Question 977825
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Assume that there are a total of *[tex \Large n] samples comprising this distribution, of which *[tex \Large p] have a value of 10.  Then since the only other value is 20, *[tex \Large n\ -\ p] samples must have a value of 20.


The probability of a 10 value is then *[tex \Large \frac{p}{n}] and the probability of a 20 value is *[tex \Large \frac{n\ -\ p}{n}]


Then, the expected value for the distribution is the probability of a 10 value times 10 and the probability of a 20 values times 20, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10p}{n}\ +\ \frac{20(n\ -\ p)}{n}\ =\ 16]


Solve for *[tex \Large p] in terms of a fraction of *[tex \Large n], and the fraction will be *[tex \Large \frac{p}{n}], the probability of a 10 value.


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

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*[tex \LARGE \ \ \ \ \ \ \ \ \ \