1. Chebyshev and uniform distribution

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

Find $P(|Y-\sigma| \leq 2\sigma)$ 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 $Vr(X) = \sigma ^2 = \frac{(\theta_2-\theta_2)^2}{12}$ so in this case it should be $4\sigma = \frac{(\theta_2-\theta_2)^2}{12}$. 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.

2. Originally Posted by lllll
I'm having a lot of trouble with this question:

Find $P(|Y-\sigma| \leq 2\sigma)$ 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 $Vr(X) = \sigma ^2 = \frac{(\theta_2-\theta_2)^2}{12}$ so in this case it should be $4\sigma = \frac{(\theta_2-\theta_2)^2}{12}$. 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.
$\Pr(|Y - \sigma| \leq 2 \sigma) = \Pr(-\sigma \leq Y \leq 3 \sigma)$.

If Y is uniform over [a, b]:

$\bullet$ $f(y) = \frac{1}{b-a}$ for $a \leq y \leq b$ and zero otherwise.

$\bullet$ $\sigma = \frac{b-a}{\sqrt{12}}$.

Note that $- \sigma = \frac{a-b}{\sqrt{12}} < a$ for $b > a$. This means that:

$\Pr(-\sigma \leq Y \leq 3 \sigma) = \Pr(-\sigma \leq Y \leq a) + \Pr(a \leq Y \leq 3 \sigma) = \Pr(a \leq Y \leq 3 \sigma)$.

Furthermore:

$3 \sigma = \frac{3(b-a)}{\sqrt{12}} < b$ if $a > \frac{-(\sqrt{12} - 3)b}{3}$.

$3 \sigma = \frac{3(b-a)}{\sqrt{12}} > b$ if $a < \frac{-(\sqrt{12} - 3)b}{3}$. In this case the required probability is obviously equal to 1.