Chapter 18, Problem 16E

(a)

To determine

To find out the appropriate model for the distribution of $\widehat{p}$ and find out the name of the distribution, the mean and the standard deviation and also make sure to verify that the conditions are met.

Answer to Problem 16E

The sampling distribution model for the proportion of students in the sample who wear contact lenses is $N(0.30,0.046)$ .

Explanation of Solution

In the question, it is given that assume that $30\%$ of students at a university wear contact lenses. Thus, let us check the conditions first as:

Randomization condition: It is satisfied as we assume that the $100$ students are a representative sample of all students.

  $10\%$ condition: It is satisfied as the sample size is less than $10\%$ of the population of all students at the university.

Success/failure condition: As, $np=30,n(1-p)=70$ both are greater than ten then it is also satisfied.

Thus, all the conditions are met. And the mean and the standard deviation can be calculated as:

  $\begin{array}{l}\mu =p=0.30\\ \sigma =\sqrt{\frac{pq}{n}}\\ =\sqrt{\frac{0.30(0.70)}{100}}\\ =0.046\end{array}$

Therefore, the sampling distribution model for the proportion of students in the sample who wear contact lenses is $N(0.30,0.046)$ .

(b)

To determine

To find out what is the approximate probability that more than one-third of this sample wear contacts.

Answer to Problem 16E

The approximate probability that more than one-third of this sample wear contacts is $0.257$ .

Explanation of Solution

In the question, it is given that assume that $30\%$ of students at a university wear contact lenses. And the sampling distribution model for the proportion of students in the sample who wear contact lenses is $N(0.30,0.046)$ . Thus, the approximate probability that more than one-third of this sample wear contacts is calculated as:

  $\begin{array}{l}z=\frac{0.33-0.30}{0.046}\\ =0.652\\ P(z>0.652)=normalcdf(0.652,E99,0,1)\\ =0.257\end{array}$