60DS: there is a set of number, the mean is m, the standard deviation is R, if add a number x to this set, if r<= R?1) x=m 2) m小于x小于m+R 设 旧的集合数列mean=M, standard deviation=R 新的集合数列mean=m, standard deviation=r 插入的数是x 则 nR^2=(a1-M)^2+(a2-M)^2……+(an-M)^2 (1) (n+1)r^2=(x-m)^2+(a1-m)^2+(a2-m)^2……+(an-m)^2 (2) M-m=(a1+a2+……+an)/n-(x+a1+a2+……+an)/(n+1)=(a1+a2+……+an-nx)/n(n+1)=(nM-nx) /n(n+1)=(M-x) /n+1 (3) 由第一个条件 x=M, 得出(3)式=0即M=m 则(2) (n+1)r^2=(x-m)^2+(a1-m)^2+(a2-m)^2……+(an-m)^2=(a1-M)^2+(a2-M)^2……+(an-M)^2=nR^2 得 r^2=R^2- R^2/(n+1)< R^2 即r<R, 关键是第二个条件 M<x<M+R 由(3)M-m<0, 即M<m 看(1)-(2)=(a1-M+a1-m)(m-M)+ (a2-M+a2-m)(m-M) ……+ (an-M+an-m)(m-M)- (x-m)^2=(m-M)[2(a1+a2……+an)-nM-nm]- (x-m)^2=(m-M)(2nM- nM-nm) - (x-m)^2=n(m-M)(M-m)- (x-m)^2<0 得 (n+1)r^2> nR^2 推不出r<=R. 答案:A 欢迎讨论。 |