Thank you, Jim. Yes: the two questions are totally unrelated. And thanks for your suggestions!

Please note, though, that what you said: "In other words, Mathematica does not have a specific formula for the sample variance with a denominator of n−1." is wrong.

The Mathematica function "Variance" indeed holds "n-1" (the Bessel's variance) at the denominator, by default.....I thought there was another function with "N".

If you, for example, make a list of numbers:

list = {600, 470, 170, 430, 300} ;

and calculate the variance:

Variance[list]

it returns: 27130

While a Variance with "N" numbers would be: 21704

Therefore, for such result, the syntax would be:

N[Variance[list]*((n - 1)/n)]

The concept you are missing is that you need to define a probability distribution. The PDF is not the same thing as the distribution itself. Basically, you describe a TruncatedDistribution of an ExponentialDistribution. So you could write this as follows. First, check that my assertion is correct.

PDF[TruncatedDistribution[{0, 1}, ExponentialDistribution[1]], x]

You will see that the PDF is basically some constant multiplied by Exp[-x].

So now life is easy. You can compute the Variance.

Variance[TruncatedDistribution[{0, 1}, ExponentialDistribution[1]]]

You can plot the CDF and PDF.

Plot[CDF[TruncatedDistribution[{0, 1}, ExponentialDistribution[1]], 
  x], {x, -1, 2}]

You could do this even if you did not recognize the distribution as a truncated exponential. The code below shows how. The key is to wrap up the computed PDF inside ProbabilityDistribution.

Variance[ProbabilityDistribution[
  c Exp[-x] /. 
   First[Solve[Integrate[c Exp[-x], {x, 0, 1}] == 1, c]], {x, 0, 1}]]