Xn are iid Exponential(.5). After using the central limit theorem, Xn is Normally distributed. Use chebychev's inequality to find out how large n be so that P(|Xn-2| < .01) > .95.
I tried working on it and got the E(Xn) = 2n and Var(Xn) = 4n, and P(|Xn-2n| > .01) < (4n/(.01)^2)(1/n)= 40000...