What is the reasoning behind standardization (dividing by standard deviation)? Why does dividing a dataset by sigma make the sample variance equal to 1? Assuming a zero mean for simplicity.
What's the intuition behind this?
Dividing by the range (max-min) makes intuitive sense. But standard deviation does not.
 A: Standardizing is is just changing the units so they are in "standard deviation" units. After standardization, a value of 1.5 means "1.5 standard deviations above 0". If the standard deviation were 8, this would be equivalent to saying "12 points above 0".
An example: when converting inches to feet (in America), you multiply your data in inches by a conversion factor, $\frac{1 foot}{12 inches}$, which comes from the fact that 1 foot equals 12 inches, so you're essentially just multiplying your data points by a fancy version of 1 (i.e., a fraction with equal numerator and denominator). For example, to go from 72 inches to feet, you do $72 inches \times \frac{1 foot}{12 inches}=6feet$.
When converting scores from raw units to standard deviation units, you multiply your data in raw units by the conversion factor $\frac{1sd}{\sigma points}$. So if you had a score of 100 and the standard deviation ($\sigma$) was 20, your standardized score would be $100 points \times \frac{1 sd}{20 points}=5sd$. Standardization is just changing the units. 
Changing the units of a dataset doesn't affect how spread out it is; you just change the units of the measure of spread you're using so that they match. So if your original data had a standard deviation of 20 points, and you've changed units so that 20 original points equals 1 new standardized unit, then the new standard deviation is 1 unit (because 20 original units equals 1 new unit).
A: This stems from the property of variance. For a random variable $X$ and a constant $a$, $\mathrm{var}(aX)=a^2\mathrm{var}(x)$. Therefore, if you divide the data by its standard deviation ($\sigma$), $\mathrm{var}(X/\sigma)=\mathrm{var}(X)/\sigma^2=\sigma^2/\sigma^2=1$.
