Question: If you roll one 6 sided die 1000 times, what is probability that sum of fallen numbers will be between 3400 and 3550 ?

Answer: 
For a single die roll, the mean is $\frac72$ and the variance is $\frac{35}{12}$.

The distribution of the sum of $1000$ die rolls should approximate a normal distribution with a mean of $1000\cdot\frac72=3500$ and a variance of $1000\cdot\frac{35}{12}=\frac{8750}{3}$.

If the range includes $3400$ and $3550$, then we compute the cumulative normal distribution between $\frac{-100.5}{\sqrt{\frac{8750}{3}}}$ and $\frac{50.5}{\sqrt{\frac{8750}{3}}}$ standard deviations from the mean:
$$
0.8251260-0.0313793=0.7937467
$$
If the range excludes $3400$ and $3550$, then we compute the cumulative normal distribution between $\frac{-99.5}{\sqrt{\frac{8750}{3}}}$ and $\frac{49.5}{\sqrt{\frac{8750}{3}}}$ standard deviations from the mean:
$$
0.8203139-0.0327096=0.7876043
$$
Answer:

EV = 1000\*3.5

SD = SQRT(1000)\*1.7

P( 3400 <= SUM <= 3550) = P[((3400-3500)/53.75)<= z <=((3550-3500)/53.75)]

Find NORMSDIST and get the probability. (should be close to .7923).

Thanks

Satish