If Chebyshev's upper bound gives same value as the actual probability calculation, what can we conclude? As an example, if Chebyshev tells $P(|X-\mu|\geq k\sigma)\leq 0.25$ and the actual probability for $k=2$ is also $0.25$.
 A: Given a real number $\mu$ and a positive real number $\sigma$, consider a discrete random variable $X$ that takes on values $\mu$, $\mu+2\sigma$ and $\mu-2\sigma$ with probabilities $\frac{3}{4}, \frac{1}{8}$ and $\frac{1}{8}$ respectively. It is easy to verify that $E[X] = \mu$, $\operatorname{var}(X) = \sigma^2$. For this random variable,
$$P\{|X-\mu| \geq 2\sigma\} = P\{|X-\mu| = 2\sigma\} = 0.25$$
exactly.  
More generally, for $a > 1$, a random variable $X$ that takes on values $\mu$, 
$\mu+a\sigma$ and $\mu-a\sigma$ with probabilities $\frac{a^2-1}{a^2}, \frac{1}{2a^2}$ and $\frac{1}{2a^2}$ respectively enjoys the property that $E[X] = \mu$, $\operatorname{var}(X) = \sigma^2$,
and for this random variable,
$$P\{|X-\mu| \geq a\sigma\} = P\{|X-\mu| = a\sigma\} = \frac{1}{a^2}$$
exactly.
This is a discrete distribution for which the Chebyshev inequality is
satisfied with equality (cf. @glen_b's comment).  Can you find other
solutions? For example, can you put smaller masses further away from 
the mean than $a\sigma$ so that the variance remains $\sigma^2$ and
Chebyshev's inequality is satisfied with equality at $a\sigma$? Why or why not?
