Sure.
In general relativity time intervals depend on the observer.
The important effect here is the gravitational time dilation. Basically, the stronger the gravitational field you're in, the slower time passes for you compared to a distant observer.
If you remember, the gravitational potential of a spherical body is given by V=-GM/r, where G is Newton's constant M is the mass and r is the distance from the center.
The time interval "T" measured by a distant observer is related to the free falling time "t" by
T=t/Sqrt (1+2V/c^2),
where V is the potential above, c is the speed of light and sqrt means square root. If the person is right at the horizon, r=2GM/c^2 (the black hole radius) and the denominator goes to zero, so the time T goes to infinity. This means that the time a distant stationary observer measures is infinitely dilated with respect to the time of the freefalling person at the horizon.
Because of this, from the outside viewpoint, the person is never able to cross the horizon and enter the black hole, but instead freezes right at the horizon.
The situation is quite different from the point of view of the person that is falling into the BH though. Since he falls freely, he doesn't experience a gravitational force, so from his point of view, time flows normally. All you have to do, roughly, is calculate the length he travels divided by his speed. The length is not the black hole radius because its actually bigger on the inside (seriously) and the speed is not constant, but complications aside, in the end you get that a finite time passes equal to t=pi GM/c^3 before you reach the center.
This is all for nonspinning BHs and radial trajectories, but qualitatively it is correct. Hopefully I didn't confuse you further!