Different cross section for W and Z boson At hadron colliders the cross section for W production is about ten times larger than the production cross section for Z bosons (e.g. Figure 2 in this review article). I guess the dominant contribution is from Drell-Yan production so let's only consider this.
A factor of two probably comes from the fact, that there is a positive and a negative W. But what about the remaining difference of factor 5?
 A: The Figure 2 of this paper

http://arxiv.org/pdf/hep-ph/0611148v1.pdf

doesn't show a factor-of-ten difference at all! Extract the ratio properly on the log scale and you will see it is less than four, just slightly greater than 1/2 of the height corresponding to the decade. Note that 1/2 of the weight corresponds to the factor of $\sqrt{10}\sim 3.16$.
The ratio of the cross sections is close to four rather than two mostly because of the Weinberg angle $\theta_W$. The production of neutral $W^0 W^0$ boson pair would have cross section close to one-half of the inclusive charged $W^+W^-$ cross section. However, a $W^0$ only contains $\cos\theta_W$ times $Z^0$, and this must be used for both copies of the $W^0$. So some of the processes produce $\gamma Z$ or $\gamma\gamma$. Only the fraction $\cos^2\theta_W$ produces $Z^0 Z^0$  and $\cos^2\theta_W\sim 1-0.23\sim 0.77$ is a new factor in the amplitude.
Note that we are neglecting the production of $B^0 B^0$ – which also splits to four contributions $ZZ,Z\gamma,\gamma\gamma$ – because it's suppressed by $g_Y^2$ and it's much smaller than the coupling constant for the $SU(2)$.
So the estimated ratio of the $W/Z$ inclusive cross sections is $2/0.77^2\sim 3.37$, very close to what the graph shows. I had to square the amplitude to get the probability which is why the factor $1/0.77$ appeared twice.
Other potential asymmetries that raise $W^+W^-$ relatively to $Z^0Z^0$ probably include the fact that the different $u\bar d/\bar u d$ quarks are more likely to be found inside the hadrons than $u\bar u$ and $d\bar d$. Also, if a $Z$ appears in a propagator, the diagram has a greater (in the $s$-channel) suppression $m_Z^2$ in the denominator than $m_W^2$.
