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Your plot corresponds to the probability $p_i$ of occurence of the $i$-th microscopic state whose energy is $E_i$. At the macroscopic level, you cannot distinguish the different microscopic states but you can measure macroscopic quantities as the energy E. The probability to measure an energy $E$ is $$P(E)=\sum_i p_i\delta(E-E_i)=\Omega(E){e^{-\beta E}\over{\cal Z}}$$ where $$\Omega(E)=\sum_i \delta(E-E_i)$$ is the number of microscopic states whose energy is $E$. Usually, you have very few microscopic states with a low energy but this number increases with $E$. Multiplying by the Boltzmann weight which behaves in the opposite way, you get a curve which typically increases as $\Omega(E)$ for $E$ small and then decreases exponentially.

In the particular case of the Maxwell-Boltzmann distribution, i.e. for a classical ideal gas, we have for a single particle $$\eqalign{ P(E)&={1\over z}\int \delta\Big(E-{p^2\over 2m}\Big) e^{-\beta{p^2\over 2m}} {d^3\vec r d^3\vec p\over h_0^3}\cr &={V\over zh_0^3}\int_0^{+\infty} \delta\Big(E-{p^2\over 2m}\Big) e^{-\beta E} 4\pi p^2dp }$$ where $d^3\vec p$ has been written in spherical coordinates. Setting $u=p^2/2m$ and $p=\sqrt{2mu}$ so that $du=pdp/m$, we get $$P(E)={4\pi V\over zh_0^3}e^{-\beta E}\delta(E-u)\sqrt{2mu} \times mdu={4\pi V\over{\cal Z}h_0^3}m^3/2\sqrt Ee^{-\beta E}$$ Plugging now the expression of the partition function, we get $$P(E)={4\pi\over (2\pi k_BT)^{3/2}}\sqrt Ee^{-\beta E}$$

However, this is only for a single particle. The same calculation should be done for a gas of $N$ particles. $$P(E)={1\over {\cal Z}}\int \delta\Big(E-\sum_{i=1}^{3N}{p_i^2\over 2m}\Big)e^{-\beta\sum_{i=1}^{3N}{p^2\over 2m}} {d^{3N}\vec r d^{3N}\vec p\over h_0^{3N}}$$ The equivalent of spherical coordinates in a space of dimension $3N$ leads to ${\rm Cst}\ \!p^{3N-1}dp$ after integrating over the $3N-1$ angles. Therefore, I guess that we should expect something like $$P(E)={\rm Cst}\ \!E^{3N/2-1}e^{-\beta E}$$