# Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

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2. ### UltraworldEmperor

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For a company the following data is given.
Code:
```Quarter|Employees|Total costs
----------------------------------
1        342     \$  4540
2,3,4      721     \$ 37221
```
(so for the months april - december you have 721 people who work for you and you have costs of \$ 37221)

How do you find the 'Average Cost per Employee per Year'?

My answer: 1/(1/4) * 1/342 * 4540 + 1/(3/4) * 1/721 * 37221 = 121.9

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4. ### Gabryel KarolinGammelgädda

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Data is not clear, does "2,3,4" mean those three combined or that the three have equal values in terms of employees and total costs.

5. ### Miseisle of lucy

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Yeah, it's a terrible table. To me the answer is (1/4 * 4540/342 + 3/4 * 37221/721)*4 = \$168

6. ### Gabryel KarolinGammelgädda

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yepp, I got the same. Either that or 65, if you count it the other way I mentioned.

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combined

8. ### El_MachinaeColour vision since 2018Retired Moderator

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Hmm I think the original method is correct, you divide the 3 quarterly total by 3 to get it per quarter, then multiply it by 4 to get yearly.

Mise is just multiplying the 3 quarterly total by 3 (effectively).

That's not multidimensional data anyway. A multidimensional mean is a vector.

10. ### UltraworldEmperor

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You can visualise it by thinking about 2 cuboids right next to each other

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Yeah it's a good way to visualise statistics as areas of a histogram.

I wonder why this thread has been moved to OT?

I'll report this post to see if it will get moved back

12. ### UltraworldEmperor

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Personally I am uncertain about my own answer, despite having a good background in quantitative reasoning. I am hoping people at CFC could help me.

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I'm not totally sure either, but I prefer your answer to Mise's since you have a divide by 3 rather than a multiply for the 3 quarter table entry.

You are effectively normalising each entry to cost/employee/quarter, then multiplying the sum by 4 to get yearly.

14. ### oagersnapEmperor

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You are overcomplicating it. It's

(4540/342)+(37221/721) = 64.9

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That depends whether there are 721 employees for each quarter in the 3 quarterly total, or 721 in total for the 3 quarters.

The table is unclear about which is the case though. Now I think you may be right!

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17. ### Gabryel KarolinGammelgädda

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I have a cold and haven't done much college math but I'm curious why the answer isn't simply 4580/342 + 37221/721. Or 4580/342 + (37221/721)*3.

You already have the total cost per year, its 4580+37221, why are you even dividing with 1/4, 3/4, it shouldn't be neccesary.

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Calm down!

I take it the example is from a book - are there not similar examples showing what the usual method is?

EDIT: I'm in the oagersnap/Gabryel camp now.

19. ### oagersnapEmperor

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Yes, that's right. I just naturally assumed that there would be 721 employees for each quarter.. But the table is quite unclear.

Exactly my point. The second solution you suggest would be correct if the 721 employees are a combined total for the last 3 quarters.

20. ### Miseisle of lucy

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Right, so now you have two "Cost Per Year" figures, one that is based on 1 quarter and has 342 employees and one that is based on 3 quarters and has 721 employees. So you divide Cost Per Year by employees to get 2 different "Avg Cost per Year per Employee" figures. Then you take a weighted average: the company spends 1 quarter using one avg cost, and 3 quarters using the other avg cost, so you weight one by 1 and the other by 3. The number you get will be the same as the number I have