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Q: Need Solution to Weighted Average Problem (\$20.00) ( No Answer,   1 Comment )
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 Subject: Need Solution to Weighted Average Problem (\$20.00) Category: Science > Math Asked by: vijayd-ga List Price: \$20.00 Posted: 02 Oct 2006 19:18 PDT Expires: 01 Nov 2006 18:18 PST Question ID: 770318
 ```Need Solution to Weighted Average Problem (\$20.00) The Problem: I have 3 scores s1, s2, s3 that are weighted at w1, w2, and w3 respectively to give an average score of s = (w1)(s1)+(w2)(s2)+(w3)(s3). Note: w1+w2+w3=1. s1, s2, s3 are all greater than 0 and between 0 and 100. Now, I lose the original scores s1, s2, and s3. All, I have is the average score s and the weights w1, w2, and w3. Given these four values, I want to find out what the average score would have been if I had weighted each score equally, i.e. w1=w2=w3=(1/3)``` Request for Question Clarification by livioflores-ga on 02 Oct 2006 19:57 PDT ```Hi!! If I do not misunderstood the statement of your problem, it has no a single solution but an infinite number of them. You know that: w1=w2=w3=(1/3); s1, s2, s3 are all greater than 0 and between 0 and 100 but unknown; You said you know the value of s, then: s = (w1)*(s1)+(w2)*(s2)+(w3)*(s3) = (1/3)*(s1+s2+s3) From the above the only thing you can say about s1, s2, and s3 is that: s1+s2+s3 = 3*s Since s is known you know the value of the sum of the scores s1, s2, and s3, but nothing else can be said. There are infinite different combinations of 3 numbers between 0 and 100 that satisfy: s1+s2+s3 = X = 3*s For example if s is equal to 24 then: s1+s2+s3 = 72 ; possible solutions are: s1 = 20; s2 = 10 and s3 = 42 s1 = 12.5 ; s2 = 25.3 and s3 = 34.2 etc. For a more accurate result you need to find more info related to s1, s2 and s3. Please let me know if this is what you are asking for or not. Regards, livioflores-ga``` Clarification of Question by vijayd-ga on 03 Oct 2006 04:18 PDT ```To clarify let me give an example with numbers. Imaginge I give a student 3 tests. His scores were s1=87, s2=79, s3=93. And I initially decided to weight each test as w1=.25, w2=.35, and w3=.40 respectively. Now, the final score would be s = (w1)(s1)+(w2)(s2)+(w3)(s3) = 86.6. In my grade book I make note of the final score s = 86.6 and throw away the actual scroes. Now, I realize that it might have been better if I had weighted all the tests equally, but all I have is the score s and the weights I had initially used. I am not necessarily interested in getting the values of s1, s1, and s3. All I want is given s and the initial weights I used, is there an equation that I can plug them into that will transform s into what would have been an equally weighted average. In this case the answer would be 86.33. I hope this clarifies the question.```
 ```sorry, can't be done. for proof, note that given the w1, w2, w3, and s in your clarification, both of the following sets of scores are possible: s1=87, s2=79, s3=93. mean=86.33 or s1=81, s2=81, s3=95. mean=85.67 there are, as livioflores says, an infinite number of possible scores. finite, but still more than a couple, if you know the original scores were integers. now, depending on why you want to know the mean, if you're interested in finding an upper and lower bound on the mean score, you might re-ask the question that way. assuming w1, w2, and w3 are all relatively close to each other, i suspect you could bound it within a reasonably tight range. -cab```