Is the standard deviation of set A is greater than standard deviation of set B ?

Thinking this way, it becomes easy to answer this questions.
As questions asks to find which SD is greater then first thing that must pop is that there must be relationship between the two sets - A and B.
So...

(1) Set A consists of consecutive multiples of 10
Clearly, not sufficient.

(2) Set B consists of consecutive multiples of 2
Clearly, not sufficient.

Hence answer must be either C or E. Here, it looks like details of two sets are available and we might be tempted to pick C which is a trap. Here's why

Together 1 and 2.
\(SD(σ) = \sqrt{variance} = \sqrt{\frac{∑(x_i - x_{avg})^2}{n}}\)

As you can see SD depends on three parameters. So, \(SD_A < SD_B\) or \(SD_A > SD_B\) because it depends on 'n - number of elements in the set'. For example
Let Set A = {10, 20, 30}, \(SD_A = \frac{2*10^2}{3}\)
Set B = {2, 4, 6}, \(SD_B = \frac{2*2^2}{3}\)
Then \(SD_A > SD_B\)

But if
Set A = {10, 20, 30}, \(SD_A = \frac{2*10^2}{3}\)
Set B = {2, 4, 6, 8, .... , 98}, \(SD_B = \frac{2(48^2 + 46^2 + ... + 2)}{49}\)
Then \(SD_A < SD_B\)

Answer E.
_________________

IMO E

(1) Set A consists of consecutive multiples of 10

No information about Set B.
Not Sufficient

(2) Set B consists of consecutive multiples of 2

No information about Set A.
Not Sufficient

Together:
Standard deviation depends on number of terms.
Since we have no idea about number of terms in both set.
So Not Sufficient.

Note: Range Rule of Determining SD states that the SD of a set is equal to 1/4 of its range.

(1) Set A consists of consecutive multiples of 10
No information about set B.
(Insufficient)

(2) Set B consists of consecutive multiples of 2
No information about set A.
(Insufficient)

Statement 1&2 together:

Case 1:
Let Set A = {10,20,30,40}
Set B = {10,20,30,40}
S.D. will be equal.

Case 2: SET A = {10,20,30,40}
Range = 40-10 = 30
SD = 30/4 = 7.5

Set B = {2,8,14,20,26,32,38}
Range = 38-2 = 36
SD = 36/4 = 9

SD of Set B is greater

Case 3: SET A = {10,20,30,40}
Range = 40-10 = 30
SD = 30/4 = 7.5

Set B = {2,8,14,20}
Range = 20-2 = 18
SD = 18/4 = 4.5

SD of Set A is greater

(Insufficient)

IMO E

Posted from my mobile device

In theory, the standard deviation of either set could be zero if one of the sets contains only one element. If we allow that possibility, the answer is immediately E.

If we disallow that possibility, then if the set of consecutive multiples of 10 is {10, 20}, its standard deviation is 5 (every element in the set is 5 away from the mean, so that is automatically the standard deviation). If you take a very large set of consecutive multiples of 2, say all the multiples of 2 between 0 and 1000, almost every value in this set will be more than 5 away from the mean, and the standard deviation will be much larger than 5. So the answer is still E.

If we knew the sets were the same size, then both statements would be sufficient together, since then the elements of A would be a greater distance apart than the elements of B (in fact, the standard deviation of A would be precisely five times as big as that of B, because the distances within A would be five times as big).
_________________

Ans - E

Standard deviations is under root of variance, which is diff of term and the av, the more this deviation is the more is SD

Statement 1
Given info on set A, but no B. Not sufficient

Statement B
Gives info on set B, but no A, Not sufficient

With Statement 1 and 2
Even if Set A is consecutive multiple of 10, and set B is consecutive multiple of 2.

It is unsure what would the variance be depending of the starting number. Set B can still be a multiple of set A, which will prove otherwise.

Posted from my mobile device

Standard deviation depends on the variance and also number of terms in the set.Here in both the statements we don't have number of terms mentioned so we can't conclude from any of the statements.

Answer E

Similar question- https://gmatclub.com/forum/which-data-s ... 93817.html

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________