http://i235.photobucket.com/albums/e...untitled-7.jpg

$\displaystyle \bar x_{n+1} = \frac{1}{n+1} \sum^{n+1}_{k=1} x_k$ ...and then, I'm stuck. No idea what to do next. (Headbang)

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- Apr 24th 2008, 08:56 PMcirrus74Sample mean and variance
http://i235.photobucket.com/albums/e...untitled-7.jpg

$\displaystyle \bar x_{n+1} = \frac{1}{n+1} \sum^{n+1}_{k=1} x_k$ ...and then, I'm stuck. No idea what to do next. (Headbang) - Apr 25th 2008, 04:02 AMmr fantastic
(a) $\displaystyle \bar x_{n} = \frac{1}{n} \sum^{n}_{k=1} x_k \Rightarrow \sum^{n}_{k=1} x_k = n \bar x_{n}$.

Therefore:

$\displaystyle \sum^{n+1}_{k=1} x_k = \left( \sum^{n}_{k=1} x_k \right) + x_{n+1} = n \bar x_{n} + x_{n+1}$.

Therefore $\displaystyle \bar x_{n+1} = \frac{n \bar x_{n} + x_{n+1}}{n+1}$.

You should have another go at (b) now ....