Think about the mean as a line going through a space showing the constant average of a given set of numbers.
Looking at the sets, you can clearly see that set || is not scattered around this mean, while set ||| is scattered the most.
Note that the farther a set is scattered around the mean, the higher the standard deviation.
--> Answer C is correct

Good Job!!!

The following sets each have a mean of 10 and the standard deviations are given as variables.

Set I = {7, 8, 9, 11, 12, 13}, standard deviation = P
Set II = {10, 10, 10, 10, 10, 10}, standard deviation = Q
Set III = {6, 6, 6, 14, 14, 14}, standard deviation = R

Rank these three standard deviations from least to greatest.

A. P, Q, R

B. P, R, Q

C. Q, P, R

D. Q, R, P

E. R, Q, P

Set II has the lowest deviation from the mean because all the numbers equal the mean.

Set I the mean is 10 and the deviation 3,2,1,1,2,3 (10-7,10-8) etc..

Set III is 6,6,6,14,14,14 so it means 4,4,4,4,4,4

So Q, P, R

Answer choice C

without calculating SD, there is any other way....

No because the question arounds the SD property

Since the numbers in set II are the same, the standard deviation of that set is 0.
In other words, Q = 0.

Since all standard deviations are greater than or equal to 0, we know that Q is the smallest standard deviation, which means the correct answer is either C or D.

For the remaining two sets, it's sufficient to think of Standard Deviation as the Average Distance from the Mean (see the video below for more on this)

For set I, we can see that:
7 is 3 away from the mean of 10.
8 is 2 away from the mean of 10.
9 is 1 away from the mean of 10.
11 is 1 away from the mean of 10.
12 is 2 away from the mean of 10.
13 is 3 away from the mean of 10.

For set III, we can see that:
6 is 4 away from the mean of 10.
6 is 4 away from the mean of 10.
6 is 4 away from the mean of 10.
14 is 4 away from the mean of 10.
14 is 4 away from the mean of 10.
14 is 4 away from the mean of 10.

At this point, we can see that the average distance from the mean for set I will be smaller than the average distance from the mean for set III.
This means, the standard deviation of set I (aka P) will be less than the standard deviation of set III (aka R)

So the ordering is as follows: Q < P < R

Answer: C

RELATED VIDEO

For the purposes of the GRE, we can use an informal definition of standard deviation where we consider standard deviation to be the average distance each element is away from the mean of the set.
See the video beneath my solution above.

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