For GRE purposes, Standard Deviation (SD) can often be thought of as "the average distance the data points are away from the mean."

So, with {0, 2, 4, 6, 8}, the mean is 4.
0 is 4 units away from the mean of 4.
2 is 2 units away from the mean of 4.
4 is 0 units away from the mean of 4.
6 is 2 units away from the mean of 4.
8 is 4 units away from the mean of 4.

So, the SD can be thought of as the average of 4, 2, 0, 2, and 4. The average of these values is 2.4, so we'll say that the SD is about 2.4

Note: This, of course, isn't 100% accurate, but it's all you should really need for the GRE.

Okay, so which pair of new numbers, when added to the original 5, numbers will yield a new SD that is closest to 2.4?

Well, to begin, it's useful to notice that each pair consists of numbers that are equidistant from the original mean of 4.
For example, in answer choice A, -1 is 5 units less than 4, and 6 is 5 units more than 4.
As such, add the two values in each answer choice will yield a mean of 4.

Okay, let's see what happens if we add -1 and 9 (answer choice A).
Well, -1 is 5 units away from the mean of 4, and 9 is 5 units away from the mean of 4. So, 5 and 5 will be added to 4, 2, 0, 2, and 4 to get a new SD. As you can see, this will result in a much larger SD.

Now, let's examine D (2 and 6)
Well, 2 is 2 units away from the mean of 4, and 6 is 2 units away from the mean. So, we'll be adding 2 and 2 to the five original differences of 4, 2, 0, 2, and 4. Since the average of 4, 2, 0, 2, and 4 is 2.4, adding differences of 2 and 2 should have the least effect on the original SD.

As such, the correct answer must be D

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