Improve Your Understanding of College Level Statistics Problems

Confidence interval for parameter
${\displaystyle X_1,X_2,\cdots ,X_n}$- i.i.d observations
${\displaystyle X_1=\xi +\eta }$ where $\xi ~N({\theta }^2,{\theta }^2+1),\eta =\{\begin{array}{ll}0,& 1/2\\ 4\theta ,& 1/2\end{array},\xi \text{ and }\eta$ and ${\displaystyle \eta }$ are independent.
Find ${\displaystyle \alpha }$-confidence interval.
The first that I need to do is to find some estimate for ${\displaystyle \theta }$. The only one that I find is ${\displaystyle \frac{5\bar{X}-S^2+1}{10}}$ but it is difficult to find distribution.
Is it possible to do something else?

Dottie’s Tax Service specializes in federal tax returns for professional clients, such as physicians, dentists, accountants, and lawyers. A recent audit by the IRS of the returns she prepared indicated that an error was made on 7 percent of the returns she prepared last year. Assuming this rate continues into this year and she prepares 80 returns, what is the probability that she makes errors on: a. More than six returns? b. At least six returns? c. Exactly six returns?

Assume a binomial probability distribution with n = 50 and p =0.25. Compute the following: a. The mean and standard deviation of the random variable. b. The probability that X is 15 or more. c. The probability that X is 10 or less.

Given you have an independent random sample ${\displaystyle X_1,X_2,\dots ,X_n}$ of a Bernoulli random variable with parameter $p$, estimate the variance of the maximum likelihood estimator of $p$ using the Cramer-Rao lower bound for the variance
So, with large enough sample size, I know the population mean of the estimator ${\displaystyle \hat{P}}$ will be $p$, and the variance will be:
$Var\left[\hat{P}\right]=\frac{1}{nE\left[\right((\partial /\partial p) \ln \hspace{0.17em}{f}_x\left(X\right){)}^2]}$
Now I'm having some trouble calculating the variance of ${\displaystyle \hat{P}}$, this is what I have so far:
since the probability function of ${\displaystyle \bar{X}}$ is binomial, we have:
$f_x\left(\overline{X}\right)=\left(\genfrac{}{}{0ex}{}{n}{\sum _{i=1}^n{X}_i}\right)*p^{\sum _{i=1}^n{X}_i}*(1-p{)}^{n-\sum _{i=1}^n{X}_i}$
so: $\ln \hspace{0.17em}{f}_X\left(X\right)=\ln \left(\left(\genfrac{}{}{0ex}{}{n}{\sum _{i=1}^n{X}_i}\right)\right)+\sum _{i=1}^n{X}_i\ln \left(p\right)+\hspace{0.17em}(n-\sum _{i=1}^n{X}_i)\ln (1-p)$
and: $\frac{\partial \ln \hspace{0.17em}{f}_X\left(X\right)}{\partial p}=\frac{{\sum }_{i=1}^n{X}_i}{p}-\frac{(n-{\sum }_{i=1}^n{X}_i)}{(1-p)}=\frac{n\overline{X}}{p}-\frac{(n-n\overline{X})}{(1-p)}$
 

and: $(\frac{\partial ln\hspace{0.17em}{f}_X\left(X\right)}{\partial p}{)}^2=(\frac{n\overline{X}}{p}-\frac{(n-n\overline{X})}{(1-p)}{)}^2=\frac{n^2{p}^2-2n^{2}p\overline{X}+n^2{\overline{X}}^2}{p^2(1-p{)}^2}$
since ${\displaystyle E\left[{\bar{X}}^2\right]={\mu }^2+\frac{{\sigma }^2}{n}}$, and for a Bernoulli random variable ${\displaystyle E\left[X\right]=\mu =p=E\left[\bar{X}\right]}$ and ${\displaystyle Var\left[X\right]={\sigma }^2=p(1-p)}$:
$E\left[(\frac{\partial \ln \hspace{0.17em}{f}_X\left(X\right)}{\partial p}{)}^2\right]=\frac{n^2{p}^2-2n^{2}pE\left[\overline{X}\right]+n^{2}E\left[{\overline{X}}^2\right]}{p^2(1-p{)}^2}=\frac{n^2{p}^2-2n^2{p}^2+n^2(p^2+\frac{p(1-p)}{n})}{p^2(1-p{)}^2}=\frac{np(1-p)}{p^2(1-p{)}^2}=\frac{n}{p(1-p)}$
Therefore, $Var\left[\hat{P}\right]=\frac{1}{nE\left[\right((\partial /\partial p) \ln \hspace{0.17em}{f}_x\left(X\right){)}^2]}=\frac{1}{n\frac{n}{p(1-p)}}=\frac{p(1-p)}{n^2}$
However, I believe the true value I should have come up with is ${\displaystyle \frac{p(1-p)}{n}}$.