All the probabilities add up to 1, so the table does indeed describe a distribution.
The mean is given by
\(\mathbb E(X)=\displaystyle\sum_x xp(x)=0\times0.250+1\times0.441+2\times0.259+3\times0.050=1.109\)
The standard deviation is given by \(\sqrt{\mathbb E(X^2)-\mathbb E(X)^2}\). You have
\(\mathbb E(X^2)=\displaystyle\sum_xx^2p(x)=0^2\times0.250+1^2\times0.441+2^2\times0.259+3^2\times0.050=1.927\)
so the standard deviation is approximately \(\sqrt{1.927-1.109^2}\approx0.835\).
