Saturday, May 26, 2007

An Interesting Effect

Don't anyone have a heart attack that I'm posting only days after my last post ;-).

So, I had an interesting effect while grading one of my students' finals. Let's see if anyone can figure out this paradox. Names and actual grades changed to protect the innocent ;-).

So, I have a student who we will call Sally. Sally was hovering on the C/D range before the final exam (70-79 is a C). She had just less than a 70 before the final exam, but rounded to the nearest whole number, she had a 70 average and was passing with a C.

Enter final exam. Sally's final exam score was higher than her previous course average (so more than a 70), but her final exam made her course average drop below a 70, rounding to a 69 as the nearest whole number. So her final course grade was a D.

So question: how did this happen?

7 comments:

Adrian C. Keister said...

Reply to Susan.

There must be a difference in weights.

I can prove that given n equally weighted numbers x(n), and their average

A = [x(1)+x(2)+...+x(n)]/n,

if we add one more equally weighted number

x(n+1) > A,

then the new average

A' = [x(1)+...+x(n)+x(n+1)]/(n+1) > A.

The proof is fairly simple. Note that the new average

A' = (nA + x(n+1))/(n+1).

If A' <= A, then multiplying both sides by n+1 gives

nA + x(n+1) <= nA + A,

which implies that

x(n+1) <= A,

a contradiction. Hence, A' > A.

Moreover, if the new average is greater on the real line, then it is pretty obvious there's no way it could round lower than the old average.

Conclusion: there must be something funny going on with the weights: how the final is weighted relative to the other factors in the overall grade.

In Christ.

Adrian C. Keister said...

Or there could possibly be something going on w.r.t. rounding. Depending on when you round things, you can get some pretty strange results.

In Christ.

Adrian C. Keister said...

Yeah, I think I've got one explanation: in order to calculate your old average, you've used rounded numbers on previous tests, quizzes, homeworks, etc. So the actual average, if you were to allow infinitely many digits of precision, is higher than your reported average, especially if Sally's grades tended to round down. Enter the final exam, which grade is higher than your old reported average, but not higher than the actual unrounded average. Hence it brings the overall grade down.

So, how'd I do?

:-)]

In Christ.

Susan said...

You're very adorable. Heehee.

Very nice proof on the first one, and some nice thoughts overall. It had nothing to do with rounding. The only rounding was done at the very last. I will give this bit of information: the final was averaged in as a test grade alone, not as a separate final %. Does that help?

Joanna said...

I saw this linked from Ashley's blog, and I like math puzzles... so here's my input.

I'm operating on the key words 'course average'- this usually comprises more than just test scores. You also said that the final was averaged 'as a test grade', implying there are different weights.

Let's say Sally had a homework average of 60, and homework was weighted as 25% of the grade, and tests and quizzes at 75%. Going into the final, to have at least a 69.5 (rounds to 70) average, she'd need a test score average of about 73. Let's say she gets a 71 on her final, and there were 2 previous tests...

Before the final
(73 * .75)+(60*.25) = 69.75

After the final
New test average: (73+73+71)/3= 72.333

(72.333 * .75)+(60*.25) = 69.25

Adrian was right, it had to do with weighting. I like word problems :)

Susan said...

Nice, Joanna :-). You're absolutely correct. Isn't that an interesting paradox?

Joanna said...

It is an interesting paradox... you just have to wish that the grade was hovering between a B and a C instead of a C & D, so Sally would pass the course...