assuming that the system downtime is normally distributed with mean (mu)4.47 sec and standard deviation(sigma) 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-mu)/sigma = (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%' )

Parker Bird

Parker Bird

Answered question

2022-07-19

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%' )

Answer & Explanation

polishxcore5z

polishxcore5z

Beginner2022-07-20Added 14 answers

The worst 5% are the 5% of cases having the longest downtime. These are the 5% is the 5% at the righthand end of the distribution, so 95% are to the left of that. Thus, you want to look through the cumulative-to-the-left table for the z-score that has (as close as possible to) 95% to the left of it. Say you've found that number, and it's s standard deviations. (It should be a bit less than 2.) Now you have to figure out what s standard deviations means in terms of downtime.
One standard deviation is 0.38 seconds, so 2 standard deviations will be 2⋅0.38=0.76 seconds, and in general s standard deviations will be 0.38s seconds. Once you've found the right s from the table, you can substitute it into this formula to see how many seconds (instead of how many standard deviations) above the mean corresponds to the worst 5%. As an example, if s were 2, you'd be looking at a downtime cutoff 0.76 seconds above the mean. Since the mean is 4.47 seconds, though would be a downtimes of 4.47+0.76=5.23 seconds. In general, it will be a downtime of 4.47+0.38s seconds: 5% of the cases will be worse than that, 95% will be no worse than that, so that's the minimum downtime for the worst 5% of all cases.

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