我也考到了这个题，题目的大致意思是：A set of N numbers with a symmetric distribution, the median is M，the standard distribution is P,68% of the numbers located within the median and one standard distribution, ask what percent of the numbers located within N-(m+p)?也就是说除了包括MEDIAN 和一倍方差之外的数占多少比例？
答案好象有16% 32% 64%
我选的是16%， 但是我的数学不是满分，所以可能作错了。请高手解释解题思路。
我觉得应该是16%呀，如果按照楼上的题目的话
可以理解成正态分布吧，如果可以的话，
because 68% of the numbers located within the median and on standard distribution, the remaining percentage should be 32%, and distributed equally at the two sids. Thus for N-(m+P), it should be half of the remaining percentage, that is 16%.
大家看看，这么理解对吗？
