= Solution
The <standard normal density> is
$$
\phi(z)=\frac1{\sqrt{2\pi}}e^{-z^2/2},
\qquad z\in\mathbb R.
$$
It is nonnegative, and the <Gaussian integral> gives $\int_{-\infty}^{\infty}\phi(z)\,dz=1$, so it is a <probability density function>. Completing the square gives the <moment-generating function of a standard normal variable>
$$
M_Z(\theta)=\int e^{\theta z}\phi(z)\,dz
=e^{\theta^2/2}.
$$
Thus $\mathbb EZ=M_Z'(0)=0$ and
$$
\boxed{\operatorname{var}(Z)=M_Z''(0)-M_Z'(0)^2=1.}
$$
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