Chapter 8, Problem 15STP

a.

To determine

A scatter plot and line of fit for the following table data:

Margo’s keyboarding speed
Week	Speed
$1$	$18$
$2$	$21$
$3$	$26$
$4$	$28$
$5$	$35$

Answer to Problem 15STP

A scatter plot and line of fit is drawn as:

Explanation of Solution

Given:

Margo’s keyboarding speed
Week	Speed
$1$	$18$
$2$	$21$
$3$	$26$
$4$	$28$
$5$	$35$

Concept Used:

The slope intercept form is:

  $y=mx+b$

Where,

Slope: $m$

  $y$ -intercept: $b$

Calculation:

Consider;

Week: $x$

Speed: $y$

Now, take the data points as:

  $\begin{array}{l}A=\left(x_1,y_1\right)=\left(1,18\right)\\ B=\left(x_2,y_2\right)=\left(2,21\right)\\ C=\left(x_3,y_3\right)=\left(3,26\right)\\ D=\left(x_4,y_4\right)=\left(4,28\right)\\ E=\left(x_5,y_5\right)=\left(5,35\right)\end{array}$

Now, the slope is calculated as:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{21-18}{2-1}\\ m=\frac{3}{1}\\ m=3\end{array}$

So, the slope intercept eq. can be written as:

  $\begin{array}{l}y=mx+b\\ y=3x+b\end{array}$

Now, need to put $\left(2,21\right)$ in $y=3x+b$ :

  $\begin{array}{l}y=3x+b\\ 21=3\times 2+b\\ b=21-6\\ b=15\end{array}$

So, the final slope-intercept form is written as:

  $\begin{array}{l}y=3x+b\\ y=3x+15\end{array}$

Now, the scatter plot of the data points and its line of fit eq. is drawn as:

Conclusion:

Hence, the scatter plot and the line of fit is drawn as above.

b.

To determine

The speed value of the Margo’s keyboarding speed after the six week of class.

Answer to Problem 15STP

The speed value of the Margo’s keyboarding speed after the six week of class is $33wpm$

Explanation of Solution

Given:

  $x=6$

Concept Used:

The slope intercept form is:

  $y=mx+b$

Where,

Slope: $m$

  $y$ -intercept: $b$

Calculation:

Consider;

Week: $x$

Speed: $y$

The slope-intercept form is taken from above SUB-PART:

  $\begin{array}{l}y=3x+b\\ y=3x+15\end{array}$

Put the value of the week $x=6$

  $\begin{array}{l}y=3x+b\\ y=3x+15\\ y=3\times 6+15\\ y=18+15\\ y=33\end{array}$

So the speed is estimated as $y=33wpm$

Conclusion:

Hence, the speed is estimated as $y=33wpm$