# Calculating course averages

And the end of every semester, instructors are often asked “What do I need on the final to make a ___ in the course?”, where the desired course grade is given. (I’ve never done a survey, but A appears to be the most desired course grade, followed by C, D, and B.) Here’s the do-it-yourself algorithm that I tell my students, in which the final counts for 20% of the course average.

Let $F$ be the grade on the final exam, and let $D$ be the up-to-date course average prior to the final. Then the course average is equal to $0.2F + 0.8D$.

Somehow, students don’t seem comforted by this simple algebra.

More seriously, here’s a practical tip for students to determine what they need on the final to get a certain grade (hat tip to my friend Jeff Cagle for this idea). It’s based on the following principle:

If the average of $x_1, x_2, \dots x_n$ is $\overline{x}$, then the average of $x_1 + c, x_2 + c, \dots, x_n + c$ is $\overline{x} + c$. In other words, if you add a constant to a list of values, then the average changes by that constant.

As an application of this idea, let’s try to guess the average of $78, 82, 88, 90$. A reasonable guess would be something like $85$. So subtract $85$ from all four values, obtaining $-7, -3, 3, 5$. The average of these four differences is $(-7-3+3+5)/4 = -0.5$. Therefore, the average of the original four numbers is $85 + (-0.5) = 84.5$.

So here’s a typical student question: “If my average right now is an $88$, and the final is worth $20\%$ of my grade, then what do I need to get on the final to get a $90$?” Answer: The change in the average needs to be $+2$, so the student needs to get a grade $+2/0.2 = +10$ points higher than his/her current average. So the grade on the final needs to be $88 + 10 = 98$.

Seen another way, we’re solving the algebra problem

$88(0.8) + x(0.2) = 90$

Let me solve this in an unorthodox way:

$88(0.8) + x(0.2) = 88 + 2$

$88(0.8) + x(0.2) = 88(0.8+0.2) + 2$

$88(0.8) + x(0.2) = 88(0.8) + 88(0.2) + 2$

$x(0.2) = 88(0.2) + 2$

$x = \displaystyle \frac{88(0.2)}{0.2} + \frac{2}{0.2}$

$x = 88 + \displaystyle \frac{2}{0.2}$

This last line matches the solution found in the previous paragraph, $x = 88 + 10 = 98$.

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