Exercise 10.13.6

[Pipe-Vash]

• On part b) the correct 95% CI would be $\Phi(\overline{X}) \pm z_{.025} \frac{\phi(\overline{X})}{\sqrt{n}}$.
• On part e) As correctly expressed here, the true value of $\psi= 1-F_X(0)$. What I think is the correct procedure for the rest of the exercise is the following: By the WLLN $\overline{X}$ converges in probability to $\mu_X=\mathbb{E}(X)$ (the true mean of the RVs $X_i$). According to this, $\hat{\psi}$ tends in probability to $\Phi(\mu_X)$. To finalize, in general we have that $1-F_X(0)\neq \Phi(\mu_X)$.