Hi @gregmat,

I think you did a little error in yesterday’s class.

Here you were trying to maximize the median (which would be ten) while minimizing the average. But when calculating the average, you calculated 15x1 and then 35x6 etc. But this is not possible, since if we suppose that the median is 10 (the 48th number), then for the second group we would have just 32x6 and 3x10. Thus, the average would be even a little bigger, but the answer to the question would still be the same.

Just a minor thing, but wanted to let you know.

Take care.