Is $v=\ (r_i)^{-1}\cdot z $, a uniformly random value of a field?

Question

We consider a finite field $\mathbb{F}_q$ where $q=2p+1$ and $q$ and $p$ are prime numbers.

Let $r_i$ be a value picked uniformly at random from the field such that $r_i>\frac{q}{2}$. Let $z$ be a fixed field element.

Question: Is $v=(\ (r_i)^{-1}\cdot z) \bmod q$, a uniformly random value of the field?

Answer 1

No, since the restriction $r_i > \frac{q}{2}$ means $v$ can only assume $2p + 1 - (p + 1) = p$ different values.