Chapter 18, Problem 43E

(a)

To determine

To find the mean and the standard deviation of the scores.

Answer to Problem 43E

The mean is $2.859$ and the standard deviation is $1.324$ .

Explanation of Solution

It is given in the question, the table for all the students who took AP statistics exam. Thus, the mean and the standard deviation of the scores is calculated from the values given in the table as:

  $\begin{array}{l}\mu =E(X)\\ =5(0.126)+4(0.222)+3(0.253)+2(0.183)+1(0.216)\\ =2.859\end{array}$

  $\begin{array}{l}\sigma =\sqrt{{(5-2.859)}^2(0.126)+{(4-2.859)}^2(0.222)+{(3-2.859)}^2(0.253)+{(2-2.859)}^2(0.183)+{(1-2.859)}^2(0.216)}\\ =1.324\end{array}$

(b)

To determine

To explain would you expect their scores to follow a Normal model if we select a random sample of $40$ AP statistics students.

Answer to Problem 43E

No, we do not expect.

Explanation of Solution

It is given in the question, the table for all the students who took AP statistics exam. And the mean is $2.859$ and the standard deviation is $1.324$ . Thus, if we select a random sample of $40$ AP statistics students so the distribution would resemble the population, mostly uniform for scores $1-4$ ,

with about half as many $5$ s. Therefore, we would not expect their scores to follow a Normal model.

(c)

To determine

To describe the sampling model for these means.

Explanation of Solution

It is given in the question, the table for all the students who took AP statistics exam. And the mean is $2.859$ and the standard deviation is $1.324$ . Thus, let us first check the appropriate conditions and assumptions for the following data as:

Random condition: It is satisfied as the scores are selected randomly.

Independent condition: It is reasonable to think that the randomly selected scored are independent of one another.

$10\%$ condition: The sample size of $40$ scores represents less than $10\%$ of all scores.

Large enough sample condition: A sample of $40$ scores is large enough thus, it is satisfied.

Thus, all the conditions are met. Now, let us find the mean and the standard deviation as:

  $\begin{array}{l}{\mu }_{\overline{x}}=2.859\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{1.324}{\sqrt{40}}\\ =0.209\end{array}$

Now, since the conditions are satisfied the sampling distribution model for the mean of $40$ randomly selected AP statistics scores is Normal, with ${\mu }_{\overline{x}}=2.859$ and standard deviation ${\sigma }_{\overline{x}}=0.209$ .