Chapter 7.3, Problem 102E

a

To determine

To Find: The least squares regression parabola

Answer to Problem 102E

The equation of parabola is $y=-0.015x^2+3.25x-106$

Explanation of Solution

Given: The number of femalesx and the percent of females y

Number $x$	Percent $y$
$120$	$68$
$140$	$55$
160	30

Concept Used:

Gaussian Elimination is the process of using valid row operations on a matrix until it is in reduced row echelon form

Calculation:

Let least squares regression parabola

  $y=a{x}^2+bx+c$

When $x=120,y=68$

Substituting $x=120,y=68$ on the equation of parabola

  $\begin{array}{l}$ 68={(120)}^{2}a+120b+c$68=14400a+120b+c \end{array}$

Substituting $x=140,y=55$ on the equation of parabola

  $\begin{array}{l}$ 55={(140)}^{2}a+140b+c$55=19600a+140b+c \end{array}$

Substituting $x=160,y=30$ on the equation of parabola

  $\begin{array}{l}$ 30={(160)}^{2}a+160b+c$30=25600a+160b+c \end{array}$

This produces the following system of linear equations

  $\begin{array}{l}x=160,y=30\text{ Equation }1\\ 19600a+140b+c=55\text{ Equation }2\\ 25600a+160b+c=30\text{ Equation }3\end{array}$

Adding $-1$ times the first equation to the second equation

  $\begin{array}{l}-14400a-120b-c=-68 \\ 19600a+140b+c=55 \end{array}$

  ________________________________.

  $5200a+20b=-13$

Adding $-1$ times the first equation to the second equation

produces a new second equation

  $\begin{array}{l}14400a+120b+c=68 \text{ Equation }1\\ 5200a+20b=-13 \text{ Equation }2\\ 25600a+160b+c=30\text{ Equation }3\end{array}$

Adding $-1$ times the first equation to the third equation

  $\begin{array}{l}-14400a-120b-c=-68 \\ 25600a+160b+c=30\end{array}$

   _________________________________.

  $1200a+40b=-38$

Adding $-1$ times the first equation to the third equation produces a new third equation

  $\begin{array}{l}14400a+120b+c=68 \text{ Equation }1\\ 5200a+20b=-13 \text{ Equation }2\\ 1200a+40b=-38 \text{ Equation }3\end{array}$

Adding $-2$ times the second equation to the third equation

  $\begin{array}{l}-10400a-40b=26 \\ 11200a+40b=-38 \end{array}$

  ______________________.

  $800a=-12$

Adding $-2$ times the second equation to the third equation produces a new third equation

  $\begin{array}{l}14400a+120b+c=68 \text{ Equation }1\\ 5200a+20b=-13 \text{ Equation }2\\ 800a=-12 \text{ Equation }3\end{array}$

Multiplying the third equation by $\frac{1}{800}$ produces a new third equation

  $\begin{array}{l}14400a+120b+c=68 \text{ Equation }1\\ 5200a+20b=-13 \text{ Equation }2\\ a=-0.015 \text{ Equation }3\end{array}$

From $\text{Equation 3}$ we know the value of a . To solve for b , substitute $a=-0.015$ into equation $2$ to obtain

  $b=3.25$

Finally, substitute $a=-0.015$ and $b=3.25$ into $\text{Equation 1}$ to obtain $c=-106$

Substitute $a=-0.015$ , $b=3.25$ , $c=-106$ on $y=a{x}^2+bx+c$

The equation of parabola is $y=-0.015x^2+3.25x-106$

b

To determine

Graph of:parabola $y=-0.015x^2+3.25x-106$

Explanation of Solution

Given information: Equation of parabola $y=-0.015x^2+3.25x-106$

Graph:

  

c

To determine

To Find: percent of females that had offering when there were $170females$

Answer to Problem 102E

percent of females is $13\%$

Explanation of Solution

Given information: equation of parabola is $y=-0.015x^2+3.25x-106$

Percent at $x=170$

Calculation:

  $y=-0.015x^2+3.25x-106$

Percent when

  $\begin{array}{l}x=170\\ $ y=-0.015{(170)}^2+3.25\times 170-106 $y=13\end{array}$