Chapter 10.2, Problem 34E

Solution

a.

To explain: The reason behind circle graph will not work for these data.

The reason behind circle graph will not work for these data is the variable is not categorical.

Given:

The average wind speeds for one year at 44 climatic data centers around the United States are as follows:

  $\begin{array}{l}\text{9}\text{.0, 6}\text{.9, 9}\text{.1, 9}\text{.2, 10}\text{.2, 12}\text{.5, 12}\text{.0, 11}\text{.2, 12}\text{.9, 10}\text{.3, 10}\text{.6, 10}\text{.9, 8}\text{.7, 10}\text{.3, 11}\text{.0,}\\ \text{7}\text{.7, 11}\text{.4, 7}\text{.9, 9}\text{.6, 8}\text{.0, 10}\text{.7, 9}\text{.3, 7}\text{.9, 6}\text{.2, 8}\text{.3, 8}\text{.9, 9}\text{.3, 11}\text{.6, 10}\text{.6, 9}\text{.0, 8}\text{.2,}\\ \text{9}\text{.4, 10}\text{.6, 9}\text{.5, 6}\text{.3, 9}\text{.1, 7}\text{.9, 9}\text{.7, 8}\text{.8, 6}\text{.9, 8}\text{.7, 9}\text{.0, 8}\text{.9, 9}\text{.3}\end{array}$

Concept used:

Categorical variables place the individual data into a category.

Calculation:

The given data are all numerical.

Circle graphs are only appropriate for the categorical variables.

Here the variable is not categorical so, it is not appropriate to use a circle graph.

Conclusion:

The reason behind bar graph will not work for these data is the variable is not categorical

b.

To draw: A stem plot for this data set.

The stem plot is:

  $\begin{array}{cc}\text{Stem}& \text{Leaf}\\ 6& \text{2 3 9 9}\\ 7& \text{7 9 9 9}\\ 8& \text{0 2 3 7 7 8 9 9}\\ 9& \text{0 0 0 1 1 2 3 3 3 4 5 6 7}\\ 10& \text{2 3 3 6 6 6 7 9}\\ 11& \text{0 2 4 6}\\ 12& \text{0 5 9}\end{array}$

Given:

The average wind speeds for one year at 44 climatic data centers around the United States are as follows:

  $\begin{array}{l}\text{9}\text{.0, 6}\text{.9, 9}\text{.1, 9}\text{.2, 10}\text{.2, 12}\text{.5, 12}\text{.0, 11}\text{.2, 12}\text{.9, 10}\text{.3, 10}\text{.6, 10}\text{.9, 8}\text{.7, 10}\text{.3, 11}\text{.0,}\\ \text{7}\text{.7, 11}\text{.4, 7}\text{.9, 9}\text{.6, 8}\text{.0, 10}\text{.7, 9}\text{.3, 7}\text{.9, 6}\text{.2, 8}\text{.3, 8}\text{.9, 9}\text{.3, 11}\text{.6, 10}\text{.6, 9}\text{.0, 8}\text{.2,}\\ \text{9}\text{.4, 10}\text{.6, 9}\text{.5, 6}\text{.3, 9}\text{.1, 7}\text{.9, 9}\text{.7, 8}\text{.8, 6}\text{.9, 8}\text{.7, 9}\text{.0, 8}\text{.9, 9}\text{.3}\end{array}$

Concept Used:

Draw a vertical line.

Put the digits of the tens to the left of the vertical line and the digits of the ones of every data value to the right of the vertical line

Calculation:

The stem plot is:

  $\begin{array}{cc}\text{Stem}& \text{Leaf}\\ 6& \text{2 3 9 9}\\ 7& \text{7 9 9 9}\\ 8& \text{0 2 3 7 7 8 9 9}\\ 9& \text{0 0 0 1 1 2 3 3 3 4 5 6 7}\\ 10& \text{2 3 3 6 6 6 7 9}\\ 11& \text{0 2 4 6}\\ 12& \text{0 5 9}\end{array}$

c.

To describe: The shape of the distribution

The shape of the distribution is unimodal and symmetric.

Given:

The average wind speeds for one year at 44 climatic data centers around the United States are as follows:

  $\begin{array}{l}\text{9}\text{.0, 6}\text{.9, 9}\text{.1, 9}\text{.2, 10}\text{.2, 12}\text{.5, 12}\text{.0, 11}\text{.2, 12}\text{.9, 10}\text{.3, 10}\text{.6, 10}\text{.9, 8}\text{.7, 10}\text{.3, 11}\text{.0,}\\ \text{7}\text{.7, 11}\text{.4, 7}\text{.9, 9}\text{.6, 8}\text{.0, 10}\text{.7, 9}\text{.3, 7}\text{.9, 6}\text{.2, 8}\text{.3, 8}\text{.9, 9}\text{.3, 11}\text{.6, 10}\text{.6, 9}\text{.0, 8}\text{.2,}\\ \text{9}\text{.4, 10}\text{.6, 9}\text{.5, 6}\text{.3, 9}\text{.1, 7}\text{.9, 9}\text{.7, 8}\text{.8, 6}\text{.9, 8}\text{.7, 9}\text{.0, 8}\text{.9, 9}\text{.3}\end{array}$

Calculation:

From part $\left(b\right)$ ,it can be seen that the most of the data values are in the middle of the stem plot.

Also, there is no gap in the stem plot which implies that there is no outlier in the data.

Conclusion:

The shape of the distribution is unimodal and symmetric.