Hydrodynamic perturbations = change in pressure due to a flow velocity (particles don't return to equilibrium positions).
Acoustic perturbations = change in pressure due to the fact the particles undergo an elastic restoring force (for a compressible fluid) which causes perturbations to travel at the speed of sound.
Any change in the pressure/velocity field will experience acoustic perturbations. The hydrodynamic fluctuating velocity therefore causes acoustic perturbations. Lighthill's analogy explains the equivalent source term in the wave equation, which is a weakly radiating quadrupole source for turbulence, depended on the magnitude of the fluctuating hydrodynamic velocity. The acoustic perturbations are of much smaller magnitude than the hydrodynamic perturbations.
If you define sound as the change in pressure at a receiver, then sound = hydrodynamic + acoustic perturbations. The hydrodynamic part only exists within the source region.
To first order, the speed of sound is not affected by pressure. Pressure waves can be shown to fulfill the D'Alembert wave equation $(c_S^2\,\nabla^2 - \partial_t^2)\psi=0$ where the wavespeed $c_S$ is given by:
$$c_S = \sqrt{\frac{K}{\rho}}$$
where $K$ is the bulk modulus of the medium in question and $\rho$ its density. Now, for an ideal gas, the bulk modulus $K$ is in most conditions proportional to the pressure; if the compression is adiabatic (good approximation for high frequency sound, as there is little time for heat to shuttle back and forth in the gas), then $K=\gamma\,P$, where $\gamma$ is the Heat Capacity Ratio or Adiabatic Index. However, from the ideal gas law $P\,V=n\,R\,T$ we have:
$$\rho = \frac{n\,M}{V} = \frac{P\,M}{R\,T}$$
where $M$ is the mean molar mass of the gas in question in kilograms. Thus the pressures cancel out in the speed of sound:
$$c_S = \sqrt{\frac{\gamma\,R\,T}{M}}$$
Thus we see that the speed is also weakly affected by the humidity - more water in the air lowers the mean molecular mass. If we put $M=0.029$, $T=300K$ and $\gamma = 1.4$ for air, we get $c_S=347{\rm m\,s^{-1}}$.
Best Answer
The sound intensity is
$I = \xi^2 \omega^2 c \rho$
where $\xi$ is the particle displacement, $\omega$ the frequency, $c$ the speed of sound and $\rho$ the density of the medium.
What does this mean?
You're in a room where the pressure is lower (and so the density of air is also lower). Say that your sound source (string, speaker or whatever) is vibrating with the same amplitude and frequency, so $\xi$ and $\omega$ don't change. The speed of sound is roughly independent of the pressure. So your sound intensity will be roughly proportional to the density of the air, or proportional to the pressure.