How Does the Central Limit Theorem Estimate Error in Area Calculations?

[nolita_day]

Your help/input is much appreciated! Thanks in advance!

Object A is in the unit square 0<x<1, 0<y<1. Consider a random point distributed uniformly over the square; let Y=1 if the point lies inside A and Y=0 otherwise. How could A be estimated from a sequence of n individual points unif. distr. on the square?

I got this part just taking Y as ~Bernoulli(p=A). Let Y_{i} = f(X_{i}), then A≈\frac{1}{n}\sum{Y_{i}}

How could the CLT be used to gauge the probable size of the error of the estimate? Denoting the estimate by \hat{A}, if A=0.2, how large should n be so that P(|\hat{A} - A| &lt; .01) ≈ .99?

I can't figure out this one. Am I supposed to take \hat{A} as the random variable or |\hat{A}-A| as the RV? What is the standard deviation of \hat{A} vs. |\hat{A}-A|? My attempt below is only for the first part of the question in the previous paragraph.

Taking \hat{A} as the RV, I started out like this...

E[\dfrac{Y_{1}+...+Y_{n}}{n}]=\hat{A}
Then Var(\hat{A})=Var( E[\dfrac{Y_{1}+...+Y_{n}}{n}] )=\frac{1}{n^{2}}Var( nE[Y_{i}] ) = Var( E[Y_{i}] ) = Var(A) (since Y_i ~ Bernoulli(A) = A)

To estimate A based on \hat{A}, pick a small error and a high probability of getting that small error and you could use CLT as follows:

P(|\hat{A} - A| &lt; .01) = .99
P(-.01 &lt; \hat{A} - A &lt; .01) = .99
P(\dfrac{-.01}{σ_{\hat{A}}} &lt; \dfrac{\hat{A} - A}{σ_{\hat{A}}} &lt; \dfrac{.01}{σ_{\hat{A}}}) = .99
P(\dfrac{-.01}{\sqrt{A}} &lt; Z &lt; \dfrac{.01}{\sqrt{A}}) = .99
\Phi(\dfrac{.01}{\sqrt{A}}) - \Phi(\dfrac{-.01}{\sqrt{A}}) = .99
2\Phi(\dfrac{.01}{\sqrt{A}}) - 1) = .99
\dfrac{.01}{\sqrt{A}}= 2.58
A = .000015023?

I'm definitely sure this is wrong... how the heck did I come up with an actual number for A?

 

[haruspex]

This is a bit hard to follow because you confuse A with A-hat.
\hat{A} = \sum Y_i/n, A = E(\hat{A}).