greets
See the answer to your other question about the distribution of the number
of times a 5 will come up.
Now 600 times is a large number of rolls so we may as well use the normal
approximation to the binaomial distribution, in this case it will have mean:
$\displaystyle m=Np = 600 \times (1/6) = 100$
and standard deviation:
$\displaystyle
\sqrt{Np(1-p)} = \sqrt{600(1/6)(5/6)} \approx 9.13
$
So you would expect the number of 5's to be $\displaystyle 100 \pm k \times 9.13$, where
the value of $\displaystyle k$ is chossen by you depending on what confidence you want
that the actual number lies in the interval, often a value of $\displaystyle k=1.96 $ is used.
RonL