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标题: TS-3_2 [打印本页]

作者: xiaoyue 时间: 2005-9-26 13:00 标题: TS-3_2

a certain characteristic in a large population has a distribution that is symmetric about the mean m. if 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m+d?

AK: 84%

how to figure out?

作者: sammen 时间: 2005-9-26 13:32

distribution that is symmetric 关键是这句话.数目这个数列的分布是关于中位数对称的.D就是偏差,所以,小于中位数+正偏差的数目的百分比就是50%+(68%/2)=84%

作者: luoye1 时间: 2005-9-27 06:11

你画一个水平直线，A-------B-----M------C--------D.

68%在标准差里面，也就是在BC中间，因为是均匀分布,M是中位,所以 BM=MC=34%, AB=CD=16%.

原题问得是what percentage小于标准差的范围,就是求AC的百分比,就是84%

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