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| − | Let A=total amount of wages | + | Let A=total amount of wages |
| − | W=avg wage | + | W=avg wage |
Wd=avg difference between individual's wage and the avg wage | Wd=avg difference between individual's wage and the avg wage | ||
Latest revision as of 16:52, 16 October 2008
Let A=total amount of wages W=avg wage Wd=avg difference between individual's wage and the avg wage
E[Wi]=W1+W2+...+Wn=W E[Wdi]=[(W1-W)+(W2-W)+...+(Wn-W)]/n=Wd
B/c A=W1+W2+...Wn= $ \sum_{i=1}^n W_i $ and A=$ \sum_{i=1}^n W_i + W_di $ So I conclude it is a fact that Wdi=0 and $ \sum_{i=1}^n W_di = 0 $ therefore there are two cases when $ \sum_{i=1}^n W_di = 0 $ holds 1.all Wdi=0 and 2.not all Wdi=0 but $ \sum_{i=1}^n W_di = 0 $
Case1: Wdi all be zero, which imply that there's no difference between avg wage and each individual's wage because it said in the quote that "people have wage that above avg" "lower than avg", therefore this case doesn't apply to the situation in the quote.
Case2: as it said in the quote that no on below avg, but its ok for people above avg, which imply that all Wdi < 0 therefore $ \sum_{i=1}^n W_di < 0 $, which is not equal to zero. since it contradicts to the fact proved earlier, it does not apply to the situation in the quote as well.
Because the both cases fail to hold, i conclude the situation mentioned in the quote doesnt exist
