Question 908695
<font face="Times New Roman" size="+2">


If a mean of *[tex \Large n] scores is *[tex \Large \mu], then the sum of the *[tex \Large n] scores is *[tex \Large \mu\cdot n].  The new sum if *[tex \Large x_{n+1}] is added is *[tex \Large \mu\cdot n\ +\ x_{n+1}], and the new mean, *[tex \Large \mu_{n+1}\ =\ \frac{\mu\cdot n\ +\ x_{n+1}}{n\ +\ 1}].


In other words, multiply 68 times 5, add 20 to the product, and divide the final sum by 6.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank">
<img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>