Explanation

For a Poisson random variable, the probability mass function is:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

Given:

• $P(X = 1) = 0.09 = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-\lambda} \lambda$ ...(i)
• $P(X = 2) = 0.0045 = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^2}{2} = 0.0045$

From equation (ii): $\frac{e^{-\lambda} \lambda^2}{2} = 0.0045$ $e^{-\lambda} \lambda^2 = 0.009$ ...(ii)

Now, divide equation (ii) by equation (i): $\frac{e^{-\lambda} \lambda^2}{e^{-\lambda} \lambda} = \frac{0.009}{0.09}$ $\lambda = 0.1$

For a Poisson distribution, the variance equals the mean: $\text{Var}(X) = \lambda = 0.1$

Therefore, the correct answer is D. 0.1