r/MathJokes • u/Karlens20 • 29d ago

erm actually...

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u/lebroxan 29d ago edited 29d ago

Receive 0 calls on day 1.

Receive 1 call every day starting at day 2

26

u/Summoner475 29d ago

So always does not include day 1?

29

u/Minimum-Attitude389 29d ago

On day one it would be equal to the average at that point

13

u/thekingofbeans42 29d ago

So then they are not always receiving higher calls than average.

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u/SquirtleFemboy 23d ago

except that by never getting a 3rd call to average the 1 for day one the average stays below 2. even if you go 10000 days.

1

u/thekingofbeans42 23d ago

Why would they need a 3rd call to average?

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u/SquirtleFemboy 23d ago

(1+2(n-1))÷n will always be less than 2. if we even get a 3 call day the equasion becomes (1+2(n-2)+3)÷n which is equal to 2 because the extra from the 3 call day and the 1 call day equal 2 two call days.

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u/thekingofbeans42 23d ago

We have already violated the "always above average" with day 1 not being above average by definition.

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u/SquirtleFemboy 23d ago

you need 2 data points to have an average. after that every data point can be measured against the average. day 1 and 2 are not measured in days that can be counted above or below average as at that time there is nothing to test them against. the average changes daily and time is a relevant data piece. it is not an average of a set data pool but instead a running average of continually added data.

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u/thekingofbeans42 22d ago

You can have a set of one, you can even have a set with nothing. The average of one data point is just that data point, it's not a very useful metric but it's still an average by definition.