Chapter 1.7, Problem 24E

(a)

To determine

To graph: The scatter plot of the data with $t=9$ corresponding to $2009$ .

Explanation of Solution

Given information:

The populations $P$ (in thousands) of Wyoming from $2009$ through $2013$ are shown in table.

Year	Population, $P$ (in thousands)
$2009$	$560$
$2010$	$564$
$2011$	$568$
$2012$	$577$
$2013$	$583$

Graph:

To graph the points on scatter plot, follow the steps using graphing calculator.

First press the $\boxed{2\text{nd}}$ key and then $\boxed{\text{Y}=}$ and make sure that Plot $1$ is ON and choose the scatter plot icon in type.

Go to ${\text{Y}}_1$ and then press the key $\boxed{\text{STAT}}$ and then in edit enter the data in ${\text{L}}_1$ and ${\text{L}}_2$ .

Now, press the $\boxed{\text{ZOOM}}\boxed{9}$ key and $\boxed{\text{TRACE}}$ key to produce the graph in standard window as shown in Figure (1).

  

Figure (1)

(b)

To determine

To find: The linear model for the data with the help of regression of graphing utility.

Answer to Problem 24E

The linear model for the data is equal to $P=5.9t+505.5$ .

Explanation of Solution

Given information:

The populations $P$ (in thousands) of Wyoming from $2009$ through $2013$ are shown in table.

Year	Population, $P$ (in thousands)
$2009$	$560$
$2010$	$564$
$2011$	$568$
$2012$	$577$
$2013$	$583$

Calculation:

For the regression linear model of data follow the steps below:

First press the $\boxed{2\text{nd}}$ key and then $\boxed{\text{Y}=}$ and make sure that Plot $1$ is ON and choose the scatter plot icon in type.

Go to ${\text{Y}}_1$ and then press the key $\boxed{\text{STAT}}$ and then in edit enter the data in ${\text{L}}_1$ and ${\text{L}}_2$ .

Now enter the keystrokes $\boxed{\text{STAT}}\boxed{\to }\boxed{4}$ and press the $\boxed{\text{ENTER}}$ key for the regression.

  

Figure(2)

Therefore, the linear model for the data is equal to $P=5.9t+505.5$ .

(c)

To determine

To find: The graph of model and the scatter plot of data in same viewing window and check the model is best fit or not.

Answer to Problem 24E

The model is best fit for the data as all the points of the data lies on the graph of the model.

Explanation of Solution

Given information:

The populations $P$ (in thousands) of Wyoming from $2009$ through $2013$ are shown in table.

Year	Population, $P$ (in thousands)
$2009$	$560$
$2010$	$564$
$2011$	$568$
$2012$	$577$
$2013$	$583$

Calculation:

To graph the points on scatter plot, follow the steps using graphing calculator.

First press the $\boxed{2\text{nd}}$ key and then $\boxed{\text{Y}=}$ and make sure that Plot $1$ is ON and choose the scatter plot icon in type.

Go to ${\text{Y}}_1$ and then press the key $\boxed{\text{STAT}}$ and then in edit enter the data in ${\text{L}}_1$ and ${\text{L}}_2$ .

Now press the key $\boxed{\text{Y}=}$ and go to ${\text{Y}}_2$ and enter the right hand side of the function $y=5.9t+505.5$ . Enter the keystrokes $\boxed{5.9}\boxed{\text{X}}\boxed{+}\boxed{505.5}$ .

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph in standard window as shown in Figure (1).

  

Figure (3)

(d)

To determine

To find: The table that shows the actual values of $P$ and the values of $P$ given by the model and compare them.

Answer to Problem 24E

The table that shows the actual values of $P$ and the values of $P$ is given below:

Year	Actual value, $P$ (in thousands)	Model value, $P$ (in thousands)
$2009$	$560$	$558.6$
$2010$	$564$	$564.5$
$2011$	$568$	$570.4$
$2012$	$577$	$576.3$
$2013$	$583$	$582.2$

Explanation of Solution

Given information:

The populations $P$ (in thousands) of Wyoming from $2009$ through $2013$ are shown in table.

Year	Population, $P$ (in thousands)
$2009$	$560$
$2010$	$564$
$2011$	$568$
$2012$	$577$
$2013$	$583$

Calculation:

As found in part(b), the model for the best fit of data is equal to $P=37.28t+8416.33$ .

Substitute $t=9,10,11,12,13$ in the model for population. The observed model values for population are given below:

Year	Actual value, $P$ (in thousands)	Model value, $P$ (in thousands)
$2009$	$560$	$558.6$
$2010$	$564$	$564.5$
$2011$	$568$	$570.4$
$2012$	$577$	$576.3$
$2013$	$583$	$582.2$

The actual population is sometimes $1.5$ thousands smaller than the model population and sometimes greater.

(e)

To determine

To find: The population of Wyoming in $2050$ and check whether the answer is reasonable or not.

Answer to Problem 24E

The population of Wyoming in $2050$ is equal to $800.5$ thousands.

Explanation of Solution

Given information:

The populations $P$ (in thousands) of Wyoming from $2009$ through $2013$ are shown in table.

Year	Population, $P$ (in thousands)
$2009$	$560$
$2010$	$564$
$2011$	$568$
$2012$	$577$
$2013$	$583$

Calculation:

As found in part(b), the model for the best fit of data is equal to $P=5.9t+505.5$ and the year $2050$ is represented by $t=50$ .

Substitute $50$ for $t$ in the model.

  $\begin{array}{c}P=5.9t+505.5\\ =5.9\left(50\right)+505.5\\ =800.5\end{array}$

Therefore, the population of Wyoming in $2050$ is equal to $800.5$ thousands.

The population of New Jersey in $2050$ looks reasonable as the value is increasing on a constant rate or linearly.