Answer: 0.1

For a Poisson distribution, the probability mass function is given by: \[ p(x) = \frac{e^{-\lambda} \lambda^x}{x!} \] Given: \[ p(1) = \frac{e^{-\lambda} \lambda^1}{1!} = 0.09 \] \[ p(2) = \frac{e^{-\lambda} \lambda^2}{2!} = 0.0045 \] We can solve for \(\lambda\) by taking the ratio: \[ \frac{p(2)}{p(1)} = \frac{\frac{e^{-\lambda} \lambda^2}{2!}}{\frac{e^{-\lambda} \lambda}{1!}} = \frac{\lambda}{2} \] So: \[ \frac{0.0045}{0.09} = \frac{\lambda}{2} \] \[ 0.05 = \frac{\lambda}{2} \] \[ \lambda = 0.10 \] For a Poisson distribution, the variance equals the mean: \[ \text{Var}(X) = \lambda = 0.10 \] Therefore, the correct answer is 0.1.

Author: Nikitesh Somanthe