Kaplan challenge problem!

Edit: Removed due to crappiness of solution.

why is that?
from the very beginning we must assume that the data (U,V,W,X,Y,Z) is not distributed abnormally here. We must assume that the samples selected are also samples within the arrayed (ascending or descending) interval for the mean of population of two samples to be equal to the mean of sampling distribution, otherwise the solution offered below is devoid of any sense.

next to assuming so many parametric factors about central tendency, we must assume also that the population with which we compare two samples - the population of A,B,C is also not abnormally distributed!

this question is total mess what's the source?

GmatMathPro wrote: Statements 1&2. F/3<E/3 and G/3<E/3 implies that (F+G)/3<2E/3 (adding the inequalities). But F+G=D, so making a substitution gives:
D/3<2E/3

dividing both sides by 2 gives

D/6<E/3

taking the square root of both sides gives

√(D/6)<√{E/3)

This is the statement we were trying to prove, so sufficient.

Ans: C