Chebyshev's Theorem and probability? I have a question from the last year's Statistics exam and I could not answer it. I hope you can help me, thanks from now :)
"A company owns $100$ televisions. Each television has an $50\%$ probability of not working. According to the Chebyshev's Theorem, with what probability can we asset that between $0$ and $80$ televisions will not be working?"
 A: For a random variable $X$ with mean $\mu$ and standard deviation $\sigma$,
$$ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} $$
that is, the probability that $X$ is at most $k$ standard deviations from the mean is at least $\frac{1}{k^2}$. We have pieces to kill in.
Let's take these televisions to follow the binomial distribution with $n = 100$ and $p = 0.5$. Now, you should know how to find the mean and standard deviation of a binomial random variable: we have $\mu = 50$ and $\sigma = 5$. I'll answer a variant of this question so we can really use the inequality: what is the probability that at least 80 televisions do not work, or at most 20? Well, 80 is six standard deviations away from the mean (as is 20), so Chebyshev says that this probability is at most $\frac{1}{36}$ (this is a weak bound).
You can't quite use Chebyshev in your question as stated, as you need information about the "tails." The question asserts a "body" probability - stuff in the middle.
I highly suggest reading some examples to see how Markov and Chebyshev are used to estimate probabilities.
