OT - Calc of Avg temperature

Somewhat embarassed to ask this. Helping daughter with h/work. How do you calc the avg when the temperature values range from below zero to above? And do you tell if the result is +ve or -ve after the division?

ag -3, -2, -5 , 0, 0, 3, 4, 5

Thanks.

Reply to
artbag10
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No change to the rules add the values and divide by the number of values

total of 8 values = +2

average = 0.25 degrees

Reply to
Bob Minchin

Sum them (answer +2)

Divide by the number of readings (8)

Answer +2/8 = 0.25

Reply to
charles

On Sunday 02 February 2014 18:55 charles wrote in uk.d-i-y:

Yes - that is the definition of average. and just to claify in the general case, if you end up with a -ve number, that's fine and should be stated (to covert to +ve would be an error).

Reply to
Tim Watts

Just checking :) And of course if the total is -ve then the avg will be

-ve.

Thanks.

Reply to
artbag10

Others have given the answer, which you can calculate in your head.

Reply to
Tim Streater

In the context of such a mixture of positive and negative data point an answer of +0.25C would be slightly better.

Reply to
Anthony R. Gold

In article , Anthony R. Gold writes

Nah, I'd say it's reasonable to assume positive unless stated otherwise.

Reply to
fred

Never mind the average, that obviously positive proof of global warming, but as your daughter provided the data, definitely not man-made.

Reply to
nemo

This is homework. The answer should follow the format of the question, which does not use plus signs.

Colin Bignell

Reply to
Nightjar

Excel is more fun though :-)

Reply to
stuart noble

If we?re being particular about whether to include the positive or negative sign to clarify, should we not have a discussion about:

average: mean, median or mode average: arithmetic or geometric (but it can?t be the latter as the latter doesn?t permit negative numbers)

Reply to
Allan

Mean, mode or median ?

Reply to
Jethro_uk

On Monday 03 February 2014 10:22 Jethro_uk wrote in uk.d-i-y:

Arithmetic Mean - in common parlance, that is what "average" means...

Reply to
Tim Watts

The answer expected is probably the 0.25 as many have calculated.

However, in pedant mode, that's the average of these numbers. You can't use them to work out an average temperature - there's not enough data provided with just a few points at unknown times. At best, you could assume they are the average temperatures during equal sized sampling periods. If they are snapshot temperatures, then there's not enough data - it might have been -50 at some point between the second and third readings.

Reply to
Andrew Gabriel

It is an arithmetic homework question. Homework questions work within the confines of the artificial world created for the purposes of the question, not in the real world. It is a long time since I needed to do school homework, but I still recall that trying to think outside the restraints of the course work was a very good way to lose marks. About the only exception was English, where a certain amount of creativity was encouraged.

Colin Bignell

Reply to
Nightjar

Convert each to old money Fahrenheit using nearest available ready reckoner tables, add up values (all plusitive) then grab either log tables or a slide rule and divide by eight, using long division on a sheet of paper to verify accuracy. Convert back to celcius/centigrade figure or what not. Send kid to skool with an apple for teachie, laced with brandy.

Reply to
Adrian C

It's a question about simple arithmetic. Trouble is these days even such trivial questions have to be wrapped up in a cuddly "real world" question so the little darlings don't suffer major psychological trauma, an infringement of their human rights.

Notwithstanding what Caprimulgus said, IME even having done such questions doesn't help children work out how to solve real-world problems; they still need to learn or be taught how to *apply* their knowledge.

Reply to
Tim Streater

I think you are both right. I had much the same thoughts as Mr Gabriel but also see we really need to know the context. Eg if the daughter is preparing for Oxford Admissions Tests I'd hope a little more than arithmetic *might* be involved ;)

Reply to
Robin

Or do the same using Kelvin - no negative values to worry about - then go back to Celcius.

Reply to
PeterC

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