Chapter 5.8, Problem 31PPSE

Solution

To calculate : A polynomial function that best fits the given data.

The model that best fits the data is $y=0.11x^4-1.91x^3+10.05x^2-15.82x+77.2$ .

Given information :

The data is:

Year	On-time Flights (%)
$\begin{array}{l}1998\\ 2000\\ 2002\\ 2004\\ 2006\end{array}$	$\begin{array}{l}77.20\\ 72.59\\ 82.14\\ 78.08\\ 75.45\end{array}$

Formula used:

  $\left(n+1\right)$ Principle: For any set of $n+1$ points in the coordinate plane that pass the vertical line test, there is a unique polynomial of degree at most $n$ that fits the point perfectly.

Calculation :

Let $x$ represent the years after 1998 and $y$ represent the On-time Flights (%).

$x$	$y$
$\begin{array}{l}0\\ 2\\ 4\\ 6\\ 8\end{array}$	$\begin{array}{l}77.20\\ 72.59\\ 82.14\\ 78.08\\ 75.45\end{array}$

Here, there are five set of points. From the $\left(n+1\right)$ principle, there is a unique polynomial of degree at most $4$ that fits the points perfectly.

So, the quartic model best fits the data.

Let $y=a{x}^4+b{x}^3+c{x}^2+dx+e$ .

Substitute the point $\left(0,77.20\right)$ .

  $\begin{array}{c}y=a{x}^4+b{x}^3+c{x}^2+dx+e\\ 77.20=a{\left(0\right)}^4+b{\left(0\right)}^3+c{\left(0\right)}^2+d\left(0\right)+e\\ 77.20=e\end{array}$

Substitute the point $\left(2,72.59\right)$ .

  $\begin{array}{c}y=a{x}^4+b{x}^3+c{x}^2+dx+e\\ 72.59=a{\left(2\right)}^4+b{\left(2\right)}^3+c{\left(2\right)}^2+d\left(2\right)+e\\ 72.59=16a+8b+4c+2d+e\end{array}$

Substitute the point $\left(4,82.14\right)$ .

  $\begin{array}{c}y=a{x}^4+b{x}^3+c{x}^2+dx+e\\ 82.14=a{\left(4\right)}^4+b{\left(4\right)}^3+c{\left(4\right)}^2+d\left(4\right)+e\\ 82.14=256a+64b+16c+4d+e\end{array}$

Substitute the point $\left(6,78.08\right)$ .

  $\begin{array}{c}y=a{x}^4+b{x}^3+c{x}^2+dx+e\\ 78.08=a{\left(6\right)}^4+b{\left(6\right)}^3+c{\left(6\right)}^2+d\left(6\right)+e\\ 78.08=1296a+216b+36c+6d+e\end{array}$

Substitute the point $\left(8,75.45\right)$ .

  $\begin{array}{c}y=a{x}^4+b{x}^3+c{x}^2+dx+e\\ 75.45=a{\left(8\right)}^4+b{\left(8\right)}^3+c{\left(8\right)}^2+d\left(8\right)+e\\ 75.45=4096a+512b+64c+8d+e\end{array}$

The augmented matrix for the above system of equations is:

  $\left[\begin{array}{llllll}0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 77.20\hfill \\ 16\hfill & 8\hfill & 4\hfill & 2\hfill & 1\hfill & 72.59\hfill \\ 256\hfill & 64\hfill & 16\hfill & 4\hfill & 1\hfill & 82.14\hfill \\ 1296\hfill & 216\hfill & 36\hfill & 6\hfill & 1\hfill & 78.08\hfill \\ 4096\hfill & 512\hfill & 64\hfill & 8\hfill & 1\hfill & 75.45\hfill \end{array}\right]$

Use calculator to solve the system of equations.

  $a=0.11,b=-1.91,c=10.15,d=-15.82,e=77.2$

Therefore, the model that best fits the data is $y=0.11x^4-1.91x^3+10.05x^2-15.82x+77.2$ .