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For a poisson distribution with parameter $ \lambda=0.25$ the value of the $2^{nd}$ moment about the origin is

\[\begin{array}{1 1}(1)0.25&(2)0.3125\\(3)0.0625&(4)0.025\end{array}\]

1 Answer

For Poisson distribution $\mu_1'=\mu_2= \lambda$
$\therefore \mu_2= \mu_2'-\mu_1'^{2}$
$\lambda= \mu_2' -\lambda^2$
$\mu_2'= \lambda^2+\lambda$
$\qquad= (0.25)^2+0.25$
$\qquad= 0.0625+0.25$
$\qquad= 0.3125$
Hence 2 is the correct answer.
answered May 23, 2014 by meena.p
 

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