While it is not necessarily true, and usually isn't, in a list of values,

the median of a normal distribution with mean μ and variance σ2 is also μ.

So in your problem, the median is also ~ 16.
 
{In fact, for a normal distribution, mean = median = mode. The median of a
uniform distribution in the interval [a,b] is (a+b)/2, which is also the mean.}

Edwin