Chapter 7.3, Problem 95E

(a)

To determine

To calculate: The system of linear equations with the help of given data and then find the least square regression parabola.

Answer to Problem 95E

The equation of regression parabola is $y=\frac{33}{200}{x}^2-\frac{131}{20}x+103$ .

Explanation of Solution

Given information:

During the testing of a new automobile barking system, the speeds $x$ and the stopping distances $y$ were recorded in the table.

  

Formula Used:

Use the below formulas for system of equations

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

Calculation:

As per given table the points are $\left(30,55\right),\left(40,105\right),\left(50,188\right)$ .

Now calculate

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i}=x_1+x_2+x_3\\ =30+40+50\\ =120\end{array}$ , $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^2}=x_1^2+x_2^2+x_3^2\\ ={\left(30\right)}^2+{\left(40\right)}^2+{\left(50\right)}^2\\ =900+1600+2500\\ =5000\end{array}$ , $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^3}=x_1^3+x_2^3+x_3^3\\ ={\left(30\right)}^3+{\left(40\right)}^3+{\left(50\right)}^3\\ =27000+64000+125000\\ =216000\end{array}$

Also calculate

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^4}=x_1^4+x_2^4+x_3^4\\ ={\left(30\right)}^4+{\left(40\right)}^4+{\left(50\right)}^4\\ =810000+2560000+6250000\\ =9620000\end{array}$ , $\begin{array}{c}{\displaystyle \sum _{i=1}^n{y}_i}=y_1+y_2+y_3\\ =55+105+188\\ =348\end{array}$

Also calculate

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i{y}_i}=x_1{y}_1+x_2{y}_2+x_3{y}_3\\ =30\times 55+40\times 105+50\times 188\\ =1650+4200+9400\\ =15250\end{array}$ ,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^2{y}_i}=x_1^2{y}_1+x_2^2{y}_2+x_3^2{y}_3\\ ={\left(30\right)}^2\times 55+{\left(40\right)}^2\times 105+{\left(50\right)}^2\times 188\\ =49500+168000+470000\\ =687500\end{array}$

Now substitute all the values in the formula

  $\begin{array}{l}3c+120b+5000a=348\\ 120c+5000b+216000a=15250\\ 5000c+216000b+9620000a=687500\end{array}$

The system of equations is equivalent to

  $\begin{array}{l}3c+120b+5000a=348\dots \left(1\right)\\ 12c+500b+21600a=1525\dots \left(2\right)\\ 10c+432b+19240a=1375\dots \left(3\right)\end{array}$ .

Now solve the system of equations.

Solve equation $\left(1\right)$

  $\begin{array}{l}3c+120b+5000a=348\\ 3c=348-5000a-120b\\ c=\frac{348-5000a-120b}{3}\end{array}$

Substitute the value of $c$ in equation $\left(2\right)$

  $\begin{array}{l}12c+500b+21600a=1525\\ 12\left(\frac{348-5000a-120b}{3}\right)+500b+21600a=1525\\ 1392-20000a-480b+500b+21600a=1525\\ 1600a+20b=133\dots \left(4\right)\end{array}$

Substitute the value of $c$ in equation $\left(3\right)$

  $\begin{array}{l}10c+432b+19240a=1375\\ 10\left(\frac{348-5000a-120b}{3}\right)+432b+19240a=1375\\ 3480-50000a-1200b+1296b+57720a=4125\\ 7720a+96b=645\dots \left(5\right)\end{array}$

Solve equation $\left(4\right)$

  $\begin{array}{l}1600a+20b=133\\ 20b=133-1600a\\ b=\frac{133-1600a}{20}\end{array}$

Now substitute the value of $b$ in equation $\left(5\right)$

  $\begin{array}{l}7720a+96\left(\frac{133-1600a}{20}\right)=645\\ 77200a+6384-76800a=6450\\ 400a=66\\ a=\frac{33}{200}\end{array}$

Substitute the value of $a$ in equation $\left(4\right)$

  $\begin{array}{l}1600\left(\frac{33}{200}\right)+20b=133\\ 264+20b=133\\ 20b=-131\\ b=-\frac{131}{20}\end{array}$

Now substitute both the value of $b,c$ in equation $\left(1\right)$

  $\begin{array}{l}3c+120b+5000a=348\\ 3c+120\left(-\frac{131}{20}\right)+5000\left(\frac{33}{200}\right)=348\\ 3c-786+825=348\\ 3c=309\\ c=103\end{array}$ .

Therefore, the least square regression parabola

  $\begin{array}{l}y=a{x}^2+bx+c\\ y=\frac{33}{200}{x}^2-\frac{131}{20}x+103\end{array}$ .

Therefore, the equation of regression parabola is $y=\frac{33}{200}{x}^2-\frac{131}{20}x+103$ .

(b)

To determine

To Graph: The least square regression parabola with the points marked.

Answer to Problem 95E

The graph of the regression parabola with points marked is shown in Figure(1).

Explanation of Solution

Given information:

The equation of regression parabola is $y=\frac{33}{200}{x}^2-\frac{131}{20}x+103$ .

Graph:

The graph of the regression parabola is given below:

  

Figure(1)

(c)

To determine

To calculate: The estimate of the stopping distance for a speed of $70\text{mi}/\text{h}$ .

Answer to Problem 95E

The stopping distance is $453\text{mi}$ for the speed of $70\text{mi}/\text{h}$ .

Explanation of Solution

Given information:

The relation of stopping distance with speed $y=\frac{33}{200}{x}^2-\frac{131}{20}x+103$ .

Calculation:

As the relation of stopping distance with speed is $y=\frac{33}{200}{x}^2-\frac{131}{20}x+103$ .

Now substitute $70$ for $x$ in above equation

  $\begin{array}{c}y=\frac{33}{200}{\left(70\right)}^2-\frac{131}{20}\left(70\right)+103\\ =808.5-458.5+103\\ =453\end{array}$ .

Therefore, the stopping distance is $453\text{mi}$ for the speed of $70\text{mi}/\text{h}$ .