The table shows the total cost c of buying t movie tickets. Write an equation to represent the relationship between c and t.t | c1|72|143|214|28

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 ${\displaystyle Var\left[X\right]={\sigma }^2=p(1-p)}$:
${\displaystyle E\left[{\left(\frac{\partial \ln :\left\{f_X\left(X\right)\right\}}{\partial p}\right)}^2\right]=\frac{n^2{p}^2-2n^{2}pE\left[\bar{X}\right]+n^{2}E\left[{\bar{X}}^2\right]}{p^2{(1-p)}^2}=\frac{n^2{p}^2-2n^2{p}^2+n^2(p^2+\frac{p(1-p)}{n})}{p^2{(1-p)}^2}=\frac{np(1-p)}{p^2{(1-p)}^2}=\frac{n}{p(1-p)}}$
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 ${\displaystyle \frac{p(1-p)}{n}}$.

Here are summary stastistics for randomly selected weights of newborn girls: ${\displaystyle n=224,\bar{x}=28.3\text{hg},s=7.1\text{hg}}$. Construct a confidence interval estimate of mean. Use a 98% confidence level. Are these results very different from the confidence interval ${\displaystyle 26.5\text{hg}<\mu <30.7\text{hg}}$ with only 14 sample values, ${\displaystyle \bar{x}=28.6}$ hg, and ${\displaystyle s=2.9}$ hg? What is the confidence interval for the population mean ${\displaystyle \mu }$?

$27,2hg<\mu <29,4hg?$ (Round to one decimal place as needed).

Are the results between the two confidence intervals very different?