Answer: even with a contrived example yielding minimal mean for a given median necessarily between 4 and 6, the mean is always greater than the median.
Step-by-step explanation:
Total number of students: 20
c(1 through 6) < 2
4 <= c(7 through 11) <= 6
10 < c(12 through 20)
Median is sum of students 10 and 11 divided by 2.
Median is <= 6
Minimal mean = (6×0+5×4+9×11)/20 = 5.95
Maximal median for that mean = (4+4)/2=4, smaller.
Maximal median for any mean = (6+6)/2=6
Minimal mean for that median = (6×0+3×4+2×6+9×11)/20 = 6.15
General case: given
median = (c(10)+c(11))/2
mean = (6×0+3×4+2×median+9×11)/20
= (12+99+2×median)/20
= (12+99)/20 + 2×median/20
= 111/20 + median/10
Let x = median.
mean > median when
111/20+x/10 > x
111/20 > 9x/10
111×10/(9×20) > x
6.1667 > x
But the median has to be <= 6, so the mean is necessarily greater than the median.
