# math problem

• posted

I come to the experts on everything I can think of help. My teacher-wife asked me if she was calculating this correctly. I didn't think so. What's the answer and why/how?

Chapter tests count for 60% of the final grade. Mid-term and final tests count for 40% of the final grade. Scores for chapter tests were

80, 50 and 69. Mid-term and final tests were 80 and 90. What is the final grade. One of us says 73.8 and the other says 80.7

Thanks

• posted

Okay, let's assume the chapters are *equally* weighted (with each other), as are mid-term and finals. These are assumptions. And all out of 100, natch.

1So there are 3 chapter tests, each would count for 60/3 = 20%, and two exams, which would count for 40/2 = 20% each as well. That's easy.

Just multiply each score by 0.2 and add, or to put it another way, it's less calculation to just add up all the scores and divide by 5.

I get 73.8% for the example given.

Best regards, Spehro Pefhany

• posted

It sounds like a bullshit way to grade a course, when the mid-term & final each count the same as a chapter test. But then....... Yesterday I was talking to a guy w/ an engraving shop. I commented that Little League must be good for business. He replied, "Yes, it's very good. EVERYONE who plays gets a trophy...... not like before when only 1st & 2nd place teams got them and the rest of us got, "try harder next year."

• posted
73.8 is correct, how did you get 80.7?
• posted

I get 73.8%. Here's how I got it:

First, find the average of the test scores: (80+50+69)/3=66.3%

Then, find the average for the midterm and the final: (80+90)/2=85%.

Now, calculate the weighted percentage: (66.3%)*0.6+(85%)*0.4= 73.8%

How does the math go to get 80.7%??

• posted

I think that your answer of "add it up and divide by 5" only works because the number of tests is the same ratio as the weighting. You can see this by taking an extreme case. Suppose there were 100 chapter tests and only two term tests, this would not be right.

The formula should be:

grade = ((sum(chapter tests)/number of chapter tests)*weight)+((sum(term tests)/(number of term tests)*weight)

You get the same answer but this equation works for any number of tests and any weighting.

73.78= (((80+50+69)/3)*.60)+(((80+90)/2)*.4)

I hope I didn't just do some high schoolers homework:-)

Al

• posted

oops, finger check

73.8 not 73.78

Al

• posted

I sure hope this wasn't a math class. As others have said, 73.8.

The average of the chapter test scores count for 60% of the total, so take the average of them and multiply that average by 0.6.

Then average the midterm and final grades, and multiply that average by 0.4.

Then add those two weighted averages together. 73.8

Ed Huntress

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Which one says which?

• posted

Yes, that is correct, as I hope I explained. Being inherently lazy and concerned with entry errors, I try to simplify things before crunching numbers (and by using an RPN calculator, of course).

This only works if the chapter tests and the exams are weighted equally with each other.

n --- \ G= | Wi * Si / --- i=0

Where Wi is the weight of each individual score (0 You get the same answer but this equation works for any number of tests and any

With the above assumption. It's also possible, for example, that the 3 chapter exams add up to 60%, but they are not equal.

Oh, well.

Best regards, Spehro Pefhany

• posted

Around here, all the kids get "participation" trophies (thanks, China), a couple get "most improved" and "most sportsmanlike" trophies and all members of the first, second place teams get much nicer (or gaudier/bigger trophies, depending on the taste of the people doing the selection). On tournaments, the winners get a trophy (first place), second and third get crummy medallions.

Believe me, the kids still know the difference between winning and not winning, and they REALLY want to win.

Best regards, Spehro Pefhany

• posted

The problem is incompletely specified (method of averaging not given) and there are many "correct" answers.

By weighting the separate averages:

(80+50+69)*0.60/3 + (80+90)*0.40/2 = 73.8

But because the number of samples (3 and 2) happen to match the 60/40 weights, this is simply to overall average:

(80+50+69+80+90)/5 = 73.8

It would also be reasonable to use RMS averaging (75.0 here), or different weights to the extreme score values (e.g., drop the lowest, giving it a weight of zero). These would yield different result which were less sensitive to sampling errors (one poor performance on a bad day). But that's asking a lot from a shop teacher, I suppose.

All of which just shows that grades are full of noise and bias, and are unreliable detectors of intelligence.

• posted

actually the terms chapter, mid-term and final were made up. I don't know exactly what they were, just their weight.

• posted

I knew I shouldn't have put that last sentence in there ;-)

• posted

According to these conditions, each test of whatever sort counts for 20%.

73.8 is correct for the conditions given (60% chapter, 40% mid/final). 77.5 would be correct if the weighting was 60% mid/final, 40% chapter

I can't figure out where 80.7 came from, other than that place where most statistics come from, which I'd not care to investigate due to the smell.

The closest I can get would be 80.8 from dropping the lowest chapter test score, and using 40% weight chapter, 60% mid/final.

Someone needs to go back to math class - about High School Algebra 1 or so.

• posted

The explanation answers the original question -

Why is the final grade 73.8 rather than 80.7?

No question asked for a formula for the general case of N chapter tests where N may not match the ratio of the test weights. I agree with the reasoning applied however.

I am more curious as to how a final grade of 80.7 was computed, though it was not documented. Tom

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I must be lazier than you. I don't like to have to simplify things or worry about entry errors so I use an AE calculator, of course.

Richard Coke

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As credentialed teacher in California, I would say that it depends if the group of grades were first averaged and then percentage taken or were the grades totaled and then perenctages taken. And were the perenctages averaged? Personally, I have never gone for the averaging angle. If I see that the total class in scoring low, I beleive in looking in the mirror to find the problem!

Neal

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There are grading calculators made just for teachers that simplify these things. My exwife had one and it worked great. Probably something for the computer nowadays. I know its not hard math but when you just need to get the grades in on time....

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I worked for a trophy shop for years and that was a large part of the business just cranking out the same cheap junk that I was certain the kids would know they didnt deserve. THe leagues have budgets and now that women run most of this stuff its all touchy feely.

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