We roll a symmetrical die $200$ times. $X$ is a random variable representing the number of the 6 face appearing. Using Chebyshev's inequality find $c>0$ so that the probability $$Pr(X\in(a-c, a+c))$$ is at least $0.85$.
My attempt: $$Pr(a-c<X<a+c)\geq0.85$$ $$1-Pr(a-c<X<a+c)\leq1-0.85$$ $$1-Pr(|X-a|<c)\leq 0.15$$ $$Pr(|X-a|\geq c)\leq 0.15$$
$E[X]=200\cdot1/6, \sigma^2=200\cdot5/36$
Now: $$0.15=\frac{\sigma^2}{c^2}$$ And from that we get $c>0$.
The thing is my colleague got an answer with $c$ being an interval. Now I'm not sure which one of our solutions is correct (or maybe neither is).
Define variable $$ X_n = \Big\{ \begin{array}{cc} 0 & else\\ 1 & dice=6 \end{array} $$ Then $\mathbf{E}X_n = P(X_n=6) = \frac{1}{6} \ \mathbf{E}X_n^2 = \frac{1}{6}, \mathbf{Var}X_n = \frac{1}{6} - \frac{1}{6^2} = \frac{5}{36}$. Now define $$ S_n = \sum_{k=1}^{200}X_k $$ Combine them. and from this you can find $\mathbf{E}S_n$ and $\mathbf{Var}S_n$. Now, I assume $a$ in your example is $\mathbf{E}S_n$. In such case, $$ P(c + a \leq S_n \leq -c +a) = P(|S_n - a|<c) = 1-P(|S_n - a|>c) \geq 1- \frac{\mathbf{Var} S_n}{c^2} \geq 0.85 $$
Now solve for $c$, keep in mind $c>0$.