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(D) The probability mass function of a Poisson distribution is given by $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!}$.Given the equation $2 P(X=1) = 5

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$\sqrt{2}$

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(D) The probability mass function of a Poisson distribution is given by $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!}$.Given the equation $2 P(X=1) = 5 P(X=5) + 2 P(X=3)$.Substituting the formula: $2 \frac{e^{-\lambda} \lambda^1}{1!} = 5 \frac{e^{-\lambda} \lambda^5}{5!} + 2 \frac{e^{-\lambda} \lambda^3}{3!}$.Dividing both sides by $e^{-\lambda}$ (since $e^{-\lambda} 
eq 0$):$2\lambda = \frac{5 \lambda^5}{120} + \frac{2 \lambda^3}{6}$.$2\lambda = \frac{\lambda^5}{24} + \frac{\lambda^3}{3}$.Dividing by $\lambda$ (assuming $\lambda > 0$):$2 = \frac{\lambda^4}{24} + \frac{\lambda^2}{3}$.Multiply by $24$: $48 = \lambda^4 + 8\lambda^2$.$\lambda^4 + 8\lambda^2 - 48 = 0$.Let $u = \lambda^2$,then $u^2 + 8u - 48 = 0$.$(u + 12)(u - 4) = 0$.Since $\lambda^2$ must be positive,$u = 4$,so $\lambda^2 = 4$,which implies $\lambda = 2$.The standard deviation of a Poisson distribution is $\sigma = \sqrt{\lambda}$.Therefore,$\sigma = \sqrt{2}$.

$X = x_i$	$1$	$2$	$3$	$5$
$P(X = x_i)$	$2k^2$	$k$	$k$	$k^2$

If $X$ is a Poisson variate such that $2 P(X=1)=5 P(X=5)+2 P(X=3)$,then the standard deviation of $X$ is

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