Devious Deviation

[Alexander]

Hi -

I don't understand the explanation. I understand Set B standard dev is 0. Why is having a 0 in the 'new' set critical?

Thanks,
Alexander


Set A: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Set B: 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Set C: –10, –9, –8, –7, –6, –5, –4, –3, –2, –1

Quantity A

The Standard Deviation of the set formed from the combination of Sets A and C

Quantity B

The Standard Deviation of Set B

If we were to combine Sets A and C, we would have one set of 20 integers: –10, –9, –8, –7, –6, –5, –4, –3, –2, –1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The arithmetic mean of this new set is 0. However, none of the numbers is actually equal to 0. Thus, the Standard Deviation in this new set must be positive.

Because all of the numbers in Set B are the same, the Standard Deviation in this set is equal to zero.

The correct answer is A.

[tommywallach]

Hey Alexander,

Not sure what you mean. SD is just a function of range. When you put the two sets together they have a pretty big range. Set B has no range at all. That's all there is to the question...

-t