A student measures times for 20 oscillations of a simple pendulum...

Mean time $t_{mean} = \dfrac{30 + 32 + 35 + 31}{4} = \dfrac{128}{4} = 32\text{ s}$

Absolute errors in each measurement are:
$\Delta t_1 = |30 - 32| = 2\text{ s}$
$\Delta t_2 = |32 - 32| = 0\text{ s}$
$\Delta t_3 = |35 - 32| = 3\text{ s}$
$\Delta t_4 = |31 - 32| = 1\text{ s}$

Mean absolute error $\Delta t_{mean} = \dfrac{2 + 0 + 3 + 1}{4} = \dfrac{6}{4} = 1.5\text{ s}$

Since the minimum division (least count) of the measuring clock is $1\text{ s}$, the mean absolute error must be rounded off to the nearest integer to match the precision of the instrument.

Rounding $1.5\text{ s}$ to the nearest integer gives $2\text{ s}$.

Therefore, the correct mean time within error limits is $(32 \pm 2)\text{ s}$.

Answer: $(32 \pm 2)\text{s}$