(Fisrt, note that the book's solution is extremely unconvincing. The outcome would depend on the ratios that we use. See the note below to see why.)
The three sisters are names Anna, Carla, and Maria. Denote Anna's speed as As, Carla's speed as Cs, and Maria's speed as Ms. Similarly, denote Anna's thrift as At, Carla's as Ct, and Maria's as Mt. Lastly, denote Anna's warmth as Aw, Carla's as Cw, and Maria's as Mw.
Anna claims that she made 5 shirts while Carla made 2. Then Carla says that while Anna made 3 shirts, Maria made 4. Normalizing As = 1, we get Cs = 2/5 and Ms = 4/3.
Normalize At = 1. Next we are informed that Mt = 1/5 and Ct = 3 * (1/5) / 5 = 3/25.
Normalizing Aw = 1, we also find out that Cw = 1/3 and Mw = (1/3) * (1/4) = 1/12.
We are not given any directions regarding how to weigh these scores, but I assumed that for each individual, you take the inverse of her thrift score and add her warmth and speed scores to it. For example, Maria's score would be Ms + (1/Mt) + Mw = 4/3 + 1/(1/5) + 1/12 = 4/3 + 5 + 1/12 = 6.xxx.
Similarly, Carla's score would be Cs + (1/Ct) + Cw = 2/5 + 1/(3/25) + 1/3 = 136/15 = 9.xxx.
Anna's score is As + (1/At) + Aw = 1 + 1 + 1 = 3, which is actually the lowest.
So, according to my scoring guideline...
Answer: Not true
Note:
As mentioned above, the book offers not only an erroneous solution, but an extremely simplistic one at best. First, it states that the ratio of warmth is (Maria : Anna : Carla) = 3.75 : 60 : 15. This is wrong. Anna's score should be 3 times that of Carla's, but 60 is clearly not 15*3.
More importantly, the winner depends on how you take the ratios. Instead of 15 : 3 : 25 for lightness, suppose we multiply each score by 100. Then the ratio would be 1500 : 300 : 2500. Now calculate each person's total score. Maria's score is 20 + 1500 + 3.75 = 1523.75. Anna's score is 15 + 300 + 60 = 375, which clearly indicates that she performed worse than Maria.
The book's solution seems to have resulted from a misguided understanding of ratios. We cannot arbitrarily standardize ratios, add up the individual values, and use the outcome as a standard for comparison.
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