Bunuel wrote:
5. Set A={3-2x, 3-x, 3, 3+x, 3+2x}, where x is an integer. Is the standard deviation of set A more than the standard deviation of set B={3-2x, 3-x, 3, 3+x, 3+2x, y}
(1) The standard deviation of set A is positive. We know that the standard deviation of any set is more than or equal to zero. The standard deviation of a set is zero only when the set consists of identical elements. So, this statement implies that set A does NOT consists of identical elements or that x does not equal to zero. Still this statement is not sufficient to answer the question.
(2) y=3. The mean of set A is 3. Now, if \(x\neq{0}\) for example if x=1, then the standard deviation of B would be smaller that the standard deviation A, since the elements of B would be less widespread than the element of set A. But if x=0, then A={3, 3, 3, 3, 3} and B={3, 3, 3, 3, 3, 3}, so both will have the standard deviation of zero. Bot sufficient.
(1)+(2) Since from (1) \(x\neq{0}\), then adding a new element which equals to the mean will shrink the standard deviation, thus SD(A)>SD(B). Sufficient.
Answer: C.
I have a doubt that in set A if i substitute x as 1 and y=3 i get (1,2,3,4,5) and for set B it is (1,2,3,3,4,5) now which will have a greater standard deviation or will it be same?
Like the elements in B are more and the deviation in both sets from mean is same so B should have bigger SD but If use the logic adding nos same as mean decreases SD then Set A will have bigger SD