Hi!

could you post OA?

IMO C

We have two sets with the same number of terms:
A = {x1,x2,...,xn}
B = {y1,y2,...,yn}

sample standard deviation


We want to know if standard deviation of set A is greater than standard deviation of set B.
S(A) > S(B) ?

(1) R(A) > R(B)
The range R of a set of data is the difference between the highest and lowest terms in the set.

let us plug in numbers
for example B = {-50,-40,-30,-20,-10} and A = {-51,-31,-30,-29,-10}
R(A) = 41, R(B) = 40
S(A) < S(B)
second example B = {-50,-40,-30,-20,-10} and A = {-51,-31,-30,-29,-10}
R(A) = 41, R(B) = 40
S(A) > S(B)
INSUFFICIENT

(2) sets A and B are both evenly spaced sets

for example B = {-50,-40,-30,-20,-10} and A = {1,2,3,4,5}
S(A) < S(B)

second example B = {-50,-40,-30,-20,-10} and A = {100,200,300,400,500}
S(A) > S(B)

so it is INSUFFICIENT

but

statements 1 and 2 are together SUFFICIENT
example B = {-50,-40,-30,-20,-10} and A = {100,200,300,400,500}
R(A) > R(B)
S(A) > S(B)

hope it helps!

Can you explain part '1' explanation again? I think the #s are off?

OA: C

If they are evenly spaced, the mean is always going to be equal no matter what the range.....

But what I don't get is why the SD can't be the same.

Lets say Set A is 2,4,6,8,10 range, 8, mean 6, SD (about 2 from the mean)
Set B is 2,4,6 range = 4, mean =2, SD still about 2 from the mean.

Am I missing that SD changes when you change the values (i.e. Set A would really be sqrt of 2 over 5, and Set B would be sqrt of 2 over 3) in which case, Set A has greater SD.

Is this correct?

I guess key to this problem is trying both negative and positive numbers. I realize now that "n" is the same so my previous comment doesn't make sense. Can you shed some light on how you picked this set of numbers?

Also, for the negative values, do you calculate SD in absolute terms? So for Set {-50, -40, -30, -20, -10) The mean is -30 and the SD is 20.

missrochelle asked me to respond so here's my analysis:

The key to understanding this question is knowing what Standard Deviation (SD) is. In simple language, SD = how much does X vary from the AVERAGE of the set that X belongs to. if the Avg of a Set = 3 and X = 5. X = 2 SD away from the mean. Alright now the question.

# of terms in Set A = # of terms in Set B

Question: SD A > SD B?

(1) Range of A > Range of B

This is insufficient! Range = highest value - lowest value. In this case, we are looking out for SD. SD is computed by knowing EVERY single value.

So knowing Range alone is not helpful!

Set A & B = evenly spaced
This one MIGHT look attractive because you might think that since they are evenly spaced, SD = 0. But no! The values of both sets can be VERY different since we don't know anything about the numbers within each sets.

Set A could be like 100,1000,10,000
Set B could be 2, 4, 6

or viceversa.

In each case the SD is very different!

Evenly spaced sets can be like 2,4,6 OR like that in Set A. Set A is a multiple of 100, which is evenly spaced!

Together:

Together, both statements solve our dilemma.

Since the range of A > B, we know A has bigger numbers than B.

Set A could be like 100, 1000, 10000
Set B could be 2, 4, 6

So only a set like this is true, not the other way around. This set is evenly spaced and the range of A > B.

Therefore, we know for sure that the SD of A > SD of B.

I hope this helps a little!

YES THAT HELPS SO MUCH! I was having trouble grasping these data sufficiency questions on SD and I heard they are becoming every more popular nowadays. This really helps put it into perspective (what the range, median, mean) relate to SD. The fact that you need to know ALL values helps a ton.... Key to this question is noting that they said N is the same in both. If N isn't the same (no restriction), the answer would be E, correct?