State the estimated distribution of $P\prime .$Construct a $92\%$Confidence Interval for the true proportion of girls in the ages $8to12$beginning ice-skating classes at the Ice Chalet.

$\hat{p}=x/n$is the point estimate of $p$, where $x$ is the number of favour able events and $n$ is the total number of events.

Given that there were $64$girls and $16$boys in that class.

So , $x=64andn=64+16=80$.

So that, $\hat{p}=64/80=0.8$.

For a sample size of $80$, the mean of the sampling distribution of the sample proportion is,

${\mu }_{\hat{p}}=p=0.8\text{.}$

The distribution of $\hat{p}$is approximately normal with mean $0.8$and standard deviation of $0.044721359.$

For the population proportion, $p$the $100(1-\alpha )\%$confidence interval is:

$\left(\hat{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)$

the sample size is $n$, and for level of significance $\alpha$the standard normal distribution critical value is $z_{\alpha 2}$.

Consider$p$, the proportion of students accepted into law school from the general population, and $n$, the sample size.

The confidence level is$92\%$.

The level of significance is $\alpha =1-0.92=0.08$. Hence, $\alpha /2=0.04$.

If of the observations must lie within an interval, the remaining must lie outside the interval.

Due to symmetry, of the population will be above the top limit, while the remaining will be below the lower limit of the interval.

Thus, the upper limit of the interval is such that, lie below it.

As a result,

The confidence interval is,

$\begin{array}{r}Cl=\left(\widehat{p}\pm z_{\frac{a}{2}}\sqrt{\left.\frac{\widehat{p}(1-\widehat{p})}{n}\right)}\right.\\ =\left(0.8\pm \left(1.751\right)\left(\sqrt{\frac{\left(0.8\right)(1-0.8)}{80}}\right)\right)\\ =(0.8\pm (1.751\left)\right(0.044721359\left)\right)\\ =(0.8\pm 0.0783071)\\ \approx (0.7217,0.8783)\end{array}$

the result of confidence Interval for the true proportion of girls in the ages to beginning iceskating classes at the lce Chalet is.