The velocity profile that is shown is parabolic. It is characteristic of laminar flow in tubes or channels. A development of the equation is shown in this link. An exposition about flow in open channels is at this link.
The maximum velocity over the entire profile is at the surface (see Equation 4.7 and statement 17 in the first reference). The maximum velocity at an other range of positions as measured vertically from the bottom of the stream is always found at the top of the range (i.e. at the point closest to the surface of the stream).
Let's propose a velocity profile from the bottom to top of the channel as this.
$$ v(z) = Cz^2 $$
This satisfies the boundary condition that $v(0) = 0$ at the bottom of the channel. The parameter $C$ is found by taking one additional measurement anywhere in the stream. With this function, we find the average velocity over a range of positions $zo$ to $zt$ is an integral over the parabolic profile ...
$$ <v> = \frac{\int_{zo}^{zt} v(z) dz}{zt - zo} = \frac{C}{(zt - zo)} \int_{zo}^{zt} z^2 dz = \frac{C\left(zt^3 - zo^3\right)}{3\left(zt - zo\right)} $$
An alternative way then to find $C$ when given an average velocity over a range of distances is to use this expression.
A distinction is required to understand the average velocity expressed above. An alternative view of average velocity is from measuring at a set of points over a range $zo$ to $zt$ and then averaging the measured values. This measured average has the expression
$$ <v>_m = \frac{1}{N} \sum {v_j} $$
With $H$ as the height of the channel, in the limit that $zt - zo << H$, we can assume $<v> \approx <v>_m$. Otherwise, the better approach is to use a one-point method with $v$ at $z$ and the source equation for the profile.
Finally, consider when you have multiple measurements at different positions where the positions are not differentiated but the velocities are differentiated. One approach here is to average the velocities and use any one value to the profile for $C$. The better method however will use non-linear regression method to fit data. By example, consider this hypothetical "measured" data set generated using $v = 4 z^2$ as its base.
z=1, v = 3, 4, 5, 6
z=2, v = 14, 15, 16, 17
z=3, v = 32, 33, 35, 45
A plot of the data and a non-linearized regression fit to $v = Cz^2$ is shown below. The result is $C = 4.0 \pm 0.2$.
