Confidence interval using Chebyshev's inequality

Question

$X_1,...,X_n$ is a sample, $X_1 = ξ + η$, where $ξ,η$ are independent random variables, $ξ ∼ R[0,θ], η ∼ Bin(1,θ)$. I want to establish a confidence interval for $θ$ confidence level $1−α$ using Chebyshev's inequality. Problem is that after calculation $D(\frac{1}{n}\sum_{i=1}^nX_i) = \frac{1}{n}(\frac{\theta^2}{12} - \theta^2+\theta)$ and applying Chebyshev's inequality $P(|\frac{1}{n}\sum_{i=1}^nX_i - \frac{3\theta}{2}| \geq \frac{3\epsilon}{2}) < \frac{\frac{\theta^2}{12} - \theta^2+\theta}{n\epsilon^2\frac{9}{4}}$ on the rigth handside we have that the variance contains not only $\theta^2$, but also the multiplier $\theta$, which does not allow us to calculate the confidence interval. I also tried to find another estimation, but didn't find anything good. I'm confused about this problem and how to approach it, could anyone help? Thank you!