Stmt 1:
Avg of the master list = \((4+6+ 8+ 10+ 12+ 14+ 16 + 18+ 20 + 22)/10 = 13\)
Sum of the elements of the master list = 130
Avg of list M = (Sum of list M) \( / 8 \)
Since, Avg of the master list = Avg of list M
Sum of list M = 13*8=104
Possibilities of elements in list M such that sum=104 i.e. we need to exclude 2 elements whose sum is 26(130-104)
1) { 6, 8, 10, 12, 14, 16, 18, 20} exclude 4, 22
2) { 4, 8, 10, 12, 14, 16, 18, 22} exclude 6, 20 and so on
we get 2 different std deviation values
Not sufficientStmt 2:
Exclude 22
Possibilities of elements in list M
1) { 6, 8, 10, 12, 14, 16, 18, 20} exclude 4, 22
2) { 4, 8, 10, 12, 14, 16, 18, 20} exclude 6, 22 and so on
we get 2 different std deviation values
Not sufficientCombining 1 and 2
Master list = { 4, 6, 8, 10, 12, 14, 16, 18, 20}
sum of master list = 108
sum of list M = 104
We need to exclude 1 element from this master list such that it is 4
only 1 possibility - { 6, 8, 10, 12, 14, 16, 18, 20}
SufficientAns. C