I'm trying to solve the following problem.
Let $W \sim \mathrm{Binomial}(100,0.5)$. Use Chebyshev's (second) inequality to get the upper bound on $P(|W-50| \geq 100)$.
My solution: $$\mathrm{Var}(W) = 100\cdot0.5(0.5) = 25 \Rightarrow \sigma = \sqrt{25} = 5$$ and $$k\sigma = 100 \Rightarrow k = 20.$$
According to Chebyshev's inequality
$$P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}$$
so in this case, the upper bound is given by $$\frac{1}{20^2} = \frac{1}{400}.$$
The solution according to my textbook, however, is $1/4$. Does someone know how they received that answer?