I was walking with a group of students and noted that there was a distribution (not unlike normal?) of the speeds at which they walked. There would obviously also be a distribution of the times at which they would arrive at the destination.
The person with the smallest speed would obviously be the person with the longest time and vice versa, but the person with the median speed would also be the person with the median time.
What are the conditions for the person with the mean speed to also be the person with the mean time?
I think we are looking for a distribution where $E(\frac 1X)=\frac 1{E(X)}$
Are there established distributions with this property?
The probability distribution of a positive-valued random variable $X$ satisfies $\operatorname{E} \left( \dfrac 1 X \right) = \dfrac 1 {\operatorname{E}(X)}$ if it assumes only one possible value, for example, if $\Pr(X=5) = 1.$ But otherwise equality fails.
On this Wikipedia page, we read that Jensen's inequality is strict unless the random variable is constant or the function they call $\varphi$ is "linear" (i.e. affine). In this case we have $\varphi(x) = 1/x$, which is not "linear".
Appendix: I momentarily considered cases in which the distributions of $X$ and $1/X$ are identical, but I realized that in that case, we would have $\operatorname{E}(1/X) = \operatorname{E}(X),$ so the desired equality would hold only if $\operatorname{E}(X)= 1/\operatorname{E}(X),$ which implies $\operatorname{E}(X)=1,$ and the average a positive number other than $1$ and its reciprocal is always more than $1.$ One would then argue like this: $$ \operatorname{E}(X) = \operatorname{E}(\operatorname{E}(X\mid \{X,\ 1/X\})) = \operatorname{E}\left( \frac{X + (1/X)} 2 \right) $$ and that is the expected value of a random variable that is always $\ge 1$ and is $>1$ with positive probability, so its expected value has to be $>1.$