The electron configuration for C's ground state is 1s^(2)2s^(2)2p_x^(1)2p_y^(1)2p_z^(0).
In order for C to be able to form 4 bonds, one electron from the 2s orbital is promoted to the empty 2p_z orbital - this is referred to as the promoted state; this leads to the formation of hybrid orbitals - one s orbital mixes with the three p orbitals to form four sp^3 hybrid orbitals.
This is what happens to the C atom in methane. Once the hybrid orbitals are formed, you can no longer refer to any of the three p orbitals. Here's a diagram of the formation of the sp^3 hybrid orbitals:
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The four sp^3 orbitals now form sigma bonds with four H atoms to produce the tetrahedral-shaped methane molecule.
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So, to answer your question, the 2p_z orbital is not drawn because the C atom only has sp^3 orbitals in the CH_4 molecule.
Good explanation, nice figures can be found here: http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Hybrid_Orbitals
