# Xn~Exponential(.5) -> Xn~Normal. Use Chebychev's inequality

#### phychoflip29

##### New Member
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.
Then i don't know what to do here. And i checked online of how to do this. Can you please help me how to solve this problem. Thank you so much for your help.

#### BGM

##### TS Contributor
Do you mean

$$\Pr\{|\bar{X}_n - 2| \leq 0.01\} \geq 0.95$$

where $$\bar{X}_n$$ is the sample mean with sample size $$n$$ ?

Next just directly apply Chebychev's Inequality which does not require you to use the Central Limit Theorem.

#### phychoflip29

##### New Member
Yes i mean $$\Pr\{|\bar{X}_n - 2| \leq 0.01\} \geq 0.95$$, but now $$X_n$$ is normally distributed. So I don't know what to do next

#### BGM

##### TS Contributor
I am confused - do you mean they have an exponential distribution as you mentioned in the original post, or have another normal distribution? Note that we are just interested in the mean and variance, not the exact distribution for applying the inequality.

#### phychoflip29

##### New Member
First It is an exponential distribution. Then using the central limit theorem on the exponential distribution, the exponential distribution is now a normal distribution.