Bunuel wrote:
Official Solution:
(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 than 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. Not 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 \(SD(A) \gt SD(B)\). Sufficient.
Answer: C
Hi Bunuel,
Can you please elaborate statement 2 a little more? Even if x is 1 or higher, shouldn't the SD of the set B = the SD of set A as Y which is 3 is the same value as mean and the spread is pretty much the same. Thanks.