A problem related to the Chebyshev inequality Let $(x_1,x_2,\cdots,x_n)$ be n independent random variables from population distributed uniformly $x_i\sim U(\theta,3\theta)$ with unknown $0.5\le \theta\le1$.
$\min_n$ will be the minimal number of samples s.t. $n\ge \min_n$, that results directly from Chebyshev inequality, and for which it exists 
$$P(2\theta-0.05\le\hat{X}\le2\theta+0.05)\ge0.97$$
$$\hat{X}=\frac{1}{n} \sum_{i=1}^n x_i$$ 
What is the domain of $\min_n$?
I've tried to play with the inequality to get the Chebyshev form, but I got stuck and trying to understand how to answer this question in the formal way.
Thanks!
 A: Since it is a uniform variable : $E(x) = 2\theta$ and $\sigma_x^2 = \theta^2/3 $ .
Taking $n$ samples you have : 
$$E(\hat{x}) = \frac{1}{n}\cdot\sum_{i=1}^{n} E(x_i) = 2\theta $$
$$\sigma_{\hat{x}}^2 = \frac{1}{n^2}\cdot\sum_{i=1}^{n}\sigma_x^2 = \frac{1}{n^2}\cdot\sum_{i=1}^{n}\frac{\theta^2}{3} = \frac{\theta^2}{3n}\implies \sigma_{\hat{x}} = \frac{\theta}{\sqrt{3n}} $$.
Therefore you have that the Chebishev inequality is :
$$P(2\theta-\lambda\frac{\theta}{\sqrt{3n}} \leq \hat{x}  \leq 2\theta+\lambda\frac{\theta}{\sqrt{3n}} ) > 1 - \frac{1}{\lambda^2}$$
Remembering $1/2\leq\theta\leq1$  and since you can put $ \lambda\frac{\theta}{\sqrt{3n}} = k$, then :  $$\frac{\lambda^2}{12k^2} \leq n = \frac{\lambda^2\theta^2}{3k^2} \leq \frac{\lambda^2}{3k^2}$$
if you take $1 - \frac{1}{\lambda^2} = 0.97 \implies \lambda^2 = 100/3$ , this is the minimum value of $\lambda^2$ so that the second inequality given is true. You take then also $k = 0.05 = \frac{5}{100}$ because otherwise you can't be sure if you increment $k$ that the second inequality remains true .
It results in $1112\leq n \leq 4445$.
(I'm not sure that the last steps are correct, if someone finds a mistake please point it out)
