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 i$f$: all scores are 1; or one score is 2 and t$h$e rest is 1; or all scores are 5; or one score is 4 and t$h$e rest is 5. In all other cases, you can change two scores two obtain a di$f$$f$erent score set wit$h$ the same avera$g$e. – 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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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 $\{1,2,3,4,5\}$?