If X is a Poisson variable such that 3 P(X=4) | vedclass

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(B) Given that $X$ follows a Poisson distribution with parameter $\lambda$. The probability mass function is given by $P(X=x) = \frac{e^{-\lambda} \la

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$2$

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(B) Given that $X$ follows a Poisson distribution with parameter $\lambda$. The probability mass function is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$ for $x = 0, 1, 2, \dots$.Given the equation: $3 P(X=4) = \frac{1}{2} P(X=2) + P(X=0)$.Substituting the formula: $3 \frac{e^{-\lambda} \lambda^4}{4!} = \frac{1}{2} \frac{e^{-\lambda} \lambda^2}{2!} + \frac{e^{-\lambda} \lambda^0}{0!}$.Dividing both sides by $e^{-\lambda}$: $\frac{3 \lambda^4}{24} = \frac{\lambda^2}{4} + 1$.Simplifying: $\frac{\lambda^4}{8} = \frac{\lambda^2}{4} + 1$.Multiplying by $8$: $\lambda^4 = 2 \lambda^2 + 8$,which gives $\lambda^4 - 2 \lambda^2 - 8 = 0$.Let $u = \lambda^2$. Then $u^2 - 2u - 8 = 0$.Factoring the quadratic: $(u - 4)(u + 2) = 0$.This gives $u = 4$ or $u = -2$.Since $\lambda^2 = u$ and $\lambda^2$ must be non-negative,we have $\lambda^2 = 4$,which implies $\lambda = 2$ (as $\lambda > 0$).The mean of a Poisson distribution is $\lambda$,so the mean is $2$.

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x)$	$a$	$a$	$a$	$b$	$b$	$0.3$

If $X$ is a Poisson variable such that $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$,then the mean of $X$ is

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