Solution
= Solution
For any $t>0$, the <exponential Markov bound> and part (b) give
$$
\mathbb P(X\geq v)
\leq e^{-tv}M_X(t)=e^{-tv+t^2/2}.
$$
The minimum occurs at $t=v$, so the <Gaussian tail bound> is
$$
\boxed{\int_v^\infty\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx
\leq e^{-v^2/2}.}
$$