The situation you are describing is an example of Fresnel diffraction (or near-field diffraction).
In general, when a wave propagates every point of the wave front can be thought of as its own source of waves traveling in all directions (called Huygens construction). It turns out that neighboring point sources along an infinite straight wave front reinforce the "forward" direction only, but if you put an obstacle in the way you can see this diffraction.
The mathematics needed is simplified when you look at the effect of this diffraction "far away" (far compared to the wavelength of the wave). In the case of sound, a frequency of 55 Hz (low end of the range of sounds you hear) has a wavelength of about 6 m, so on the scale of your drawing diffraction would occur.
This explains why you can hear the thumping bass of a loud car stereo before the car turns the corner, and only make out the song when the car is in sight.
The calculation of relative sound level into the room as drawn is tricky - it involves an integral that is usually evaluated using a graphical technique called the Cornu spiral, and strongly depends on dimensions and frequency. But as a rule of thumb, "high frequencies travel straighter". And "sound barriers" do work (somewhat) to reduce nuisance noise (for example the noise of cars speeding along a highway).
If you want to estimate the attenuation, you will find this link has some helpful equations and graphs.
UPDATE
There is a problem with the link given: it defines the Fresnel number as
$$N = \frac{2d}{\lambda}$$
But has a confusing definition of $d$. To make things work, you need to set the straight line distance from source to receiver to $D$ (not $d$). If you do that, then
$$d = A + B - D$$
and the Fresnel number is
$$N = 2\;\frac{A+B-D}{\lambda}$$
In an elastic solid, particles return to their equilibrium position after the sound wave has passed: because they are attached to each other they have a definite "sense of place".
Air molecules don't have a fixed position: the jostle around, and the average effect of their motion is experienced as pressure (number of particles hitting an area multiplied by their change of momentum normal to the surface = force per area = pressure).
Now when you increase density, you increase the number of molecules per unit volume; this increases the number that hit a unit area per unit time, and thus the pressure. But since the molecules "in the next slab of air over" don't have an equally high pressure, more molecules will move from the high pressure area to the low pressure area than vice versa, so the high pressure "wave" moves on. If you lower the density in a region temporarily, the opposite happens: molecules from adjacent higher pressure areas will move in "to fill the void".
All this is possible because, if you imagine a box full of air with an imaginary plane running down the middle, a roughly equal number of molecules will move across the plane from left to right, and from right to left, every second. If you start with one color of gas on one side of the membrane, and another color of gas on the other, very quickly the two will mix - this is called diffusion.
All of which is a long winded way of saying: there is a microscopic picture of molecules moving that explains pressure, diffusion, and sound waves. This picture tells us molecules have no fixed position; but they do have a tendency to move from high pressure to low pressure.
I also recommend reading the question/answer that I wrote previously. It has a possibly helpful picture.
Best Answer
The sound box couples the vibration of the instrument to the air around it, so that it makes more noise that a vibrating string would alone. The string alone displaces very little air, and so makes very little sound. The much larger surface area of the body means that it can displace much more air, and thus make much more sound when it vibrates.
The material the body is made out of is very important, as it must be light and thin enough to vibrate well, but also strong enough to not give out under the tension of the strings.
You can see essentially this effect by considering an acoustic violin (which has a sound box) to an electric violin (which does not). An electric violin has a solid body, which does not vibrate much and so does not produce much sound, and so requires external amplification.