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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!

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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 $2\cdot0.38=0.76$ 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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Let the random variable $X$ be the downtime. You want to find the number $x$ such that the downtime will be $\gt x$ "only in $5\%$ of the cases," or more precisely, with probability $5\%$. Equivalently, you want the number $x$ such that the probability that the downtime is $\le x$ is $0.95$. In symbols, we want to find $x$ such that $P(X\le x)=0.95$.

Note that if $X$ is normally distributed with mean $\mu$, standard deviation $\sigma$, then the random variable $\dfrac{X-\mu}{\sigma}$ has standard normal distribution. Because tables are only available for the standard normal, we reduce the equation $P(X \le x)=0.95$ to an equation that involves the standard normal. We have $$P(X\le x)=P(X-\mu\le x-\mu)=P\left(\frac{X-\mu}{\sigma}\le \frac{x-\mu}{\sigma}\right)=P\left(Z\le \frac{x-\mu}{\sigma}\right),$$ where $Z$ is standard normal.

Now using tables for the standard normal, we find the $z$ such that $P(Z\le z)=0.95$, set $\dfrac{x-\mu}{\sigma}=z$, and solve for $x$.

In our specific case, the $z$ such that $P(Z\le z)=0.95$ is approximately $1.645$. (Here I used a table for the standard normal from Wikipedia, looked in the body of the table until I found a number close to $0.95$. Your normal table may be set up differently, or you may be using software to find the relevant number.) We have $\mu=4.47$ and $\sigma=0.38$, so you will be solving the equation $$\frac{x-4.47}{0.38}=1.645.$$

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