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Five street vendors sell oranges. Four of them ask \$1 each, and one asks \$3.

The first approach to a price average would be ( 1 + 1 + 1 + 1 + 3 ) / 5 = 1.4.

The second approach, only considering unique prices, would be ( 1 + 3 ) / 2 = 2.

What are the correct terms for these two different averages, and in what situations is one a better approach than the other?

Average Joe
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I do not think there is a one unique 'correct way'. Different ways to calculate the mean tell you different things. The first one tells you about the mean price over the sample of shops/vendors. The second one tells you about the mean price level.

There are other possibilities. Say each of the cheap vendors sells $\frac{1}{4}$ of all sold oranges. Then $(1\frac{1}{4}+1\frac{1}{4}+1\frac{1}{4}+1\frac{1}{4})$ is the mean price at which oranges are transacted.

Jan
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The first one is the only one that makes sense. It's just the average. The second one doesn't have a name I know, and doesn't deserve one.

You could rewrite the first one as $$ \frac{4 \times 1 + 1 \times 3}{4 + 1} $$ and call it a weighted average.

This is the average vendor price. If the vendors had different numbers of oranges you could compute the weighted average price per orange.

Ethan Bolker
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  • How do you know that there is no venue of research whatsoever where the second kind of average is relevant? Because if it is useful, however marginal and obscure, then it probably has a name, and definitely deserves one. – Arthur Feb 03 '17 at 17:17
  • The first is the "average price paid, for all oranges sold", assuming all vendors each get an equal share of customers. I don't think that rules out eventual use of the value in the second approach? – Average Joe Feb 03 '17 at 17:21
  • @Arthur Yes, there may possibly be a context in which the second kind of average makes sense. Even if so I think it's unusual, so I would not imagine a widely known name. just one for that context. – Ethan Bolker Feb 03 '17 at 17:41