we are told that mean = median in 2 cases. 1 when it’s evenly spaced out and 2nd when there’s symmetry. In the above case, none of that exists ,yet median = mean. I was going ahead with the approach to check if there’s symmetry/even case, but none of that exists yet both are same. Pls help clear this doubt.
I think it’s instructive to think about how conditional statements work to resolve your query.
Consider the two sentences:
S_1: It is raining outside.
S_2: The ground will be wet.
We claim that if it is raining outside, then the ground will be wet. Basically, the occurrence of S_1 leads to S_2. In other words, it is sufficient for it to be raining outside for the ground to be wet.
But does the occurrence of S_2 automatically imply S_1 must have occurred? No, because the ground could be been wet for another reason. It could’ve been the case that someone washed their car, and thus the ground is wet.
In a similar vein, if you think about your query logically like above (as sentences/ premises if you like) , you’ll realize that a symmetric distribution implies that the mean = median, but the reverse isn’t true. The fact that a dataset has the property that the mean = median doesn’t necessarily imply that the data must be symmetric.
The counterexample being the very question u posted.