1. ## standard deviation

If the standard deviation is a value between 0 and 1, will the variance be less than the standard deviation?

2. ## Reply to the question

Yes as far as I can tell.

If it was to be between 0 and 1 then it would be in decimal form.

Using the fact that the Standard Deviation^2 = Variance.

multiplying a decimal number by itself will end up making it less than its original value.

Not really conclusive proof at all.

3. Hint: $0 < x < 1\quad \Rightarrow \quad x^2 < x < \sqrt x$

4. I figured as much since the variance is the square of the SD... its just strange to picture the SD being greater than the variance.