Question

The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, for ages $8$to $12$, the beginning ice-skating class was picked. In that class were $64$girls and $16$boys. Suppose that we are interested in the true proportion of girls, ages $8$to $12$, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

State the estimated distribution of $P\text{'}$. Construct a $92\%$ Confidence Interval for the true proportion of girls in the ages $8-12$ beginning ice-skating classes at the Ice Chalet.

Short answer

With a mean of $0.80$and a variance of $0.002$, follows the normal distribution. is the confidence interval for the given condition.

Step-by-step solution

Given information

Total number of children $=80$

Total number of girls $=64$

Total number of boys $=16$

Explanation

There are $80$children in this question, $64$of them are girls and $16$of whom are boys.

$p\text{'}=0.80$and $n=80$are the values we have.

Will follow a Normal distribution with a mean of $\mu$equal to $\text{'}p\text{'}$and

a variance of ${\sigma }^2=\frac{p\text{'}(1-p\text{'})}{n}$

Now, according to the specifications of the query, there is a normal distribution with a mean of $0.80$and a variance of $\frac{0.80(1-0.20)}{80}=0.002$.

Confidence interval $=\left[\mathrm{p}\pm {\mathrm{Z}}_{\frac{a}{2}}\times \sqrt{\frac{p(1-p)}{n}}\right]$

$$\begin{gathered} =\left[0.80\pm Z_{\frac{0.08}{2}}\times \sqrt{\frac{0.80\times 0.20}{80}}\right] \\ =\left[0.80\pm 1.7507\times \sqrt{\frac{0.80\times 0.20}{80}}\right] \\ =[0.7217,0.8783] \end{gathered}$$