Look up
http://talk.collegeconfidential.com/...78#post1113278
and
http://talk.collegeconfidential.com/...19#post1113919
Just to illustrate Ziggy's approach:
It's often helpful to visualize
1.numbers
and
2.their average
as
1.straight glasses with different levels of water
and
2.same glasses with equal volume of water in each,
total volume of water in these glasses being equal to
total volume of water in the original configiration of glasses.
This is actually an analogy to bar graphs.
In order to have all glasses filled equally we need to redistribute the water:
add to those where the level is below the average,
taking water out of those wth the level above the average.
You can see that "added volume" has to be equal to "removed volume".
So, if the average of p students' scores is 70, and the average score of another group of n students is 92, you can draw
p glasses with the level of 70 and n glasses in the same row with the level of 92.
Now draw the line through all the glasses at level 86.
In order to get a volume in each glass equal to 86, we'll pour out what's above 86 level into what's below it.
There are n glasses with the level above 86 by 6
(86+6=92), required loss = 6n,
and p glasses with the level below 86 by 16
(70+16=86), needed gain = 16p.
6*n = 16*p,
p/n = 6/16.
THE END.
This is a simplified situation (each of p glasses have the same volume of water; same is true for n glasses), but it visually confirms what Ziggy showed in his diagram:
"...CHANGES in scores
70................86......92
...........16...........6
... the gain of p students HAS to equal the loss of students of n students (no points vanish when averaging the scores)."