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What is the difference?For what kind of purpose do we use the weighted average method?Why don't we instead always use the normal average method?and also,how to get the weighted average of some numbers?Also,do we have first to get the mean(normal average) of numbers in order to get their weighted average?

Thank you so much.

  • I am under the impression this question is a duplicate of something but I cannot find a post. – Lost1 Jul 21 '14 at 12:20

2 Answers2

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Let's say that you are flipping a coin, and with heads you win a dollar, tails you lose a dollar. Then you'd expect that the average amount of money you'd win is zero.

Now, let's say it turns out the coin was weighted, so that heads came up half as often as tails. Then the amount you'd expect to win is:

$$\frac{2\cdot(-1)+1\cdot 1}{3} = -\frac{1}{3}$$

This is a weighted average of $-1$ and $1$, based on their frequencies.

Thomas Andrews
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    Would you be kind enough to explain how you got 3 as the denominator? Does it imply 3 flips? – Mohamad Mar 30 '16 at 12:58
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    It's more accurate to write it as $\frac{2}{3}\cdot(-1)+\frac{1}{3}\cdot 1$, where the $2/3$ is the probability that you get tails and $1/3$ is the probability that you get heads. – Thomas Andrews Mar 30 '16 at 13:00
    1. This answer does not provide a clear distinction between the weighted and arithmetic averages. Furthermore, computing them would give the same result, as explained in Patrick A. answer. In this case the arithmetic average would be given as $\frac{(-1) + (-1) + 1}{3} = -1/3$. It is just that the weighted average is more convenient for data sets where a certain event repeats more than once.
  • – bruno Apr 10 '20 at 17:47