I am given the speeds of eight objects, and I have to find the average speed and the RMS speed of these objects. I tried taking the actual average of these speeds: $\frac{1}{8}\Sigma(v_i)$, but I get $7.4$ $m/s$ instead of $7.5$.

I know that $v_{average}=2 \sqrt{\frac{2 k_B T}{\pi m}}$ and $v_{rms}=\sqrt{\frac{3 k_B T}{m}}$, but I do not see how these formulas could help me in any way as I do not know the mass of the given objects, just their velocities and their number.

I also know that $f(v)=(\frac{m}{2 \pi k_B T})^{\frac{3}{2}} 4\pi v^2e^{\frac{-mv^2}{2k_BT}}$ is the probability per unit speed. I feel like it might have some relevance in terms of weighting in both cases but again, I do not know the masses of the objects. I tried assuming they all had the same mass and did not get the result I wanted.

I tried to find the rms speed by taking $\sqrt{(\Sigma (vi-v_{average})^2)}$ but I get 11.2 instead of 8.5.

Any suggestions on the right direction?

NOTE: The objects' velocities in m/s are 2,4,5,5,8,9,11,15 (a user asked previously. I can remove if too specific).


1 Answer 1


The average speed is indeed the simple numerical mean of the given speeds

The RMS speed is the root of the mean of the squares i.e. it is the square root of the average of the squares of the given speeds.

For a simple example. If there are 3 objects with speeds 1,2 and 3.

Then average speed = (1+2+3)/3 = 2

rms speed = sqrt((1+4+9)/3) =sqrt(14/3) = 2.16

  • $\begingroup$ In statistics one usually distinguishs between quantities calculated for the population and the sample. E.g. the standard deviation of $x_i=\{1,2,3\}$ is $\sqrt{\sum_i(x_i - \bar x)^2/3}$, if $x_i$ is the population, but $\sqrt{\sum_i(x_i - \bar x)^2/(3-1)}$, if $x_i$ is a random sample. I'm not sure, but I would use the same logic for the RMS formula. How sure are you that your formula is correct? $\endgroup$
    – Semoi
    Commented Apr 17, 2021 at 16:21
  • $\begingroup$ @Semoi From the OP's question, it seems that he is talking about a population , not a sample. If it is indeed a sample, then you would be right. $\endgroup$ Commented Apr 17, 2021 at 17:17
  • $\begingroup$ It was not specified, but I assumed it was a population as there was no mention of a "bigger group" they belonged to. Also, by using @silverrahul's method, I get values that are close to the required ones, so I think assuming it is a population should be a good enough approximation. $\endgroup$
    – Agnese
    Commented Apr 17, 2021 at 17:26

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