here's the question:
assuming that the system downtime is normally distributed with mean (μ)4.47 sec and standard deviation(σ) of 0.38 sec.
By using the cumulative (to the left) Z score table,
a) find the probability that the system downtime is more than 5 sec.
z score = (n-μ)/ σ
= (5-4.47)/0.38 = 1.39
the probability for 1.39 in the z score table is 0.9177
so, P(x≤5)= 0.9177
P(x>5)= 1- 0.9177 = 0.0823
b) what is the minimum downtime duration for the worse 5% ?
(i already got the answer for a, i just don't quite understand the meaning of 'for the worse 5%' )
thank you!