b)

In simple words

$P(z)=|ψ(z) |^2~1/p(z) ~1/v(z)$

So that the bigger the probability to find the particle is space Δz centered on z the smaller the speed of the particle is in that region of space and the longer the time spent by the particle in space Δz is.

c)

The width of the forbidden region (outside the well to the right) is Δz

$∫_{z_0}^{z_0+Δz} \sqrt{2m(mgz-E)}*dx=\sqrt{2m}*(2(mgz-E)^{3/2})/3mg (\text{from z0 to z0+Δz})=$

$=\sqrt{8/9m} (mgΔz)^{3/2}≈3ℏ$

$(8/9m)^{1/3}*mgΔz=(3ℏ)^{2/3}$

or $Δz=(3ℏ)^{2/3}/mg*(3^{2/3}*m^{1/3})/2=1/2g*(9ℏ/m)^{2/3}$

Inside the forbidden region the probability is ($p(z_0 )=0$ so at the denominator one has $p(z_0+Δz)$)

$P≈D^2/(|p(z_0+Δz)|)*exp(-2*2π/T*z_0 )*Δz$

$P≈m/T*1/\sqrt{(2m^2 gΔz)}*exp(-4π/T*E/mg)*Δz=$

$=1/(T√2g) √Δz*exp(-π*√(2E/m))$

$P≈1/T*\sqrt[3]{9ℏ/m}*exp(-π*√(2E/m))$