GRE数学难点分析
这里主要是再简单讲讲大家普遍感到困难的概率和排列组合。对于四分位数，我从来没有看见考过，尽管有人宣称在机考年代出现，但是我觉得出现可能性不大，如果要参考可以去看专著或者《数学无忧之最终幻想版》，另外还有正态分布，我这里不做引申，仅仅拿一位牛人网友的讲解给大家看看而不再做自己过分逻辑和专业化的引申和推导了。另外还可以参考上述提到的一些作品，但个人认为出到这种题的几率且难到作不出的机率几乎等于0，因此不必想的太复杂，即使在GMAT（根据最近一些时候的华东区战友汇报）也没有这么变态，所以不用紧张。但是既然是难点归类，这里就放一些个人认为还是值得拓宽一下思路的东西。而概率和组合确是不能忽视的，因为这个考点是明确列在ETS的官方指导手册（第10版）里的。
（正态分布难题——选自《正态分布解密（GRE数学满分比备良药）》）
27．说人们的题中通常是服从正态分布，标准差为1之内的有多少a(给了，忘了)
percent，标准差为2内的为95percent。问一个调查mean为18.6，则在6.8到12.6
之间为多少percent.
答(95-a)/2
From presented conditions, I guess a = 68.97%, Right? Anyone remembers -- please confirmed?
Because:
For Normal Distribution we talking about, it has certain statistical characteristics:
:: P(Probability) ={mean-stdev < X < mean+stdev} = 68.27%
:: P(Probability) ={mean-2stdev < X < mean+2stdev} = 95.45%
:: P(Probability) ={mean-3stdev < X < mean+3stdev} = 99.73%
This property indicates that for a specified Normal Distribution, the area from mean-3stdev to mean+3stdev covers 99.73% probability. Understand? (you must have learned from
lecture)
For this case, mean=18.6, stdev? Suppose the stdev=1, then, according to the conclusion from above, if a point has 99.73% probability falls in 15.6(18.6-3*1) to 21.6 (18.6+3*1); in other words, if a is within 10~12, we say “a” is not a member of this Normal Distribution, or, “a” is
just a small probability(<=0.7%) case. Right?
In general, if you wanna know the P(a<X<b), you calculate it from Phi [(b-mean)/stdev] - Phi[(a-mean)/stdev] that is the correct solution, but virtually, it is impossible for you without a
Normal Distribution Function Table.
::Conclusion
- This question may not be accurate, because for one Normal Distribution, it has only one stdev, while this question presented two? At the same time, 95% is so close to 95.45%, a ? Hence, I prefer to my guess.
- On the other hand, suppose the question is correct. The answer MUST be not correct. Anyway, (a-b)/2 cannot connect with 6.8~12.6?
- My comments are just from personal point of view. I have no right
to force you to select this explanation.
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