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 $\frac{x-4.47}{0.38}=1.645.$