How is mass conserved when a muon decays? A small disclaimer: I am a layperson and not a formal student of physics so forgive any glaring stupidity betrayed in the nature of the question.
A muon is supposed to always decay into an electron and two different neutrinos,
However, the mass of a muon is 207 times greater than an electron, a disparity that is seemingly too great for the addition of two insignificant neutrinos. 
How is conservation of mass conserved here?
 A: Mass is not conserved in that decay, but then there is no expectation that it would be.
The "law" of conservation of mass is only an approximate rule that applies to low energy events and interactions. Chemists (well, the non-nuclear ones, anyway) talk about it, but physicists do not.
The rule that does apply is the conservation of energy (mass being one kind of energy). The "missing" mass in the reaction shows up mostly in the kinetic energy of the products. The energy of a particle is given by
$$ E = \sqrt{(pc)^2 + (mc^2)^2} \,,$$
where $p$ is the particle's momentum and $m$ is its invariant mass (what you may know as the "rest mass"). And, yes, at zero momentum you can recover the one formula that everyone knows about relativity.

Aside: the mass of the two neutrinos together is not more than a few pars in a million of that of the electron.
A: I'm going to expand on dmckee's answer because this used to puzzle me in my younger days, and I think it's a fascinating part of modern physics.
It's tempting to think of a particle like a muon as a chunk of matter whizzing around - something like a tiny billiard ball. However the physicist's description of a particle is much, much stranger. Quantum field theory describes the particle as an excitation in a quantum field.
For every type of particle there is an associated quantum field that fills all of spacetime. When you add energy to this field it appears as a particle i.e. you can create a new particle by adding energy to the field. Likewise, removing energy from the field destroys a particle. When you add energy to a quantum field some of the energy goes to the rest mass energy of the particle and some to the particle's energy of motion. dmckee has given the equation for the total energy:
$$ E = \sqrt{(pc)^2 + (mc^2)^2} $$
and roughly speaking in this equation the $mc^2$ bit is the contribution from the rest mass (yes, as in the famous equation $E = mc^2$) and the $pc$ is from the motion.
But let's get back to the specific case you asked about of a muon decaying to an electron, an electron antineutrino and a muon neutrino:
$$ \mu \rightarrow e + \bar{\nu}_e + \nu_\mu $$
In the decay:


*

*an energy $E_\mu$ is removed from the muon field destroying a muon

*an energy $E_e$ is added to the electron field creating an electron

*an energy $E_{\bar{\nu}_e}$ is added to the electron antineutrino field creating an electron antineutrino

*an energy $E_{\nu_\mu}$ is added to the muon neutrino field creating a muon neutrino


The exchange of energy destroys one particle and creates three new ones. Energy is always conserved so we can say with confidence that the energy before equals the energy after:
$$ E_\mu = E_e + E_{\bar{\nu}_e} + E_{\nu_\mu} $$
But as dmckee points out, mass is not conserved. For example when we add an energy $E_e$ to the electron field some of this energy goes into the electron rest mass and some goes into the kinetic energy of the electron. Likewise for the neutrinos. In effect the missing mass gets turned into energy of motion of the three product particles.
