The respective speed of five molecules are 2, 1.5, 1.6, 1.6 and 1.2 km/s. The most probable speed in km/s will be?
Following up on user andselisk's answer
...

The problem is weird for several different reasons. 


First the solution depends on assuming that the velocity of the gas molecules follow the Maxwell-Boltzmann distribution which is weird for only 5 molecules. The problem also doesn't state that the 5 molecules are a sample from a much larger number of molecules. 
Second the answer depends on how you calculate to get a result. 


For the Maxwell-Boltzmann distribution there are relationships between the most probable speed $v_p$, the average speed $\langle v \rangle$, and the root-mean-square speed $v_{rms}$. Following the Wikipedia article:

\begin{align}
\langle v \rangle &amp;= \frac{2}{\sqrt{\pi}}v_p\\
v_\mathrm{rms} &amp;= \sqrt{\frac{3}{2}}v_p
\end{align}

Rearranging these two equations we get two different ways to calculate the most probable speed.

\begin{align}
v_p &amp;= \frac{\sqrt{\pi}}{2}\langle v \rangle\\
v_p &amp;= \sqrt{\frac{2}{3}}v_\mathrm{rms}
\end{align}

Thus using the mean (average) speed we get:

\begin{align}
\langle v \rangle &amp;= \frac{2 + 1.5 + 1.6 + 1.6 + 1.2}{5} = 1.58\\
v_p &amp;= \frac{\sqrt{\pi}}{2}\langle v \rangle = 1.40
\end{align}

However using the rms speed we get:

\begin{align}
v_\mathrm{rms} &amp;= \sqrt{\frac{2^2 + 1.5^2 + 2\cdot 1.6^2 + 1.2^2}{5}} = 1.60\\
v_p &amp;= \sqrt{\frac{2}{3}}v_\mathrm{rms} = 1.31
\end{align}

The difference is due to small sample statistics and not really a statistically significant result. However the difference is absolute proof that the sample doesn't follow the Maxwell-Boltzmann distribution.