Any mathematicians in the house?

[kev222]

How's your statistics? Maybe you can help me out.

I have a list of N integers. All of the integers are in the range 0 - 255.

I compute the standard deviation of these N integers using the following calculation (which you may not need).

stddev = square_root of ((A - ((B^2) / N)) / (N - 1))

where
A = the sum of the squares of all N integers
B = the sum of all N integers
N = the number of integers

Is it possible for the standard deviation, being the square root of the average squared deviation from the mean, to fall outside the range 0 - 255?

Any help would be much appreciated. My ability to cope with abstraction is very low.

[BWilli]

i e-mailed this over to a friend of mine who is a math teacher.  not sure when he'll get back to me because he's in class right now, but when he does, and if you havent figured it out, i'll let you know what he says about it.

[kev222]

i e-mailed this over to a friend of mine who is a math teacher.  not sure when he'll get back to me because he's in class right now, but when he does, and if you havent figured it out, i'll let you know what he says about it.

Thanks man.

No, I still haven't figured it out. I suspect that it is constrained to the range 0 - 255, but I can't prove it.

[Will]

The standard deviation can't fall outside of that range. Let's consider the worst case situation. Standard deviation is basically a measure of how spread out your data is. Consider the set, S = { 0, 0, 0, 0, 0, 0, 255, 255, 255, 255, 255, 255 }

A = 390,150
B = 1,530
N = 12

((390,150 - ((1,530^2) / 12)) / (12 - 1))^(1 / 2) = 133.17

While it's not exactly a proof, you get the general idea.

[kev222]

The standard deviation can't fall outside of that range. Let's consider the worst case situation. Standard deviation is basically a measure of how spread out your data is. Consider the set, S = { 0, 0, 0, 0, 0, 0, 255, 255, 255, 255, 255, 255 }

A = 390,150
B = 1,530
N = 12

((390,150 - ((1,530^2) / 12)) / (12 - 1))^(1 / 2) = 133.17

While it's not exactly a proof, you get the general idea.

It's a good enough 'proof' for my purposes.

Cheers dude.

[BWilli]

ahh...cool...you got it.  my friend hadn't gotten back to me yet

[Jessica]

.... UCLA psych center diagnosed me with a math disorder..

lmfao.