Chebyshev inequality, lower bound on $P(X \ge 200)$ We throw a die $100$ times. Using Chebyshev inequality find the lower bound on the probability that the sum of spots in these $100$ throws is bigger than $200$.
Let $X = $ number of spots after $100$ throws
We are looking for $t$ such that $P(X \ge 200) \ge t$
This is a Bernoulli distribution with $n=100$ but what will $p$ be in this case?
Also, Chebyshev inequality immediately gives us this estimation: $P(X \ge 200) \le t$ and consequently, $P(X < 200) \ge t$ but not $P(X \ge 200) \ge t$.
Could you help me with that $p$ and the estimation?
Thank you!
 A: Various points:


*

*As you say in your comment, the expected value of each throw is $\frac{21}{6} = 3.5$ 

*You seem to have miscalculated the variance of each throw, which is $\frac{105}{36}$ rather than $\frac{21}{36}$ 

*So with $100$ throws the expected value of the sum is $\frac{2100}{6} = 350$ while the variance of the sum is  $\frac{10500}{36} \approx 291.667$ and the standard deviation is the square root of this so about $17.08$.  

*We can say that $200$ is about $\frac{350-200}{17.08} \approx 8.783$ standard deviations below the mean, but you might find it more useful to say that $199$ is about $\frac{350-199}{17.08} \approx 8.842$ below the mean to deal with your $P(X \ge 200)=P(X \gt 199)$ point   

*One version of Chebyshev's inequality is $\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}$ and you can rewrite this as $\Pr(|X-\mu|\lt k\sigma) \geq 1-\frac{1}{k^2}$ so the probability of being strictly within $8.842$ standard deviations of the mean is at least about $1-\frac1{8.842^2}\approx 0.987$ and in a sense this is an answer to your question looking for a lower bound

*You might prefer a one-sided version of Chebyshev's inequality, such as $\Pr(X-\mu \geq k\sigma) \leq \frac{1}{1+k^2}$, though since you are looking at the lower tail you would want $\Pr(\mu - X \geq k\sigma) \leq \frac{1}{1+k^2}$ or $\Pr(\mu - X \lt k\sigma) \geq 1-\frac{1}{1+k^2}$ and here this would give about $1-\frac1{1+8.842^2}\approx 0.987$ as a lower bound, so close to the two-sided result 

*Chebyshev's inequality often a poor approximation, and it is here.  In fact $P(X \lt 200) \approx 2.5234\times 10^{-20}$ so $P(X \ge 200)$ is very much closer to $1$ than Chebyshev's inequality might suggest
