Chapter 7.3, Problem 72E

a.

To determine

Data analysis on wildlife.

Answer to Problem 72E

  $\boxed{y=-0.0075x^2+1.3024x+19.8587}$

Explanation of Solution

Given information:

A wildlife management team studied the reproductive rates of deer in three tracts of a wildlife preserve. Each tract contained 5 acres. In each tract, the number of females $x$ and the percent of females $y$ that had offspring the following year were recorded. The table shows the results.

  

Use the data to create a system of linear equations. Then find the least squares regression parabola for the data by solving the system.

Calculation:

Consider the table showing the number of females $x$ and the percent of females $y$ .

  

We have to set up the system of equations for data to find the least squares regression parabola.

Use the following system of equations.

  $\begin{array}{l}nc+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)a=\left({\displaystyle \sum _{i=1}^n{y}_i}\right)\\ \left({\displaystyle \sum _{i=1}^n{x}_i}\right)c+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^3}\right)a=\left({\displaystyle \sum _{i=1}^n{x}_i{y}_i}\right)\\ \left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)c+\left({\displaystyle \sum _{i=1}^n{x}_i{}^3}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^4}\right)a=\left({\displaystyle \sum _{i=1}^n{x}_i{}^2{y}_i}\right)\end{array}$

Now to find the least squares regression parabola $y=a{x}^2+bx+c$ solve the system for $a,b.c$

Given $n=3$ now solve the system,

  ${\displaystyle \sum _{i=1}^n{x}_i},{\displaystyle \sum _{i=1}^n{x}_i{}^2},{\displaystyle \sum _{i=1}^n{x}_i{}^3}{\displaystyle \sum _{i=1}^n{x}_i{}^4},{\displaystyle \sum _{i=1}^n{y}_i},{\displaystyle \sum _{i=1}^n{x}_i{y}_{i}and}{\displaystyle \sum _{i=1}^n{x}_i{}^2{y}_i}$

To solve ${\displaystyle \sum _{i=1}^n{x}_i}$ we will find the sum of all elements of first component given in the data table.

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{x}_i}={\displaystyle \sum _{i=1}^3{x}_i}\\ =100+120+140\\ =360\end{array}$

Similarly solve for another values,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{x}_i{}^2}={\displaystyle \sum _{i=1}^3{x}_i{}^2}\\ ={\left(100\right)}^2+{\left(120\right)}^2+{\left(140\right)}^2\\ =10000+14400+19600\\ =44000\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{x}_i{}^3}={\displaystyle \sum _{i=1}^3{x}_i{}^3}\\ ={\left(100\right)}^3+{\left(120\right)}^3+{\left(140\right)}^3\\ =1000000+1728000+2744000\\ =5472000\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{x}_i{}^4}={\displaystyle \sum _{i=1}^3{x}_i{}^4}\\ ={\left(100\right)}^4+{\left(120\right)}^4+{\left(140\right)}^4\\ =100000000+207360000+3841600000\\ =691520000\end{array}$

To solve ${\displaystyle \sum _{i=1}^n{y}_i}$ we will find the sum of all elements of second component given in the data table.

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{y}_i}={\displaystyle \sum _{i=1}^3{y}_i}\\ =75+68+55\\ =198\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{x}_i{y}_i}={\displaystyle \sum _{i=1}^3{x}_i{y}_i}\\ =75\times 100+68\times 120+55\times 144\\ =23360\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{x}_i{}^2{y}_i}={\displaystyle \sum _{i=1}^3{x}_i{}^2{y}_i}\\ =75\times {\left(100\right)}^2+68\times {\left(120\right)}^2+55\times {\left(144\right)}^2\\ =750000+979200+1078000\\ =2807200\end{array}$

Now substitute the calculated values in ${\displaystyle \sum _{i=1}^n{x}_i},{\displaystyle \sum _{i=1}^n{x}_i{}^2},{\displaystyle \sum _{i=1}^n{x}_i{}^3}{\displaystyle \sum _{i=1}^n{x}_i{}^4},{\displaystyle \sum _{i=1}^n{y}_i},{\displaystyle \sum _{i=1}^n{x}_i{y}_{i}and}{\displaystyle \sum _{i=1}^n{x}_i{}^2{y}_i}$ we get,

  $\begin{array}{l}nc+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)a=\left({\displaystyle \sum _{i=1}^n{y}_i}\right)\\ \left({\displaystyle \sum _{i=1}^n{x}_i}\right)c+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^3}\right)a=\left({\displaystyle \sum _{i=1}^n{x}_i{y}_i}\right)\\ \left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)c+\left({\displaystyle \sum _{i=1}^n{x}_i{}^3}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^4}\right)a=\left({\displaystyle \sum _{i=1}^n{x}_i{}^2{y}_i}\right)\end{array}$ system of equations.

  $\begin{array}{l}3c+360b+44000a=198\\ 360c+44000b+5472000a=23360\\ 44000c+5472000b+691520000a=2807200\end{array}$

Now solve the system for $a,b.c$

Multiply first equation by $\frac{1}{3}$

  $\begin{array}{l}c+120b+14666.667a=\mathrm{66.......}\left(1\right)\\ 360c+44000b+5472000a=\mathrm{23360.......}\left(2\right)\\ 44000c+5472000b+691520000a=\mathrm{2807200......}\left(3\right)\\ \end{array}$

Now multiply first equation by $-360$ and then add to second equation.

  $\frac{\begin{array}{l}-360c-43200b-5280000a=-23760\\ 360c+44000b+5472000a=23360\end{array}}{800b+192000=-400}$

Now new system of equations is,

  $\begin{array}{l}c+120b+14666.667a=\mathrm{66.......}\left(1\right)\\ 800b+192000=-\mathrm{400............}\left(2\right)\\ 44000c+5472000b+691520000a=\mathrm{2807200......}\left(3\right)\end{array}$

Multiply first equation by $-44000$ and add to the third equation.

  $\frac{\begin{array}{l}-44000c-5280000b-645333333a=-2904000\\ 44000c+5472000b+691520000a=2807200\end{array}}{192000b+46186667a=-96800}$

Now new system of equations is,

  $\begin{array}{l}c+120b+14666.667a=\mathrm{66.......}\left(1\right)\\ 800b+192000=-\mathrm{400............}\left(2\right)\\ 192000b+46186667a=-\mathrm{96800......}\left(3\right)\\ \end{array}$

Multiply second equation by $\frac{1}{800}$ and third by $\frac{1}{192000}$

  $\begin{array}{l}\frac{1}{800}\left(800b+192000=-400\right)\\ b+24a-0.5\\ \frac{1}{192000}\left(192000b+46186667a=-96800\right)\\ b+240.55556a=-0.50417\end{array}$

Now new system of equations is,

  $\begin{array}{l}c+120b+14666.667a=\mathrm{66.......}\left(1\right)\\ b+240a=-\mathrm{0.5..........}\left(2\right)\\ b+240.55556a=-\mathrm{0.50417.......}\left(3\right)\end{array}$

Multiply second equation by $-1$ and add to third equation.

  $\frac{\begin{array}{l}-b-240a=0.5\\ b+240.55556a=-0.50417\end{array}}{0.55556a=-0.00417}$

Now new system of equations is,

  $\begin{array}{l}c+120b+14666.667a=\mathrm{66.......}\left(1\right)\\ b+240a=-\mathrm{0.5..........}\left(2\right)\\ 0.55556a=-\mathrm{0.00417.......}\left(3\right)\end{array}$

Multiply third equation by $\frac{1}{0.55556}$

  $\begin{array}{l}\frac{1}{0.55556}\left(0.55556a=-0.00417\right)\\ a=-0.00751\end{array}$

Now substitute $a=-0.00751$ in second equation to solve for $b$

  $\begin{array}{l}b+240a=-0.5\\ b+240\left(-0.00751\right)=-0.5\\ b-1.8024=-0.5\\ b=1.3024\end{array}$

Now substitute $b=1.3024$ in first equation to solve for $c$

  $\begin{array}{l}c+120b+14666.667a=66\\ c+120\left(1.3024\right)+14666.667\left(-0.00751\right)=66\\ c+156.288-110.1467=66\\ c=66-156.288+110.1467\\ c=19.8587\end{array}$

Hence, the least squares regression parabola $y=a{x}^2+bx+c$ is,

  $\boxed{y=-0.0075x^2+1.3024x+19.8587}$

b.

To determine

Data analysis on wildlife.

Answer to Problem 72E

The result is shown on graph.

Explanation of Solution

Given information:

A wildlife management team studied the reproductive rates of deer in three tracts of a wildlife preserve. Each tract contained 5 acres. In each tract, the number of females $x$ and the percent of females $y$ that had offspring the following year were recorded. The table shows the results.

  

Use a graphing utility to graph the parabola and the data in the same viewing window.

Calculation:

The graph of the parabola and data in the same window as follows,

  

  

Hence, the result is shown on graph.

c.

To determine

Data analysis on wildlife.

Answer to Problem 72E

  $\boxed{24.5167\%}$

Explanation of Solution

Given information:

A wildlife management team studied the reproductive rates of deer in three tracts of a wildlife preserve. Each tract contained 5 acres. In each tract, the number of females $x$ and the percent of females $y$ that had offspring the following year were recorded. The table shows the results.

  

Use the model to estimate the percent of females that had offspring when there were $170$ females.

Calculation:

Model to estimate the percent of females that had offspring when there were $170$ females.

Put $x=170$ in equation $y=-0.0075x^2+1.3024x+19.8587$ and solve for $x$ .

  $\begin{array}{l}y=-0.0075x^2+1.3024x+19.8587\\ =-0.0075{\left(170\right)}^2+1.3024\left(170\right)+19.8587\\ =-216.75+221.408+19.8587\\ =24.5167\%\end{array}$

Hence, the percent of females that had offspring $\boxed{24.5167\%}$

d.

To determine

Data analysis on wildlife.

Answer to Problem 72E

  $\boxed{187}$

Explanation of Solution

Given information:

A wildlife management team studied the reproductive rates of deer in three tracts of a wildlife preserve. Each tract contained 5 acres. In each tract, the number of females $x$ and the percent of females $y$ that had offspring the following year were recorded. The table shows the results.

  

Use the model to estimate the number of females when $40\%$ of the females had offspring.

Calculation:

The model to estimate the number of females when $40\%$ of the females had offspring.

Put $y=0.4$ in equation $y=-0.0075x^2+1.3024x+19.8587$ .

  $\begin{array}{l}y=-0.0075x^2+1.3024x+19.8587\\ 0.4=-0.0075x^2+1.3024x+19.8587\\ 0.0075x^2-1.3024x-19.8587+0.4=0\\ 0.0075x^2-1.3024x-19.4587=0\\ x=\frac{1.3024\pm \sqrt{{\left(-1.3024\right)}^2-\left(4\times 0.0075\times \left(-19.4587\right)\right)}}{2\times 0.0075}\\ x=187.28\end{array}$

  $x=-13.95$

  $x=-13.95$ cannot be considered.

Hence, the number of females when $40\%$ of the females had offspring is $\boxed{187}$ .