1. **Problem statement:** Given a Poisson random variable $X$ with $P(X=0) = \frac{1}{2}$, find the expected value $E(X)$. 2. **Recall the Poisson distribution formula:** $$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $\lambda = E(X)$ is the mean (expected value) of the distribution. 3. **Use the given probability:** For $k=0$, $$P(X=0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}$$ Given $P(X=0) = \frac{1}{2}$, so $$e^{-\lambda} = \frac{1}{2}$$ 4. **Solve for $\lambda$:** Take natural logarithm on both sides, $$-\lambda = \ln\left(\frac{1}{2}\right) = -\ln(2)$$ Thus, $$\lambda = \ln(2)$$ 5. **Conclusion:** The expected value of $X$ is $$E(X) = \lambda = \ln(2) \approx 0.693$$