The classical Dvoretzky covering problem asks for conditions on the sequence of lengths $\{\ell_n\}_{n\in \mathbb{N}}$ so that the random intervals $I_n : = (ω_n -(\ell_n/2), ω_n +(\ell_n/2))$ where $ω_n$ is a sequence of i.i.d. uniformly distributed random variable, covers any point on the circle $\mathbb{T}$ infinitely often. We consider the case when $ω_n$ are absolutely continuous with a density function $f$. When $m_f=essinf_\mathbb{T}f>0$ and the set $K_f$ of its essential infimum points satisfies $\overline{\dim}_\mathrm{B} K_f<1$, where $\overline{\dim}_\mathrm{B}$ is the upper box-counting dimension, we show that the following condition is necessary and sufficient for $\mathbb{T}$ to be $μ_f$-Dvoretzky covered \[ \limsup_{n \rightarrow \infty} \left(\frac{\ell_1 + \dots + \ell_n}{\ln n}\right)\geq \frac{1}{m_f}. \] Under more restrictive assumptions on $\{\ell_n\}$ the above result is true if $\dim_H K_f<1$. We next show that as long as $\{\ell_n\}_{n\in \mathbb{N}}$ and $f$ satisfy the above condition and $|K_f|=0$, then a Menshov type result holds, i.e. Dvoretzky covering can be achieved by changing $f$ on a set of arbitrarily small Lebesgue measure. This, however, is not true for the uniform density.