91影视

The proportion of all U.S. adults admitted that they are 鈥渃ord cutters鈥� is $p=0.15$.

A random sample of size$n=500$ is selected.

Let$\hat{p}$ represents the sample proportion of adults who admitted that they are 鈥渃ord cutters.鈥�

a

The mean of the sampling distribution of$\hat{p}$ is obtained as:

$\begin{array}{rcl}E\left(\hat{p}\right)& =& p\\ & =& 0.15\end{array}$.

Since$E\left(\hat{p}\right)=p$.

Therefore,$E\left(\hat{p}\right)=0.15$ .

b.

The standard deviation of the sampling distribution of$\hat{p}$ is obtained as:

$\begin{array}{rcl}{蟽}_{\hat{p}}& =& \sqrt{\frac{p\left(1-p\right)}{n}}\\ & =& \sqrt{\frac{0.15脳0.85}{500}}\\ & =& \sqrt{\frac{0.1275}{500}}\\ & =& \sqrt{0.000255}\\ & =& 0.01596\end{array}$..

Therefore,${蟽}_{\hat{p}}=0.0160$ .

c.

Here,

$\begin{array}{rcl}n\hat{p}& =& 500脳0.15\\ & =& 75\\ & >& 15\\ & & \end{array}$,

and

$\begin{array}{rcl}n\left(1-\hat{p}\right)& =& 500脳0.85\\ & =& 425\\ & >& 15\\ & & \end{array}$.

The conditions to use Central Limit Theorem are satisfied. According to the Central Limit Theorem, the shape of the sampling distribution of the sample proportion is approximately symmetric (normal).

d.

The probability that $\hat{p}$is less than 0.12 is obtained as:

$\begin{array}{rcl}P\left(\hat{p}<0.12\right)& =& P\left(\frac{\hat{p}-p}{{蟽}_{\hat{p}}}<\frac{0.12-p}{{蟽}_{\hat{p}}}\right)\\ & =& P\left(Z<\frac{0.12-0.15}{0.0160}\right)\\ & =& P\left(Z<\frac{-0.03}{0.0160}\right)\\ & =& P\left(Z<-1.875\right)\\ & =& P\left(Z<-1.88\right)\\ & =& 0.0301\\ & & \end{array}$.

To find the probability z-table is used; the value at the intersection of -1.80 and 0.08 is the required probability.

Hence, the required probability is 0.0301.

e.

The probability that$\hat{p}$ is greater than 0.12 is obtained as:

$\begin{array}{rcl}P\left(\hat{p}>0.10\right)& =& P\left(\frac{\hat{p}-p}{{蟽}_{\hat{p}}}>\frac{0.10-p}{{蟽}_{\hat{p}}}\right)\\ & =& P\left(Z>\frac{0.10-0.15}{0.0160}\right)\\ & =& P\left(Z>\frac{-0.05}{0.0160}\right)\\ & =& P\left(Z>-3.125\right)\\ & =& 1-P\left(Z猢�-3.13\right)\\ & =& 1-0.0009\\ & =& 0.9991\end{array}$.

To find the probability z-table is used, the value at the intersection of -3.10 and 0.03 is the probability of the z-score less than or equal to -3.13.