I am not sure but is it 0 and 0 ?
as sum( sqaures of ( x-x' ) ) for 0 and 0 is the lowest.

Adding 2 numbers to the list actually changes the average unless both
numbers are 6. We don't have an option (6 and 6) but lets assume that
the average is in the vicinity of 6 (atleast looking at the choices).

Since SD = sum (square (x - avg)) / N, we need to choose the numbers
that have the least value for square (x - avg).

If we consider the average as unchanged, the numbers that yield the
least sum = 0 and 6. This would cause the numerator to increase by
just 36.

-6 and 0 increases sum by 140
0 and 0 increases sum by 72
0 and 6 increases sum by 36
0 and 12 increases sum by 72

You're right. I accidentally left out 6 and 6. JAyhawk...you're so smart

yvonne12 wrote:A CERTAIN LIST OF 100 DATA HAS AN AVERAGE OF 6 AND A STANDARD DEVIATION OF D WHERE D IS POSITIVE . WHICH OF THE FOLLOWING PAIRS OF DATE, WHEN ADED TO THE LIST , MUST RESULT IN A LIST OF 102 DATA W STANDARD DEVIATION LESS THAN D?
-6 AND 0
0 AND 0
0 AND 6
0 AND 12

We should keep in mind that the closer the data points are to the mean, the "lower" the standard deviation. Thus, since the mean is 6, when we add two additional values of 6 to the data set, the standard deviation will decrease.

Note: The answer choice C also contains one value which is equal to the mean and we might be wondering why this is not the correct answer choice. The reason is that, if all the other values in the set are very close to the mean (say 5.99), then the other value we are adding to the set (which is 0) will actually increase the standard deviation even though one of the values we are adding to the set is equal to the mean. Since we are looking for a choice which MUST decrease the standard deviation, we must choose E.

Answer: E