EgmatQuantExpert wrote:

Solve any Mean and Standard deviation question under a minute- Exercise Question #44- Set A consists of 8 different integers from the list {1,3,5,7,9,11,13,15,17,19}. What is standard deviation of the set A?

1. Set A does not include 11.

2. The average of the numbers in the given list is equal to the average of numbers in set A.

Options A. Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.

C. Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statement (1) and (2) TOGETHER are NOT sufficient.

given list = {1,3,5,7,9,11,13,15,17,19}. For this list, Mean = 10 and Median = 10, Mean=Median since the elements in this list are evenly spaced.

Now we have to choose 8 integers out of the 10 odd integers given in the list, to form another set A.

Question asks for Std Dev of set A. If set A gets fixed (i.e., if we can uniquely determine the elements of set A) then we can uniquely determine the Std Dev of set A also. But if there are multiple sets of values for set A, then Std Dev of set A might not be unique.

Lets start with the statements.

(1) So if we have to pick 8 integers from the list, but 11 cannot be included.

One possible set A = {1, 3, 5, 7, 9, 13, 15, 17}

Another possible set A = {1, 3, 5, 7, 9, 13, 15, 19}

As we can see that in the second possibility for set A, deviation among values is more than that in first possibility, so Std Dev in second case will be more. So the answer is not unique. Not sufficient.

(2) So we have to pick 8 integers to form set A such that mean of set A is also 10. Which means sum of all 8 integers in set A should be = 80.

One possible set A = {1, 3, 5, 7, 13, 15, 17, 19}

Another possible set A = {3, 5, 7, 9, 11, 13, 15, 17}

In both the above cases we can see that sum is 80, but in first case values are further apart from mean that in second case. So Std Dev in first case will be more than that in second case. So the answer is not unique. Not sufficient.

Combining the statements, sum has to be 80, but 11 cannot be included in the set. Apart from 11, the 9 integers left in the list {1, 3, 5, 7, 9, 13, 15, 17, 19}. Their sum = 1+3+5+7+9+13+15+17+19 = 89. If we have to select 8 integers such that their sum = 80, then the only way is to remove '9' from the list.

So we are left with only one possibility for set A:

Set A = {1, 3, 5, 7, 13, 15, 17, 19}.

Since its a unique set, its Std Dev will also be unique. Sufficient.

Hence

C answer