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If in a poisson distribution $P(X=0)=k$ then the variance is

\[\begin{array}{1 1}(1)\log\frac{1}{k}&(2)\log k\\(3)e^{\lambda}&(4)\frac{1}{k}\end{array}\]

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In Poisson distribution
$P(X=x) =\large\frac{e^{-\lambda}. \lambda^0}{x}$$, x=0,1,2...$
$P(X=0) =k=\large\frac{e^{-\lambda}. \lambda^0}{0}$
$k=e^{-\lambda}=> \log e^{- \lambda}=\log k$
$- \lambda \log e = \log k => -\lambda =\log k$
Varience $=\lambda =-\log k=\lambda= \log \bigg(\large\frac{1}{k}\bigg)$
Hence 1 is the correct answer.
answered May 23, 2014 by meena.p
 

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