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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!

2 Answers 2