At a very first glance this questions is intimidating and perhaps will take long time to calculate


I will approach it this way...

Since nos are small and easy to handle.. guess the mean for each set and take the difference of
the mean and the extreme nos in the set.. This way you will see the answer should be either B or
D...

Take one by one

Mean for B = 5

variance will be = [sqr (4-5) + sqr (5-5) + sqr (5-5) + sqr (6-5)]/4 = 1/2

SD = root(0.5)

For D, mean = 5

variance will be = [sqr (6-5) + sqr (4-5) + sqr (4-5) + sqr (6-5)]/4 = 1

SD = root(1) = 1.

So, answer is B

How are you arriving at (5-5) twice

because there are 2 5s for operator B

You do NOT need to actually calculate SD on the GMAT.

What you need to understand is what's required for SD (for data sufficiency purposes) and a little bit about what SD means.

Standard deviation measures how spread out the numbers in a set are. The more tightly packed, the lower the SD; the more spread out, the higher the SD.

Let's look at the sets in the choices:

(A) {3, 3, 7, 7}
(B) {4, 5, 5, 6}
(C) {2, 5, 5, 8}
(D) {4, 4, 6, 6}
(E) {3, 4, 5, 8}

(Note: for ALL standard deviation/median/mode/range questions, arrange the terms in ascending order.)

At a very quick glance, we can see that set (B) is the most tightly packed. The two middle terms are right on the mean and the 1st and last terms are only 1 away from the mean. Since (B) is the most tightly packed, it will have the least SD.

On test day, you'll be able to tell which set has the least or greatest (as relevant to the question) SD at a glance - you're not expected to use the formula.

I understood your response

I understood your response

Agree with Stuart