Let $X_1,...,X_n$ be i.i.d random variables. $X_i$ ~ $Unif(-1,1)$
Let $Y_1,...,Y_n$ be i.i.d random variables. $~~$$Y_i$ ~ $Unif(-1,1)$
Let us define a new random variable $Z_i$. $Z_i=1$ if $(X_i,Y_i)$ lies within the unit-disk. And $Z_i=0$ if $(X_i,Y_i)$ lies outside the unit disk.
. What is the mean and variance of $\bar{Z_n}$? What about $4\bar{Z_n}$?
. Is $4\bar{Z_n}$ close to $\pi$? Find $P(|4\bar{Z_9}-\pi|>.01)$. Use Chebyshev's theorem to find bound on this probability. What about $P(|4\bar{Z}_{100}-\pi|>.01)$?
. Instead of finding a bound as in the last example, find the $P(|4\bar{Z_9}-\pi|>.01)$ approximately using Central Limit Theorem.
I am having trouble using Chebyshev's theorem. I think my $Var(4\bar{Z}_{9})$ might be wrong but unsure why.
$$Z_i \sim Bern(\frac{\pi}{4})$$
$$E(\bar{Z_n})=\dfrac{\pi}{4} ,~ Var(\bar{Z_n})=(\dfrac{\pi}{4n}-\dfrac{\pi^2}{16n})$$
$$E(4\bar{Z_n})=\pi ,~ Var(4\bar{Z_n})=\dfrac{\pi}{n}(4-\pi)$$
Since $P(|X-E(x)|>k)\leq\dfrac{Var(X)}{k^2}$
$$P(|4\bar{Z_9}-\pi|>.01)\leq \dfrac{(4\pi - \pi^2)}{9(.01^2)}$$
But this number is clearly incorrect. Where did I go wrong?