For the purposes of the GRE, it's sufficient to think of Standard Deviation as the Average Distance from the Mean. Here's what I mean:

Consider these two sets: Set A {7,9,10,14} and set B {1,8,13,18}
The mean of set A = 10 and the mean of set B = 10
How do the Standard Deviations compare? Well, since the numbers in set B deviate the more from the mean than do the numbers in set A, we can see that the standard deviation of set B must be greater than the standard deviation of set A.

Alternatively, let's examine the Average Distance from the Mean for each set.

Set A {7,9,10,14}
Mean = 10
7 is a distance of 3 from the mean of 10
9 is a distance of 1 from the mean of 10
10 is a distance of 0 from the mean of 10
14 is a distance of 4 from the mean of 10
So, the average distance from the mean = (3+1+0+4)/4 = 2

B {1,8,13,18}
Mean = 10
1 is a distance of 9 from the mean of 10
8 is a distance of 2 from the mean of 10
13 is a distance of 3 from the mean of 10
18 is a distance of 8 from the mean of 10
So, the average distance from the mean = (9+2+3+8)/4 = 5.5

IMPORTANT: I'm not saying that the Standard Deviation of set A equals 2, and I'm not saying that the Standard Deviation of set B equals 5.5 (They are reasonably close however).

What I am saying is that the average distance from the mean can help us see that the standard deviation of set B must be greater than the standard deviation of set A.
More importantly, the average distance from the mean is a useful way to think of standard deviation. This model is a convenient way to handle most standard deviation questions on the GRE.

More here:

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Definition

Lets understand the formula of Standard Deviation first (σ):
Variance = (SD)\(^2\) =σ\(^2\) = [∑\(( Xi - X )^2\) / N ]

Variance is the average of the squared differences from the mean, which basically a measure of how spread out numbers are.

So lets take the first case:

Case 1

i = 3

Numbers = 10, 20 , 30.

Xm = \(\frac{10 + 20 + 30}{3}\) = 20

Variance = \(\frac{(10-20)^2 + (20-20)^2 + (30-20)^2}{3}\) = 66.667

SD = σ = \(\sqrt{Variance}\) = 8.16

Case 2

i = 7

Numbers = 10, 20 , 20, 20 ,20 ,20 , 30.

Xm = 20 (Same as the previous one)

Variance = \(\frac{(10-20)^2 + (20-20)^2 + (20-20)^2 +(20-20)^2 +(20-20)^2 +(20-20)^2 +(30-20)^2}{7}\) = 28.57

SD = σ = \(\sqrt{Variance}\) = 5.34

So quantity A is greater.
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Thanm you for the response.
Isn't there a GRE way to solve this type of problems??

Posted from my mobile device

If by GRE way you mean quick intuition, then yes. Standard deviation is simply a measure of spread of the population. In the two cases you stated above you can see in the second case many values are basically a repetition of the mean value which suggests that it will have a way lower SD. So someone who knows the concept of SD will take 5-10 secs to answer this question.
Soumya
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So if I'm getting you right,
Instead of above two sets,
if we have

Set A : 10,20,30
Set B : 10,12,13,20,26,29,30

The SD of Set B will be greater then SD of Set A.
Please correct me if I'm wrong.

Nope. No conclusion can be derived from the above sets because in this case you have to do all the calculation to get it right, there is no place of applying any intuition and this type of question has very small chance of appearing in actual GRE exam. However, the first question you posted looks like a good GRE question (albeit an easy one).
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I agree totally with soumya regarding the example just poste by the student.

In this latter scenario you can not say which SD is higher unless you perform calculation

However, regarding the main topic, conceptually you can achive the solution without any calculation.

We do have set A 10,20,30

Set B 10,20,20,20,20,20,30

Now if you consider this simple concept



From this you can infer that the SD of the first set is a little bit higher of the second one because it has LESS 20' in there. Considering that in both sets 10 and 30 are equal, because just present the gist of the problem boils down to the presence of the 20'. In the second set we have MORE 20'. As such, the SD is more "diluted", thinner.

The first set has a SD higher. For this reason A is the answer.

Hope this helps
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There is no shortcut to look at a set and get its SD. You have to use the formula.

But if two sets have the same mean and range as you example is then, you can look at how many values are nearer to mean in each set, more the numbers nearer indicates smaller SD.

Great explanation @GreenlightTestprep
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What a way to throw light on this concept. U turned my light green about SD.

see the video above by Brent from Greelighttestprep

Regards
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