In this problem we look at two general 'random inequalities'. 

Part 1
Markov's inequality tells us that the probability that the modulus of a random variable $X$ exceeds any random positive number $a$ is given by a universal inequality as follows:

$$ P(|X|\geq a) \leq \frac{E(|X|)}{a^{??}} $$

Part 2
Another important general statistical result is Chebyshev's inequality, which says that

$$ P(|X-\mu|\geq k\sigma)\leq \frac{1}{k^2} $$ where $\mu$ and $\sigma$ are the mean and standard devitation of the distribution $X$ respectively. This is true for any distribution and any positive number $k$. Can you make a probability distribution for which the inequality is exactly met when $k=2$? In other words, use the distribution maker to create a distribution $X$ for which $$ P(|X-\mu|\geq 2\sigma)=\frac{1}{4} $$