The same question came out today again, for your info. [b][b]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. [b]答(95-a)/2 [b][b][b][b]Hey pal, [b][b]Sorry for delay 'cause I just saw you question a few minutes ago. [b]For no.1, I just can tell you the key is absolutely NOT true. [b]From presented conditions, I guess a = 68.97%, Right? Anyone remebers -- pls confirmed? [b]Because: [b][b]For Normal Distribution we talking about, it has certain statistical characteristics: [b]:: P(Probability) ={mean-stdev < X < mean+stdev} = 68.27% [b]:: P(Probability) ={mean-2stdev < X < mean+2stdev} = 95.45% [b]:: P(Probability) ={mean-3stdev < X < mean+3stdev} = 99.73% [b][b]This property indicates that for a specified Normal Distribution, the area from mean-3stdev to mean+3stdev [b]covers 99.73% probability. Understand? (you must have learned from lecture) [b][b]For this case, mean=18.6, stdev? Suppose the stdev=1, then, according to the conclusion from above, [b]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, [b]we say "a" is not a member of this Normal Distribution, or, "a" is just a small probability(<=0.7%) case. [b]Right? [b][b]In general, if you wanna know the P(a

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