Central Limit Theorem

askazy

New Member
#1
If X is a Gamma(n,1). For which values of n we have
\(P(|\frac{X}{n}-1|>0.01)<0.1\)
If \(n=n_0\) is the first value of n that satisfies the above equation, so find the n values for
i) \(n=n_0\)
ii) \(n=n_0+1\)
iii)\(n=n_0+2\)
Use chi-square distribution or Central Limite Theorem
 

askazy

New Member
#2
What I think
Supose that \(X_i = exp(1) \rightarrow \sum_{i=1}^nX_i \rightarrow Gamma(n,1)=X\) now
\(P(\frac{X}{n}-1>0.01) = 1 - P(X-n<0.01n)=1-P(X<1.01n)\)
We have \(E[X_i]=1\rightarrow\sum_{i=1}^nE[X_i]=n\) analogously \(Var(X_i)=1\rightarrow\sum_{i=1}^nVar(X_i)=n\)
applying Central Limit Theorem
\(1 - P(\frac{\sum_{i=1}^nX_i - \sum_{i=1}^n\mu_i}{\sqrt{\sum_{i=1}^n\sigma_i^2}}\leq\frac{1.01n-n}{\sqrt{n}})=1-P(Z\leq0.01\sqrt{n})<0.1\)\(=-P(Z\leq0.01\sqrt{n})<-0.9\)*

I do not know if I applied the CLT correctly, and the last inequality marked with * do not remember how to solve it.