Chapter 18, Problem 18E

To determine

To use the $68-95-99.7$ rule to describe the sampling distribution model for the proportion of students in a randomly selected group of $\text{200}$ college students who engage in binge drinking and explain do you think the appropriate conditions are met.

Answer to Problem 18E

Yes, all the appropriate conditions are met.

Explanation of Solution

It is given in the question that a national study found that $\text{44\%}$ of college students engage in binge drinking. And a random selected group of $\text{200}$ college students is taken. Thus, using the rule $68-95-99.7$ , we get,

  $\begin{array}{l}\mu =p=0.44\\ \sigma =\sqrt{\frac{pq}{n}}\\ =\sqrt{\frac{0.44(0.56)}{200}}\\ =0.035\end{array}$

Thus, about $68\%$ of the sample proportions are expected to be between $\text{0}\text{.405 and 0}\text{.475}$ and about $95\%$ are expected to be between $\text{0}\text{.370 and 0}\text{.510}$ and about $99.7\%$ are expected to be between $\text{0}\text{.335 and 0}\text{.45}$ , by using the rule $68-95-99.7$ to describe the sampling distribution model. And also, the appropriate conditions necessary for your analysis are as:

Random condition: It is satisfied as we have a random sample of $\text{200}$ college students.

  $10\%$ condition: The sample size less than $10\%$ of the population size.

Success/failure condition: as $np=88,n(1-p)=112$ and both are greater than ten then this is also true.

Thus, the conditions are all met. So, the sampling distribution model for the proportion of students in a randomly selected group of $\text{200}$ college students who engage in binge drinking is $N(0.44,0.035)$ .