Originally Posted by

**lllll** I'm having a lot of trouble with this question:

Find $\displaystyle P(|Y-\sigma| \leq 2\sigma)$ for the uniform random variable distribution. Compare with the corresponding probabilistic statement given by Chebyshev's theorem and the empirical rule.

according to the solution in the book the answer is 1; I have no idea what this number represents.

The variance of a uniform distribution is $\displaystyle Vr(X) = \sigma ^2 = \frac{(\theta_2-\theta_2)^2}{12}$ so in this case it should be $\displaystyle 4\sigma = \frac{(\theta_2-\theta_2)^2}{12}$. This is where I get stuck, plus I don't think that I'm even going in the right direction. Any help would be appreciated.