Confidence interval using Central Limit Theorem I've been trying to find this information online, but have not had much success so far. I want to approximate the 95% confidence interval for the geometric distribution with the following parameters:


*

*maximum likelihood: $\hat{\theta}=1/4.9\approx 0.204$

*sample size: $n=100$
How does one approximate the confidence interval using the Central Limit Theorem? For instance, the Wikipedia article has many versions of this theorem, so I don't even know which one I should apply.
 A: We will use this form of CLT
$X_1, X_2,... , X_n$ denote the observations of a random sample of size $n$ from
any distribution having finite variance $\sigma^2 > 0$ (and hence finite mean $\mu$), then the random variable $$\frac{\sqrt{n}(\bar{X} -\mu)}{\sigma} \sim N(0,1)\tag{1} $$i.e converges in distribution to a random variable having a standard normal distribution.
Let denote the pmf of geometric distribution as 
$$f(x_i)=(1-\theta)^{x_i-1}\theta  $$ 
$ 0< \theta < 1$ and $x_i=1,2...n$
Then you can find that 
$$\mu=E(X)=\frac{1}{\theta}$$
$$\sigma^2=\frac{1-\theta}{\theta^2}$$
Now let us plug everything to $(1)$
$$Z=\frac{\sqrt{100}(\bar{X}-\frac{1}{\theta})}{\sqrt{\frac{1-\theta}{\theta^2}}}\sim N(0,1)$$
Then $P(-1.96<Z<1.96)=0.95 \Rightarrow P(-1.96<\frac{\sqrt{100}(\bar{X}-\frac{1}{\theta})}{\sqrt{\frac{1-\theta}{\theta^2}}}<1.96)=0.95$
Now what you get are
$$\frac{\sqrt{100}(\bar{X}-\frac{1}{\theta})}{\sqrt{\frac{1-\theta}{\theta^2}}}\le 1.96 \tag{2}$$
and
$$\frac{\sqrt{100}(\bar{X}-\frac{1}{\theta})}{\sqrt{\frac{1-\theta}{\theta^2}}} \ge -1.96 \tag{3}$$
Solve inequations $(2)$ and $(3)$ you can find the 95% CI of $\theta$
Note you can find $\bar{X}$ from the maximum likelihood estiamtion
The likelihood function of the geometric distribtion is
$$L(\theta)={\theta}^{n}{\left(1-\theta \right)}^{\sum_{1}^{n}{x}_{i}-n}$$
Take log and derivative in term of $\theta$ and set to zero, we can find that
$$\hat{\theta}=\frac{\sum_{i=1}^nx_i}{n}=\frac{1}{\bar{X}}$$
Since $\hat{\theta}=\frac{1}{4.9}$ therefore, $\bar{X}=4.9$ then you plug in this number to $(2)$ and $(3)$ to solve the inequations.
