Let $X_1...X_9 \sim X$, with X exponential of parameter $\lambda$ all independent. Suppose we know that $\sum_{j=1}^0 X_j = 9.15$
I want to find a confidence interval of level $95$% of $\lambda$. I used this formula:
$$ \overline X \pm Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$$
Where $\overline X$ is just 9.15/9, $Z_{\frac{\alpha}{2}}$ is the quantile of the normal distribution with $1 - \alpha = 0.95$. However I do not have the square of the variance, so I am not sure how to solve this.
Moreover, I need to do find the interval of the probability of X being greater than 1. I suppose I need to use a similar method but I am still stuck.