正态分布解密（数学满分比备良药）

作者：野火 2002-4-20 1:09:52 www.MicroEdu.com The same question came out today again, for your info.

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

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作   者：野火
标   题： Re:求教一道几经正态分布题
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Hey pal,

Sorry for delay 'cause I just saw you question a few minutes ago.
For no.1, I just can tell you the key is absolutely NOT true.

From presented conditions, I guess a = 68.97%, Right? Anyone
remebers -- pls 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% probabilty 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 Noraml
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 explaination.

- Generally speaking, I suggest you make a drawing, locate the mean
and stdev. That may be helpful for you to quickly solve the problem.

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No question is too silly to ask, but anserw all the questions is too
silly.
　

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