= Solution
At each fixed time, $|\psi(x,t)|^2$ is a bell-shaped <Gaussian distribution>[Gaussian] centred at $x=h(t)$, with width determined by $\operatorname{Re}F(t)>0$. Completing the square as in part (d) makes the density symmetric about $h(t)$. Therefore its position <expectation value> is
$$
\boxed{\langle\widehat x\rangle_\psi=h(t)}.
$$
The sketch is a Gaussian peak whose centre oscillates between the two turning positions found in part (e), while its width varies periodically through $\operatorname{Re}F(t)$.
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