Banaketa binomial: berrikuspenen arteko aldeak
Ezabatutako edukia Gehitutako edukia
77. lerroa:
::<math>\begin{
var[Y] & = var[X_1]+var[X_2]+\ldots+var[X_n]\\
& = & pq+pq+\ldots+pq\\
& = & npq\\
\end{
</math>
}}{{kaxa zabalkorra
94. lerroa:
::<math>\begin{
E[X(X-1)] & = \sum_{i=0}^nx(x-1)p(x)\\
& = & \sum_{x=0}^nx(x-1)\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}\\
& = & \sum_{x=0}^nx(x-1)\frac{n!}{(x-2)!(n-x)!}p^x(1-p)^{n-x}\\
& = & \sum_{x=2}^n\frac{n!}{(x-2)!(n-x)!}p^x(1-p)^{n-x}\\
& = & n(n-1)p^2\sum_{x=2}^n\frac{(n-2)!}{(x-2)!(n-x)!}p^{x-2}(1-p)^{n-x}\ \ \ (y=x-2;\ m=n-2)\\
& = & n(n-1)p^2\sum_{y=0}^m\frac{m!}{y!(m-y)!}p^y(1-p)^{m-y}\\
& = & n(n-1)p^2(p+(1-p))^m\\
& = & n(n-1)p^2\\
\end{
</math>
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