Imagine we have a collection of 10 heights of individuals in feet. We calculate the average value as: $$ \mu=\frac{1}{10}\sum_{i=1}^{10}h_{i} $$ where $h_{i}$ is the height of an individual $i.$ In this case, is the average $\mu$ in units of feet, or units of feet/individual? From economics, the average cost of producing $q$ units is given as: $$ AC\left(q\right)=\frac{TC\left(q\right)}{q} $$ where $TC$ is total cost for producing $q$ units, $AC$ is average cost, and $q$ is total quantity. Here, the average cost is in units of say \$/unit of output. My question: in which circumstances are the units of height in units of the variable being measured- say $y,$ and in which are they two dimensional, say $y$ per $x$ (like cost per unit, or average speed per hour)? Does it matter?
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Total is sum. Average is sum/number. – herb steinberg May 29 '21 at 20:58
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Agreed. But in which cases do we account for the number having different units than the sum? – Kwame Brown May 29 '21 at 21:10
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The average has the units of the sum per individual. I don't see anything else. In terms of the title, x is simply the number of items in the sum. – herb steinberg May 29 '21 at 21:14
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Apologies - but I think I was not clear. Is there a meaningful difference between the average being in feet, or in feet per individual? – Kwame Brown May 29 '21 at 21:19
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1Strictly speaking, the units are feet/person; we just usually don't say it that way for measures such as average height, average weight, average age, and the like, but it is understood that the average is "per person". When we talk about "average wealth per capita" [Latin for "per head"], it is understood to be, say, dollars per person, and the measure indicates that, so we again generally just give the average as dollars. Incidentally, "per unit" or "per person" is dimensionless, so the average height only has dimensions of feet. – May 29 '21 at 23:10
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To make a distinction for one of your examples, average speed is a ratio of two "dimensionful" (I swear that's what physicists call them now) quantities, so the ratio of distance divided by time has a unit of the ratio of two dimensions, length and time. This is a different sort of average from that of average height. – May 29 '21 at 23:17
1 Answers
Compare:
✔ “The average height of the students is $178.3\textrm{ cm}.$”
✗ “The average height of the students is $178.3\textrm{ cm}\text{/student}.$”
✔ “The average production cost of the units is $\$459.$”
✗ “TThe average production cost of the units is $\$459\text{/unit}.$”
✔ “The average speed of the car for journey A is $75.1\textrm{ km h}^{-1}.$”
✗ “The average speed of the car for journey A is $75.1\textrm{ km h}^{-1}\textrm{/h}.$”
✔ “The average number of goals among the teams is $13.4.$”
✗ “The average number of goals among the teams is $13.4\text{ goals/team}.$”
Average height is after all still a height, and makes sense to have exactly the same dimension as individual height. That we are referring to student heights (as opposed to tree heights) belongs to the context, rather than to the quantity's specification per se.
The third example makes this even clearer: surely the average speed doesn't have units $75.1\textrm{ km h}^{-2},$ which is that of acceleration!
In the same vein:
✔ “The number of goals scored by the team is $7.$”
✗ “The number of goals scored by the team is $7\text{ goals}.$”
The gradient of a displacement-time graph is instantaneous velocity—not average displacement nor average velocity —so has units $$\textrm{km h}^{-1}$$ instead of merely $$\textrm{km}.$$
In the graph of Total Cost $(TC)$ against Quantity $(q),$ the gradient is Marginal Cost $(MC),$ not Average Unit Cost $(AC),$ which is just $\displaystyle\frac{TC}q.$
$TC, MC,$ and $AC$ all have unit $$\$.$$
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