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#1 2008-01-03 02:52:59
- oyster
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- Registered: 2008-01-03
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Average grade
I am trying to work out what my average grade would be but have no idea how to do it properly and would appreciate it if somebody could demonstrate what to do. Currently i have 4 Distinctions, 1 Merit and 3 Passes what would the average grade be?
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#2 2008-01-04 02:21:24
- Jai Ganesh
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Re: Average grade
You'd have to tell us what's a distinction, what's a merit and what's a pass in order to answer this question.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
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#3 2008-01-04 06:03:06
- oyster
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Re: Average grade
well they are grades so they do not have a numeric value i suppose in letter terms distiction is an A, Merit B & Pass C
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#4 2008-01-04 06:49:54
- Daniel123
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- Registered: 2007-05-23
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Re: Average grade
Based on that your average would probably be a merit, but the average grade may be calculated in a particular way?
Last edited by Daniel123 (2008-01-04 06:51:15)
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#5 2008-01-04 07:41:51
- mathsyperson
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Re: Average grade
Your average grade would probably be worked out by averaging all of your marks, then treating that average as if it was a single mark and seeing what grade that would get.
If you only know your individual grades, then you can only make an educated guess.
Merit seems the most likely one though.
Why did the vector cross the road?
It wanted to be normal.
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#6 2008-01-04 10:26:39
- MathsIsFun
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Re: Average grade
well they are grades so they do not have a numeric value
It may be possible to give them a numeric value. A Distinction might be 80-90%, so you could assign a nominal 85% to it, and so on. Then average those values, and find what the equivalent grade is.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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#7 2008-01-04 14:07:34
- John E. Franklin
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- Registered: 2005-08-29
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Re: Average grade
Here's one way it might be, if it is a straight line (linear) between the three grades-levels.
pass=unknown
merit = pass + 1
distinction = merit + 1
unknown = 0 for this trial.
3 zeros, 1 one, and 4 twos makes 9 total. Divide 9 by 8 records and get 1 + 1/8.
1.125 average; and 1 is merit and 2 is distinction for this example.
For a second try, lets use pass = -5, merit = -4, and distinction = -3.
Three -5's + one -4 + four -3's = -15 -4 - 12 ---> -19 - 12 ----> -31 total.
-31 / 8 records is -24/8 - 7/8 ----> -3 -7/8 ----> -3.875 average
-3.875 average and -3 is distinction and -4 is merit in this example trial run.
In both examples, we are 12.5% of the way from merit to distinction.
Also, we are 87.5% of the way from distinction to merit.
You can set up boundaries at 50% away points, which is halfway between if you want.
This is only one of many ways to do this problem.
Bye.
igloo myrtilles fourmis
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#8 2008-01-05 06:24:47
- John E. Franklin
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- Registered: 2005-08-29
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Re: Average grade
You can also do the average
by moving objects around
on a table.
Here is a diagram that
shows the objects being
moved to the center where
merit is located.
In the end, 7 objects are
in the merit center area,
and one object is left in
the distinction area to the
right.
igloo myrtilles fourmis
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