Chapter 2.5, Problem 12PPSE

Solution

(a)

To find: Use calculator to find the line of best fit.

The line that best fit in the given data is $y=39758.667+2053.1667x$ .

Given information:

The table for pork production year 2000 to 2007 is given below:

year	2000	2001	2002	2003	2004	2005	2006	2007
Production(metric ton)	40,475	42,010	43,413	45,331	47,177	50,254	52,407	54,491

Calculation:

Let the year $2000$ represent time $t=0$ , year $2001$ represent time $t=1$ and so on.

The table will be now as given below:

year	t=0	t=1	t=2	t=3	t=4	t=5	t=6	t=7
Production(metric ton)	40,475	42,010	43,413	45,331	47,177	50,254	52,407	54,491

The equation of line is $y=a+bx$ .

The equations to find $a$ and $b$ are

  $\begin{array}{l}{\displaystyle \sum y=an+b{\displaystyle \sum x}\mathrm{....}(1)}\\ {\displaystyle \sum xy=a{\displaystyle \sum x+b{\displaystyle \sum x^2}\mathrm{....}(2)}}\end{array}$

The values of $a$ and $b$ will be calculated using the following table.

$x$	$y$	$x^2$	$x\cdot y$
0	40475	0	0
1	42010	1	42010
2	43413	4	86826
3	45331	9	135993
4	47177	16	188708
5	50254	25	251270
6	52407	36	314442
7	54491	49	381437
---	---	---	---
${\displaystyle \sum x=18}$	${\displaystyle \sum y}$ =375558	${\displaystyle \sum x^2}$ =140	${\displaystyle \sum xy}$ =1400686

Substituting these values in equation $\left(1\right)$ and $\left(2\right)$

  $\begin{array}{l}8a+28b=375558\\ 4a+14b=187779\mathrm{....}\left(3\right)\\ 28a+140b=1400686\\ 4a+20b=200098\mathrm{.....}(4)\end{array}$

Solve these two equations using Elimination method:

Subtract equation $\left(4\right)$ from $\left(3\right)$ .

  $\begin{array}{l}4a-4a+14b-20b=187779-200098\\ -6b=-12319\\ b=\frac{-12319}{-6}\\ b=2053.1667\end{array}$

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

  $\begin{array}{l}4a+20\left(2053.1667\right)=200098\\ 2a+10\left(2053.1667\right)=100049\\ 2a+20531.667=100049\\ 2a=100049-20531.667\\ 2a=79517.3333\\ a=\frac{79517.3333}{2}\\ a=39758.667\end{array}$

Now substituting the values of $a$ and $b$ in the equation $y=a+bx$ :

  $y=39758.667+2053.1667x$

The line that best fit in the given data is $y=39758.667+2053.1667x$ .

(b)

To find: Use linear relation to find how many metric tons pork will be produced in year 2025.

  $91807.8345$ metric tons pork will be produced in year 2025.

Given information:

The line that best fit in the given data is $y=39758.667+2053.1667x$ .

Here $y$ represents the pork in metric tons and $x$ represents the year.

Calculation:

Let the year $2000$ represent time $t=0$ , year $2001$ represent time $t=1$ and so on.

The table will be now as given below:

year	t=0	t=1	t=2	t=3	t=4	t=5	t=6	t=7
Production(metric ton)	40,475	42,010	43,413	45,331	47,177	50,254	52,407	54,491

Year 2025 will be $t=25$

Substitute $x=25$ in $y=39758.667+2053.1667x$

  $\begin{array}{l}y=39758.667+2053.1667\cdot 25\\ y=39758.667+51329.1675\\ y=91807.8345\end{array}$

  $91807.8345$ metric tons pork will be produced in year 2025.

(c)

To find: Use linear relation to when the production will reach $100,000$ metric ton.

The pork production will reach $100,000$ metric ton in 30 years that is in the year $2030$ .

Given information:

The line that best fit in the given data is $y=39758.667+2053.1667x$ .

Here $y$ represents the pork in metric tons and $x$ represents the year.

Calculation:

Let the year $2000$ represent time $t=0$ , year $2001$ represent time $t=1$ and so on.

The table will be now as given below:

year	t=0	t=1	t=2	t=3	t=4	t=5	t=6	t=7
Production(metric ton)	40,475	42,010	43,413	45,331	47,177	50,254	52,407	54,491

Substitute $y=100,000$ in the equation $y=39758.667+2053.1667x$ .

  $\begin{array}{l}100,00=39758.667+2053.1667x\\ 100,000-39758.667=2053.1667x\\ 60241.333=2053.1667x\\ x=\frac{60241.333}{2053.1667}\\ x=29.34\\ \end{array}$

The pork production will reach $100,000$ metric ton in 30 years that is in the year $2030$ .