Question 1177825
The pdf of the distribution is represented by

{{{f(x) = 1/(0.05- -0.05) = 1/0.10 = 10}}}, since it is uniform


Hence {{{P(X < 0.01) = 10*(0.01--0.05) = 10*0.06 = 0.6}}}.


What this means is that, the chance that the arrival time measurement error is at most 0.01 microsecond is 60%.