Question 334213
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Think about how we take an average.  We add up all of the numbers and divide by the number of numbers.  The final average for our student will be the sum of all 4 test scores divided by 4.  Let's say we want to achieve an overall average of *[tex \Large \mu] for *[tex \Large n] data elements given *[tex \Large n\ -\ 1] data elements.


We know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \mu\ =\ \frac{\sum_{i=1}^n\,x_i}{n}]


So we can say that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^n\,x_i\ =\ n\mu]


But *[tex \LARGE x_n], the answer to the question, can be found by:


*[tex \LARGE x_n\ =\ \sum_{i=1}^n\,x_i\ -\ \sum_{i=1}^{n\, -\, 1}\,x_i]


Which is to say:


*[tex \LARGE x_n\ =\ n\mu -\ \sum_{i=1}^{n\, -\, 1}\,x_i]


So add up the scores for the tests already taken and subtract that sum from the product of the total number of tests multiplied by the desired overall average.  Done.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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