og math 161

Knowing that the mean of the set is four lets us say:
(3+k+2+8+m+3)/6=4
3+k+2+8+m+3=24
m+k+16=24
m+k=8

So, m and k are integers that sum to 8 and are not equal to each other. Therefore, one of them must be less than four and the other must be greater. Some such choices for m,k are 3,5; 5,3; 2,6; 6,2; 1,7; 7,1; 0,8; 8,0; -1,9; and so on.

Another way to get there is to see that the numbers 2, 3, 3, 8 have a mean of four without m and k. We can conclude, therefore, that m and k alone must also have a mean of four. But if they can't equal each other, then one must be a number above four and the other must be a number below four.

Either way, you can see that no matter which integer values you assign for m and k, the median of the set is always 3. Here are some examples (with m and k in bold):

2, 3, 3, 3, 5, 8 ==> median is 3 (average of 3rd and 4th elements)
2, 2, 3, 3, 6, 8 ==> median is 3 (average of 3rd and 4th elements)
1, 2, 3, 3, 7, 8 ==> median is 3 (average of 3rd and 4th elements)
0, 2, 3, 3, 8, 8 ==> median is 3 (average of 3rd and 4th elements)
and so on...

Rey