各位战友，小弟在网上找了许久也没有看到GWD31套数学的讨论，很多都是讨论VERBAL的，但数学里很多题目知道答案却还是不知道怎么做，（过程！！）想在此向给位高手求救！！谢谢了~~
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Q2:
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 ?
A. 16%
B. 32%
C. 48%
D. 84%
E. 92%
Answer: D
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Q28:
A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is it possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students assigned to it?
(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it.
(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: B
第一题,去找一下标准差和均值的关系的物理意义就知道了.那个68,就是指m-到m+有68%定额数据,所以很明显,比m+小的就是50%+1/2*68%=84% 我觉得这东西中国老师很少讲,反而美国老师讲得特明白.我是来美国后在一门 research里学的,汗啊...
第2题,别听他忽悠.就是已知3<m<13<n,问n能不能被m整除.然后看眼1和2,想一想就OK了.
1,已知道3n可以被m整除.可以想,如果m是6,n是2的倍数,却不是 3的倍数,那么3n是可以,但是n就不行了......
2,对比1,一看,就是3和13的区别,"江湖经验"丰富的兄弟们估计一眼就该看出美国人的小诡计了.2乘了个13,乘个质数,其实相当于没乘,13n能被m整除的话,那n肯定本身就能被m整除了.
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