then the median of the numbers in set S is the average of the two middle terms. Since the
two middle terms among 3y 1 , 3y 2 , 3y 3 , ..., 3yn are the two corresponding middle terms
among y 1 , y 2 , y 3 , ..., yn that were multiplied by 3, the average of the two middle terms among
3 y 1 , 3y 2 , 3y 3 , ..., 3yn must be the average of the two middle terms among y 1 , y 2 , y 3 , ..., yn
multiplied by 3. Thus, the median of set T must be 3(12) = 36. Statement II must be true, so
eliminate (A) and (D) which do not contain II.
Consider statement III. If each number is multiplied by 3, the numbers are 3 times more
dispersed. So the standard deviation of the numbers in set T must be 3 times the standard
deviation of the numbers in set S. The standard deviation of the numbers in set T is 3(1.8) =
5.4. Statement III must be true. Statements I, II, and III must all be true, and (E) is correct.
You can also show that statement III is true algebraically. If the numbers in set S are x 1 , x 2 , x 3 ,
..., xn, then the standard deviation of the numbers in set S is
where x̅ is the average of the numbers in set S. Thus, . You saw when
considering statement I that the average of the numbers in set T is 3x̅, where x̅ is the average
of the numbers in set S. The standard deviation of the numbers in set T is