"Need", nah. Presumably it can't hurt.
It looks like 4 midterms, 11 homeworks, and the final; call it 4% per homework, 8% per midterm, 32% final.
If that means - do I have a percent-to-letter-grade conversion table already decided upon? No.
We will be calculating interim grades before Drop Day, which is the only time that this information can really affect anyone's behavior.
For the first midterm, we'll try open paper (book, notes, HW) but no calculators. Though on the second midterm we did exactly the reverse. The third midterm, and the final - nothing: no calculators, books, or notes.
And please, people - no wagering.
The point is the following. Any one problem one faces in real life related to these will not be of the form "graph this function". Rather, it will be a succession of a few dozen steps, where a couple of them along the way are of the form "and the function behaves a certain way, only here". Any one step, sure, you can do by calculator, but this dependence will slow you way down. Enough so, that you will tend to abandon the whole question as not the sort you can reasonably do in practice.
Related story: the math department here requires the PhD students to have a reading knowledge of French. Sounds pretty weird for a math dept until you notice how many math articles have been, and continue to be, written in French. "But this is what French-English dictionaries are for, no?" Sure; in theory you could read a French article knowing no French, using a dictionary. But in practice you never would, you'd say "that's too slow and just isn't gonna happen".
Final example: if I ask you a question about parabolas, I expect you to know that a parabola has a unique minimum (or maximum). I hope this has, by now, become automatic! If you had to break in the middle of a parabola problem to type in the parabola, wait for it to be graphed, reset the range on x and y to make sure you had a reasonable range, graph again, and eventually say "um.. in this range it _seems_ to have a unique min, at least" you'd think parabola questions were hard. And facing them in real life, you'd avoid problems that involved them, which is an awful lot of problems.
Let's focus on the second midterm. The questions concerned
all of which were well represented on the homework and review problems, and discussed in class.Definitely I made a mistake on the first midterm, blithely assuming people remembered material about trigonometric functions, and I am sorry; but on this one all the necessary material was in the stuff we did in the weeks up to the midterm.
Incidentally, there was no one pattern explaining who got which question; people who got #1 and #3 missed #2, and so on. Which is how it should be.
Certainly the style of question is not the same. But I really don't see that I'm testing on concepts that are not taught by the homework, nor that one could take the tests without understanding that material on the homework (to those who describe studying the homework and review questions as "useless" for taking the exam).
Since UCB is not a trade school, and you're not in a program training you for a very specific, repetitive job, it doesn't make sense to pick a particular kind of question (like "What is the average of this exponential function?") and ask it over and over to ensure that you can successfully answer that one question. (And then give it on a test, to check.) Of course we could set the course up that way (certainly it'd be a lot less work to teach); but what dull sort of job would you then be poised to take?
Rather, this course (and certainly this book) is really most about you understanding how to read graphs, and go back and forth between them and other ways that a function might be specified. There's no one way or five particular ways to do that.
Put most bluntly: this is not a memorization class. By the time we get through present and future value stuff, say, you should be able to answer any pure "present vs. future vs. an income stream" question, not just the five or six that have come up in homeworks.
Who says they're so low? It's not like I'm going to fail everyone who gets less than 90%. That would be a really badly designed test; a small careless mistake, and you're sunk. Whereas on the second midterm you could miss an entire question out of three (which very few people did), and still do just fine on the whole.
I knew a chemistry professor who gave 100-pt tests where the average might be 10. Stupid, a waste of time for all concerned. Luck into an extra right answer and *poof* you look like a genius. But exactly the same reasoning holds for tests with an average that is too high.
We'll take the total homework grade and scale it to 44%, the total midterms and scale to 24%, and the final scaled to 32%, then add them together. This magic number will then determine your grade - higher magic number, higher grade. In particular we are NOT turning each of categories into a letter grade and then averaging the letter grades together, and we have NOT been making any guesses about final grades based on homework grades.
Our greatest effort is being put into avoiding nasty surprises at the end - if we've been telling you that you're getting B's on the tests including the final, and your homework is good, then you won't get a C in the class, even though lots of people at the lower end may have dropped.
If we had only the midterms to judge by, we'd take their sum, and the A|B dividing line would be the sum of the individual A|B cusps from the individual tests (likewise the B|C cusps, etc.). In particular, anyone doing B or better on each test would be guaranteed to do B or better on the combination.
I have looked at the distribution of homework grades and decided where the A|B|C|D|F boundaries are there too. It seems to be commonly believed in this class that everybody is getting 100% on the homeworks and "therefore they don't count", but this really isn't true. Of course the average homework grade is not like the average midterm grade (55-60%) and the grading scale on the homeworks is indeed rather more stringent. Still, high homework grades can definitely help.
So let's calculate! Add up your midterm scores and multiply by 56/300; the 56 is for the percentage of your grade due to tests, the 300 for the total of the possible midterm scores. Then add up your homework scores and multiply by 44/27, for analogous reasons. Call this combination your Magic Number. (There are a very very few people who may profit by moving the 4% from the missing homework to their tests instead, and I am happy to do this. In this case they would weight by 60/300 & 40/27.)
Guess your final grade from this table. Again, there's so much (32%) tied up in the final that this is little guarantee either way.
| If your Magic Number is 84% or more: | some kind of A |
| If your Magic Number is 71-84% | some kind of B |
| If your Magic Number is 57-71% | some kind of C |
| If your Magic Number is 44-57% | some kind of D |
| lower | looks like an F |
Horrible as its scheduling is, much like our 8 AM class in general, this date was already set in stone by the time people registered for this course. Arranging for a separate time for people to take the final in a course this big would mean (in addition to all the difficulties in making and grading a different final) scheduling a room large enough. That's difficult enough in general, and essentially impossible during finals week, so No.