Here's a suggestion:
Take a sample of $n$ values, the larger $n$ the better the estimate of the true variance will be, but if you have the option to choose $n$ very large, you might as well want to just compute the minimum and maximum for the whole set already.
Compute the sample variance:
$s^2 = \frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2 \; ,$
where $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is the sample mean.
The square root of that number is the sample standard deviation and a good estimate for the population standard deviation (i.e. the standard deviation of all your data). Now, if your data are normally distributed, over 99% of your data will be within three standard deviations of the mean. which gives you a good range that you can then scale to the interval you wish. If your data are not normally distributed, you might want to take 5 standard deviations or more.