Chapter 8.1, Problem 8.15E

(a)

To determine

To find the mean of the four members in the population.

Answer to Problem 8.15E

  $4$

Explanation of Solution

Given:

  $\begin{array}{l}n=2\\ \text{Population=1,2,4,9}\end{array}$

The mean is:

  $\begin{array}{l}\overline{x}=\frac{1+2+4+9}{4}\\ =\frac{16}{4}\\ =4\end{array}$

(b)

To determine

To find out the sample means for the six possible samples.

Answer to Problem 8.15E

All samples: $\{1,2\},\{1,4\},\{1,9\},\{2,4\},\{2,9\},\{4,9\}$

The means are: $1.5,2.5,5,3,5.5,6.5$ .

Explanation of Solution

We require all possible samples of size two. The possible samples are then all possible combinations of two elements of the population.

  $\{1,2\},\{1,4\},\{1,9\},\{2,4\},\{2,9\},\{4,9\}$

The mean is:

  $\begin{array}{l}\{1,2\}\Rightarrow \overline{x}=\frac{1+2}{2}=1.5\\ \{1,4\}\Rightarrow \overline{x}=\frac{1+4}{2}=2.5\\ \{1,9\}\Rightarrow \overline{x}=\frac{1+9}{2}=5\\ \{2,4\}\Rightarrow \overline{x}=\frac{2+4}{2}=3\\ \{2,9\}\Rightarrow \overline{x}=\frac{2+9}{2}=5.5\\ \{4,9\}\Rightarrow \overline{x}=\frac{4+9}{2}=6.5\end{array}$

(c)

To determine

To construct the dot plot that shows the distribution.

Explanation of Solution

The dot plot is as follows:

To determine

(d)

To find the mean of the six $\overline{x}$ values from (b).

Answer to Problem 8.15E

  $4$

Explanation of Solution

The mean is as:

  $\begin{array}{l}{\mu }_{\overline{x}}=\frac{1.5+2.5+5+3+5.5+6.5}{6}\\ =\frac{24}{6}\\ =4\end{array}$

We note that mean of the sampling distribution of the sample mean is $4$ while the population mean is also $4$