Chapter 11.4, Problem 60E

(a)

To determine

To graph: The scatter plot of the data and the model in the same viewing window and compare them.

Answer to Problem 60E

The model fits to the given set of data as all the points lies on the graph of model.

Explanation of Solution

Given information:

The table shows the mean annual salaries $S$ (in thousands of dollars) of a financial manager in the United States for the years $2006$ through $2012$ .

Year	Salary, $S$ (in thousands of dollars)
$2006$	$101.45$
$2007$	$106.20$
$2008$	$110.64$
$2009$	$113.73$
$2010$	$116.97$
$2011$	$120.45$
$2012$	$123.26$
$2013$	$126.66$

A model for the data is given by $S\left(t\right)=\frac{11.67+47.505t-2.6441t^2}{1+0.276t-0.0186t^2}$ where $t$ represents the year, with $t=6$ corresponding to $2006$ .

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 go to ${\text{Y}}_2$ and enter the right hand side of the function $y=\frac{11.67+47.505t-2.6441t^2}{1+0.276t-0.0186t^2}$ . Enter the keystrokes $\boxed{(}\boxed{11.67}\boxed{+}\boxed{47.505}\boxed{\text{X}}\boxed{-}\boxed{2.6441}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{)}\boxed{÷}\boxed{(}\boxed{1}\boxed{+}\boxed{0.276}\boxed{\text{X}}\boxed{-}\boxed{0.0186}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{)}$ .

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 (1)

The model fits to the given set of data as all the points lies on the graph of model.

(b)

To determine

To find: The mean salaries in $2014$ and $2015$ with the help of model.

Answer to Problem 60E

The number of ready reserves in $2014$ and $2015$ is equal to $1.0914$ and $1.099$ millions respectively.

Explanation of Solution

Given information:

The table shows the mean annual salaries $S$ (in thousands of dollars) of a financial manager in the United States for the years $2006$ through $2012$ .

Year	Salary, $S$ (in thousands of dollars)
$2006$	$101.45$
$2007$	$106.20$
$2008$	$110.64$
$2009$	$113.73$
$2010$	$116.97$
$2011$	$120.45$
$2012$	$123.26$
$2013$	$126.66$

A model for the data is given by $S\left(t\right)=\frac{11.67+47.505t-2.6441t^2}{1+0.276t-0.0186t^2}$ where $t$ represents the year, with $t=6$ corresponding to $2006$ .

Calculation:

As $t=6$ represents the year $2006$ . So, $t=14$ corresponds to the year $2014$ and $t=15$ corresponds to the year $2015$ .

Substitute $t=14$ in the model.

  $\begin{array}{c}S\left(t\right)=\frac{11.67+47.505t-2.6441t^2}{1+0.276t-0.0186t^2}\\ S\left(14\right)=\frac{11.67+47.505\left(14\right)-2.6441{\left(14\right)}^2}{1+0.276\left(14\right)-0.0186{\left(14\right)}^2}\\ S\left(14\right)=\frac{11.67+665.07-2.6441\left(196\right)}{1+3.864-0.0186\left(196\right)}\\ S\left(14\right)=\frac{11.67+665.07-518.2436}{1+3.864-3.6456}\end{array}$

Further simplify,

  $\begin{array}{c}S\left(14\right)=\frac{11.67+665.07-518.2436}{1+3.864-3.6456}\\ =\frac{158.4964}{1.2184}\\ =130.085\end{array}$

So, the mean salaries in $2014$ is equal to $130.085$ thousand dollars.

Substitute $t=15$ in the model.

  $\begin{array}{c}S\left(t\right)=\frac{11.67+47.505t-2.6441t^2}{1+0.276t-0.0186t^2}\\ S\left(15\right)=\frac{11.67+47.505\left(15\right)-2.6441{\left(15\right)}^2}{1+0.276\left(15\right)-0.0186{\left(15\right)}^2}\\ S\left(15\right)=\frac{11.67+712.575-2.6441\left(225\right)}{1+4.14-0.0186\left(225\right)}\\ S\left(15\right)=\frac{11.67+712.575-594.9225}{1+3.864-4.185}\end{array}$

Further simplify,

  $\begin{array}{c}S\left(15\right)=\frac{11.67+712.575-594.9225}{1+3.864-4.185}\\ =\frac{129.3225}{0.679}\\ =190.46\end{array}$

So, the mean salaries in $2015$ is equal to $190.46$ thousand dollars.

Therefore, the mean salaries in $2014$ and $2015$ is $130.085$ and $190.46$ thousand dollars respectively.

(c)

To determine

To find: The limit of the model as $t\to \infty$ and interpret its meaning.

Answer to Problem 60E

The limit of the model as $t\to \infty$ is equal to $142.156$ and it means that the mean salaries reaches to $142.156$ thousand dollars and becomes constant as many years passes.

Explanation of Solution

Given information:

The table shows the mean annual salaries $S$ (in thousands of dollars) of a financial manager in the United States for the years $2006$ through $2012$ .

Year	Salary, $S$ (in thousands of dollars)
$2006$	$101.45$
$2007$	$106.20$
$2008$	$110.64$
$2009$	$113.73$
$2010$	$116.97$
$2011$	$120.45$
$2012$	$123.26$
$2013$	$126.66$

A model for the data is given by $S\left(t\right)=\frac{11.67+47.505t-2.6441t^2}{1+0.276t-0.0186t^2}$ where $t$ represents the year, with $t=6$ corresponding to $2006$ .

Calculation:

Simplify the limit.

  $\begin{array}{c}\underset{t\to \infty }{\lim }S\left(t\right)=\underset{t\to \infty }{\lim }\left(\frac{11.67+47.505t-2.6441t^2}{1+0.276t-0.0186t^2}\right)\\ =\underset{t\to \infty }{\lim }\left(\frac{t^2\left(\frac{11.67}{t^2}+\frac{47.505}{t}-2.6441\right)}{t^2\left(\frac{1}{t^2}+\frac{0.276}{t}-0.0186\right)}\right)\\ =\underset{t\to \infty }{\lim }\left(\frac{\left(\frac{11.67}{t^2}+\frac{47.505}{t}-2.6441\right)}{\left(\frac{1}{t^2}+\frac{0.276}{t}-0.0186\right)}\right)\\ =\frac{\underset{t\to \infty }{\lim }\left(\frac{11.67}{t^2}+\frac{47.505}{t}-2.6441\right)}{\underset{t\to \infty }{\lim }\left(\frac{1}{t^2}+\frac{0.276}{t}-0.0186\right)}\end{array}$

Further simplify,

  $\begin{array}{c}\underset{t\to \infty }{\lim }S\left(t\right)=\frac{\underset{t\to \infty }{\lim }\left(\frac{11.67}{t^2}+\frac{47.505}{t}-2.6441\right)}{\underset{t\to \infty }{\lim }\left(\frac{1}{t^2}+\frac{0.276}{t}-0.0186\right)}\\ =\frac{0+0-2.6441}{\left(0+0-0.0186\right)}\\ =\frac{-2.6441}{-0.0186}\\ =142.156\end{array}$

Therefore, the limit of the model as $t\to \infty$ is equal to $142.156$ and it means that the mean salaries reaches to $142.156$ thousand dollars and becomes constant as many years passes.

(d)

To determine

To check: Whether the given model is best fit for the prediction of mean annual salaries in future years.

Answer to Problem 60E

No, this is not the best model for the calculation of mean annual salaries in future years as the mean salaries becomes zero at $x=18.209$ .

Explanation of Solution

Given information:

The table shows the mean annual salaries $S$ (in thousands of dollars) of a financial manager in the United States for the years $2006$ through $2012$ .

Year	Salary, $S$ (in thousands of dollars)
$2006$	$101.45$
$2007$	$106.20$
$2008$	$110.64$
$2009$	$113.73$
$2010$	$116.97$
$2011$	$120.45$
$2012$	$123.26$
$2013$	$126.66$

A model for the data is given by $S\left(t\right)=\frac{11.67+47.505t-2.6441t^2}{1+0.276t-0.0186t^2}$ where $t$ represents the year, with $t=6$ corresponding to $2006$ .

Calculation:

As observed from the graph of the model in part(a), the mean salaries at $x=18.209$ is zero.

This is not possible in real life.

Therefore, this model is not the best fit for the calculation of mean annual salaries in future years.