How do I construct a probability distribution where none of the mass lies within one σ (standard deviation) of the mean?

Question

Does this condition satisfy that the probability distribution where none of the mass lies within one σ of the mean?

$$P(X=0) =0.5$$ $$P(X=1) =0.5$$

Answer 1

Show that $$\mu = E(X) = 0P(X=0) + 1(P(X=1) = 0 + 1(1/2) = 1/2,$$ and that $\sigma^2 = Var(X) = 1/4,$ so that $SD(X) = \sigma = 1/4.$ [You are right that the value of $\sigma$ is not "given," but nothing is stopping you from computing it from the information given.] The random variable $X$ has a Bernoulli distribution with $p = 1/2.$

Then the open interval with endpoints $\mu \pm \sigma$ is $(0, 1).$ So $P(0 < X < 1) = 0.$ as in @Henry's Comment.