The answer is E??

The average (arithmetic mean) of 100 numbers is 6, and the standard deviation is D, where D is positive. Which of the following two numbers can be added to the set so that the new standard deviation for the 102 numbers must be less than D?



Answer is E because, when you add 6 and 6, the average still remains 6 and the standard deviation decreases.

Let the standard Deviation of the set = D = √ ((a1-x)^2 + (a2-x)^2 + .......)/100), where a1,a2,...a100 are the numbers of the set and x(=6) is mean of the set
i.e .100*D*D = (a1-x)^2 + (a2-x)^2 + .......

Standard Deviation of the new set = D' = √ ((a1-x)^2 + (a2-x)^2 + ...+ (6-6)^2 +(6-6)^2)/102) - (x = mean = 6)

D' = √((100*D*D + 0 + 0)/102)
D' = √(100/102) * D
D' = 0.9X D. SO D' < D

-6 and 0
0 and 0
0 and 6
0 and 12
6 and 6

neelgandham wrote:The average (arithmetic mean) of 100 numbers is 6, and the standard deviation is D, where D is positive. Which of the following two numbers can be added to the set so that the new standard deviation for the 102 numbers must be less than D?



Answer is E because, when you add 6 and 6, the average still remains 6 and the standard deviation decreases.

Let the standard Deviation of the set = D = √ ((a1-x)^2 + (a2-x)^2 + .......)/100), where a1,a2,...a100 are the numbers of the set and x(=6) is mean of the set
i.e .100*D*D = (a1-x)^2 + (a2-x)^2 + .......

Standard Deviation of the new set = D' = √ ((a1-x)^2 + (a2-x)^2 + ...+ (6-6)^2 +(6-6)^2)/102) - (x = mean = 6)

D' = √((100*D*D + 0 + 0)/102)
D' = √(100/102) * D
D' = 0.9X D. SO D' < D

-6 and 0
0 and 0
0 and 6
0 and 12
6 and 6

Thanks! Got it!
I was not sure about answer choices A, B. But I applied the above equation to test a few scenarios and realised that it may or may not be true. This is however a MUST be true question!