是牛人就进！（依然没有解决）

shiancy

proving sd <= 1/2 range

1. for 2 elements, a1 = mean - 1/2 range, a2 = mean + 1/2 range

sd = { [(a1-mean)^2 + (a2-mean)^2] / 2 } ^(1/2)

   = { [1/4 range^2 + 1/4 range ^2] / 2 } ^(1/2)

   = 1/2 range

2. if for N elements, it is proved sd(N) <= 1/2 range(N)

for N+1 elements, drop a1, get N elements from a2 to a(N+1)

sd(N) <= 1/2 range(N) <= 1/2 range(N+1), name this sd(N) as sd.a1

similarly, sd.a2 <= 1/2 range(N+1), sd.a3 <= 1/2 range(N+1), ......

sd.a1 ^2 + sd.a2 ^2 + ... + sd.a(N+1) ^2 <= 1/4 range(N+1) ^2 *(N+1)

expand left side above, we have [ (a1-mean) ^2 /N ] for N times, because it appears in all sd.ax except sd.a1

(a1-mean) ^2 /N *N + (a2-mean) ^2 /N *N + ... + ( a(N+1)-mean ) ^2 /N *N <= 1/4 range(N+1) ^2 *(N+1)

(left side above) / (N+1) <= 1/4 range(N+1) ^2

sd(N+1) = (left side above) ^(1/2) <= 1/2 range(N+1)

proof is done.

Cantaloupe

还好GMAT考得不是这种数学，要不然偶就直接放弃了...

LS很强的说...