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I have an average of 3,5 over 16 data points. If I wish to increase this average to, say, 6, how many future data points must be of a certain value? Assume all future data points are of the same value.

If the current average of 16 data points is 3,5, then four new data points each of value 7 would result in a new average of 4,2.

I know how to calculate this by hand. What I'm looking for is some formula that I can plug in to my spreadsheet.

Would it be possible to calculate how many data points are needed to reach an average of 6, given each new data point has a value of 7? Also, would it be possible to reverse the unknown(say I know there will be 20 more data points. What does each value have to be to reach a new average of 6)?

Is there a term for this type of formula/problem? I tried searching for a solution but I'm not quite sure what I'm supposed to search for.

mhrvatin
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  • The total of your $16$ points is $3.5\times 16$. If you wish to add $N$ new points, each with value $v$, then the new average will be $\frac {3.5\times 16+v\times N}{16+N}$. You can set this to whatever you like. Of course, to get numerical values, you'll ned to specify either $v$ or $N$. – lulu Jun 11 '20 at 14:49

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A simple way of thinking about this is to consider the current total sum in any situation. In the beginning, you have $16$ data points and their total sum is $3.5 \cdot 16 = 56$. Now, you wish to add some terms, let's say $n$ terms (a known number), of value $x$ so that the new average is $\mu_{new}$. This just means that the new total sum is $$ 56 + nx = (16+n)\mu_{new} $$ from which you can solve for $x$: $$ x = \frac{(16+n)\mu_{new} - 56}{n} $$ If instead you know $x$, you can solve for $n$: $$ n = \frac{ 16\mu_{new}-56 }{x -\mu_{new}} $$

Matti P.
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  • Thanks a bunch!! That makes so much sense and worked out perfectly in my spreadsheet :) – mhrvatin Jun 11 '20 at 19:24
  • Would you be able to explain to me how you go about to solve for n? I don't understand how you got rid of the parenthesis and how µnew ended up negative, nor how it ended up next to x(sorry, I don't know what the part above or below the line, or what the line is called in English). – mhrvatin Jun 12 '20 at 19:38
  • Starting with $$ 56+nx = (16+n)\mu_{new} $$ Opening the parentheses results in $$ 56+nx = 16\mu_{new} + n \mu_{new} $$ Then move the terms with $n$ to the left and other terms to the right: $$ nx - n \mu_{new} = 16\mu_{new} - 56 $$ Can you finish from here? – Matti P. Jun 14 '20 at 07:12