Answer
One-third of the time.
Work Step by Step
Suppose the pendulum starts at maximum displacement. The equation of motion is known.
$$\theta=\theta_o cos \omega t=\theta_o cos \frac{2 \pi t}{T}$$
Find the time at which the position is half the amplitude.
$$0.5\theta_o =\theta_o cos \frac{2 \pi t}{T}$$
$$cos^{-1}0.5 = \frac{2 \pi t}{T}$$
$$\frac{\pi}{3} = \frac{2 \pi t}{T}$$
$$t = \frac{T}{6}$$
This is the time it takes for the pendulum to swing from 10 degrees to 5 degrees. We know it takes $\frac{T}{4}$ for the pendulum to swing from 10 degrees to 5 degrees. Thus it takes the difference, or $\frac{T}{12}$, for the pendulum to swing from 5 degrees to zero.
During this first quarter-cycle, the fraction of time spent between 0 and 5 degrees is $\frac{T/12}{T/4}=\frac{1}{3}$.
Due to the symmetry of motion, the behavior during each of the other quarter-cycles is the same. The pendulum is between -5 and 5 degrees one-third of the time.