible solutions,thnks anywaz for answering – 2012-10-03
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0@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. – 2012-10-03
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0@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. – 2012-10-03
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0@Hagen: You too, eh? I once had great fun deconstructing a set of student ratings of courses at the school where my father taught and demonstrating that most of them were almost certainly based on tiny (voluntary) samples. – 2012-10-03
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 $\{1,2,3,4,5\}$?