I would say I only

In I u r subtracting a constant from each element which has no effect on SD

In III u r taking the absolute value but we dont know how many elements are negative or positive and this may affect SD. It may stay the same or change.

In II SD could change.

I am taking it that this i a must be true question -> A choice that will be true no matter what the individual elements are

OA?

I think the answer is D.

since:
-s, -t, -r (if +s) => 0, s-t, s-r
isn't the same as I?
and III is completely different.
so answer is D.

I think the answer is D.

Yep and u deserve a +20 or +30 unlike me.

U multiply by -1 SD remains same as the original

Add a constant(s) to the -1 multiplied set and the sd remains the same

So 0, s-t, s-r has the same SD as r,s,t

This was a real good one...... Recognizing the sequence of operations was the key and I missed it...

I was going to try 1,2,3 but since I saw the third option, I tried

r = -1
s = 0
t = 1

1) -3,-2,-1 SD =1
2) 0 , -1, 1 SD =1
3) 1, 0, 1 SD is not 1

So, 1 & 2

crmaya,
i couldnt get the ques .pls explain...this is gng above my head. wht stem wants to say here.

Vivek,

Option 1 -> Add a constant SD does not change

Option II->

When u mutiply the set by -1 the SD does not change

Take s,r,t multiply by -1 you get -s,-r,-t (same SD as s,r,t)

Now add +s (constant) to each of the three elements to get 0,s-r,s-t . This set also has the same SD as s,r,t.

Option III

SD could change since we dont know how many of the elements are negative versus posititve in the list.

Try with 2 numbers negative versus one number negative etc...

Regards,
CR