It would be mean, since it's common for the mean to be greater than the median in a normal distribution.
I hope this helps :)
Sorry if I'm wrong or anything, but I still hope I helped.

$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
Answer:
15
Step-by-step explanation:
32-19 = 13 that have both
70-57= 13 that have both again from that sample
total number of people that have both : 13 + 13 = 26
11 kids don't have either one, so :
26-11 = 15