Time for $20$ oscillations of a pendulum is measured as $t_1 = 39.6\, s$,$t_2 = 39.9\, s$,and $t_3 = 39.5\, s$. What is the precision in the measurements? What is the accuracy of the measurement?

(N/A) Given,$t_1 = 39.6\, s$,$t_2 = 39.9\, s$,and $t_3 = 39.5\, s$.
The least count of the measuring instrument is $0.1\, s$ (as measurements have only one significant figure after the decimal point).
Precision in the measurement $=$ Least count of the instrument $= 0.1\, s$.
The average time for $20$ oscillations is given by:
$t_{avg} = \frac{t_1 + t_2 + t_3}{3} = \frac{39.6 + 39.9 + 39.5}{3} = \frac{119.0}{3} = 39.666... \approx 39.7\, s$.
Absolute errors in the measurements are:
$\Delta t_1 = |t_{avg} - t_1| = |39.7 - 39.6| = 0.1\, s$
$\Delta t_2 = |t_{avg} - t_2| = |39.7 - 39.9| = 0.2\, s$
$\Delta t_3 = |t_{avg} - t_3| = |39.7 - 39.5| = 0.2\, s$
Average absolute error (Accuracy) $= \frac{\Delta t_1 + \Delta t_2 + \Delta t_3}{3} = \frac{0.1 + 0.2 + 0.2}{3} = \frac{0.5}{3} \approx 0.17\, s$.
Rounding off to one decimal place,the accuracy of the measurement is $\pm 0.2\, s$.