Given a random variable $X$ and $100$ samples $X_1, \dots, X_{100}$ chosen i.i.d., let $X^* = \sum\limits_{i=1}^{100} X_i / 100$. $\operatorname{Var}[X] =\sigma^2=9 $ and $E[X] = \mu $. $\Pr[a\le X^*-\mu\le b] \geq 0.9$
find $a$ and $b$ using Chebyshev's inequality.
I don't know how to approach this question because every question I have saw using Chebyshev's inequality one parameter but here there is two.
Update
$ \Pr[a\le X^*-\mu\le b] = \Pr[|X^∗−μ|≤(b−a)/2]$
$ \Pr[|X^∗−μ|^2≤(\frac{b-a}{2})^2] = 1-\Pr[|X^∗−μ|^2>(\frac{b-a}{2})^2] \ge 1-\frac{E[X^* -\mu]^2}{(\frac{b-a}{2})^2} = 1- \frac{Var(X^*)}{(\frac{b-a}{2})^2}$
$ Var(X^*) = \frac{(\sigma)^2}{100} = 0.09 $
$ 1-\frac{0.09}{(\frac{b-a}{2})^2} = 0.9 $