Question

What's the variance of a uniform distribution defined on [0,6]?

Answer 1

Answer:

3

Step-by-step explanation:

Let's change the 6 to b and deal with this more generality.

First, the density function is f(x) = 1 / b for x in [0, b], and it is 0 elsewhere.  That much is reasonably clear since the full area under the curve has to be 1 for a probability density function.

Since it's uniform, we can see quickly that the mean is b/2 by virtue of this being the "middle" and the distribution being symmetric about this point.  So we don't need any maths to do get that far.  Formally, if X is a random variable with this distribution, then we have said E(X) = b/2 (the expected value of X is b/2).

For the variance, we use the fact that

var(X) = E(X²) - ( E(X) )².

To get E(X²), we calculate the integral of x² times the density function:

E(X²) = ∫ x² (1/b) dx  (integrating from 0 to b)

= (1/b) ∫ x² dx  (integrating from 0 to b)

= (1/b) [ x³ / 3 ]   (evaluating from 0 to b)

= (1/b) (b³/3) = b²/3.

Now we plug the values that we have into the formula for the variance:

var(X) = E(X²) - ( E(X) )²

= b²/3 - ( b/2 )²

= b²/3 - b²/4

= b² / 12.

For our special case where b = 6, the variance is then

6² / 12 = 6 / 2 = 3.