Work Out My UMS, please? Any mathematicians who know what standard deviation is? Watch

sunburnt_note
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#21
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#21
OK, I'm doing it based on various pieces of information you've given which appear to correspond with the mark scheme:

Unit 1 - 82/90
Unit 2 - 80/90
Unit 3 - 87/120
AS total: 249/300

Unit 4 - 37/75 (raw) is 42/90 in UMS
Unit 5 - 41/75 (raw) is 56/90 in UMS
Unit 6 - 81/100 (raw) is 114/120 in UMS

Making your total 461, a high B

Since the boundary for a B is 420/600, you'd have to majorly mess up in order to get a C - it doesn't look likely
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IanDangerously
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#22
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1. 6241/01 - 82/90
2. 6242/01 - 80/90
3. 6243/01 - 87/120

4. 6244/01 - 43/90
5. 6245/01 - raw: 55/90

Total after 5 Units: 347

I'm confused as to how to work out this one though:
6. 6246/01B & 02 - raw: 118/150 [2 different papers but one unit for this one]

Theres nothing out of 150 on the markscheme :|

(Original post by sunburnt_note)
Unit 6 - 81/100 (raw) is 114/120 in UMS
Eh? I thought he said 118/150 Raw marks for Unit 6. Which confused me because Unit 6 should be out of 100, but where'd the 81 come from? :confused:
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sunburnt_note
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Eh? I thought he said 118/150 Raw marks for Unit 6. Which confused me because Unit 6 should be out of 100, but where'd the 81 come from? :confused:
In the first post he said 81/100 was what he was predicted himself for the synoptic.


Also, you and I have slightly different figures for each unit (doesn't really matter, because they ultimately work out the same), but how are you working it out? I'll illustrate an example of what I do:
Say person X has 42/60 for his history exam; the A boundary in this instance is 46/60 (going to 72/90 UMS), but the B boundary is 40 (going to 63/90 UMS). I divide 9 (the number of UMS marks between the boundaries) by 6 (the number of raw marks between boundaries), multiply it by 2 (because the person has 2 raw marks over the B boundary), and add that number to the lower boundary (in this case, 3 add 63), giving a total of 66/90 UMS.
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