Calculate mean of the distribution which is a function of normal distribution

Question

Calculate mean of random variable $\max\{exp\{-\frac{7}{800}\},exp\{\frac{Z}{4}\}\}$, where $Z$ is normally distributed with variance 1 and mean 0.

Answer 1

$$Y=\max\{\exp(-\frac{7}{800}),\exp(\frac{Z}{4})\}=\begin{cases}\exp(-\frac{7}{800}),& -\frac{7}{800}\ge\frac{Z}{4}\\ \exp(\frac{Z}{4}),& -\frac{7}{800}<\frac{Z}{4} \end{cases}$$

$-\frac{7}{800}=\frac{z}{4}\Rightarrow z=-\frac 7 {200}$. Hence, $$\begin{align}E[Y]&=\exp(-\frac{7}{800})\Pr(Z<-\frac 7 {200})+E[\exp(\frac{Z}{4}), z\ge-\frac 7 {200}] \\&=\exp(-\frac{7}{800})\int^{-\frac 7 {200}}_{-\infty}\phi(z)dz+\int_{-\frac 7 {200}}^{\infty}\exp(\frac{z}{4})\phi(z)dz\\ &= 0.9913\left(1-Q\left(-\frac{7}{200}\right)\right)+e^{1/32}Q\left(-\frac{7}{200}-\frac{1}{4}\right)\ \end{align}$$

where $\phi(z)$ is the standard Gaussian pdf and $Q$ is the complementary cdf.