The mean and standard deviation of a single roll of this die are 3.5 and 2.291..

The mean and sd of the sample mean are 3.5 and 2.291/sqrt(300)~=0.1323

Now Chebyshev's inequality tells us that for the sample mean X:

P(|X-3.5|>= k 0.1323) <= 1/k^2

Rewording this, we have:

P((X<=3.5 - k 0.1323) or (X>=3.5 + k 0.1323)) <= 1/k^2

so if put k 0.1323 = 0.5, we have k=0.5/0.1323~=3.78 then:

P((X<=3) or (X>=4)) <= 1/3.78^2 ~= 0.07,

so:

P(3<X<4) >= 0.93.

RonL