Pick a New Average And Find the Number of Units to Add I'm looking for a formula to find how many units I have to add to an existing average in order to get to a new average that I set before.
Example:
I have an average of scores 82 that includes 17 scores, Now I want to get to a new average of 90,
I set up a score that will help me get there - 98, and now I want to know how many times I have to add 98 in order to get to the new average - 90.
I tried to look for the formula on the internet but didn't get anything about this.
Thanks !
 A: With your particular example, you have an average score of $82$ from $17$ scores, so then your sum of all scores is $82 \times 17 = 1394$. Assume that you want to add $k$ scores of $98$ so your average score would be $90$. You thus have a new total sum of $1394 + 98k$, so dividing by $17 + k$ should give you the desired average of $90$, i.e., you have
$$\begin{equation}\begin{aligned}
\frac{1394 + 98k}{17 + k} & = 90 \\
1394 + 98k & = 90(17 + k) \\
1394 + 98k & = 1530 + 90k \\
8k & = 136 \\
k & = 17
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, you would need to add another $17$ scores.
In this particular case, since the difference between your average score, i.e., $82$ and the score you're trying to get, i.e., $90$, is the same as from $90$ to the amounts you're adding, i.e., $98$, this is why you get the same number of values to add as your original number, i.e., $17$.
You can adapt this procedure for any general case. In particular, if you have an original average score of $A_O$ from $n$ scores, and you want get a new average of $A_N$ by adding $k$ scores of $s$, with $s \neq A_N$, then the formula to solve would be
$$\begin{equation}\begin{aligned}
\frac{A_O(n) + ks}{n + k} & = A_N \\
A_O(n) + ks & = A_N(n + k) \\
A_O(n) + ks & = A_N(n) + A_N(k) \\
(s - A_N)k & = (A_N - A_O)n \\
k & = \frac{(A_N - A_O)n}{s - A_N}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
With your example, you have $A_O = 82$, $n = 17$, $A_N = 90$ and $s = 98$, with substituting those values into \eqref{eq2A} giving $k = \frac{(90 - 82)17}{98 - 90} = 17$, as determined in \eqref{eq1A} earlier.
Note that since $k$ should be an integer, you may wish to round up to ensure you at least get to your new desired average of $A_N$ (i.e., if it's larger than $A_O$, the new average should be at least as large as $A_N$, and if it's smaller than $A_O$, the new average should be at least as small as $A_N$). If so, then the actual value would be
$$k = \left\lceil\frac{(A_N - A_O)n}{s - A_N}\right\rceil \tag{3}\label{eq3A}$$
