Answer: 7
==================================================
Work Shown:
Given data set of passing scores = {9,6,7,8,4,5,8,10,9,7,5,7}
Sorted data set of passing scores = {4,5,5,6,7,7,7,8,8,9,9,10}
Let x, y, and z represent the three failing scores.
The order of which doesn't matter but I'll have x < y < z.
In other words, x is the smallest, y in the middle, and z the largest of the failing scores sub-group.
Again the order of x,y,z in themselves don't matter. What does matter is that all three scores are smaller than 4. I'll assume that 4 represents the lowest passing grade possible.
------------------
We have this updated set of scores
{x,y,z,4,5,5,6,7,7,7,8,8,9,9,10}
We've gone from n = 12 scores to n = 12+3 = 15 scores.
n/2 = 15/2 = 7.5 which rounds to 8
The median of that updated set will be in slot 8. This trick only works when n is odd.
The score in the 8th slot is 7. More specifically, it is the first copy of "7" that is the median.
------------------
Another way to find the median:
Start with this set here: {x,y,z,4,5,5,6,7,7,7,8,8,9,9,10}
Erase the first and last items to get: {y,z,4,5,5,6,7,7,7,8,8,9,9}
Repeat the last step to get: {z,4,5,5,6,7,7,7,8,8,9}
Repeat again: {4,5,5,6,7,7,7,8,8}
Keep repeating those steps until you arrive at {6,7,7} which helps us zoom in on the middle-most item, which is the first "7" of that three element subset. We can then see that 7 must be the median of the 15 element set.