Chapter 1, Problem 108CR

(a)

To determine

To Graph: The scatter plot for given data.

Explanation of Solution

Given Information:

Table of entrance exam score $x$ and grade point average $y.$

Exam score, x	Grade point y
75	2.4
82	3.0
90	3.6
65	2.0
70	2.1
88	3.5
93	3.9
69	2.0
80	2.8
85	3.3

Graph:

Scatter plot for the data:

  

(b)

To determine

To find: Linear regression of scatter plot for given data.

Answer to Problem 108CR

The linear regression of the graph is $y=0.1x-5.1$ and correlation coefficient is 0.98.

Explanation of Solution

Given Information:

Table of entrance exam score x and grade point average y

Formula used:

Slope intercept form:

  $\begin{array}{l}y=mx+C\\ m=\frac{y_2-y_1}{x_2-x_1}\end{array}$

Calculation:

The linear regression of the graph:

  $\begin{array}{l}y=mx+C\\ m=\frac{3.9-2.0}{93-69}\\ m=\frac{1.9}{24}=0.079\end{array}$

For value of C

Put (69, 2) in the equation, we get

  $\begin{array}{l}y=0.079x+C\\ 2=0.079\cdot (69)+C\\ 2=5.451+C\\ C=-3.451\end{array}$

Thus, the linear regression is $y=0.079x-3.451$

And correlation coefficient: $r=0.98$

(c)

To determine

To calculate: The grade average point for a student with an entrance exam score of 95.

Answer to Problem 108CR

The grade average point for a student with an entrance exam score of 95 is 4.05.

Explanation of Solution

Given Information:

Table of entrance exam score x and grade point average y

Formula used:

Slope intercept form:

  $\begin{array}{l}y=mx+C\\ m=\frac{y_2-y_1}{x_2-x_1}\end{array}$

Calculation:

  

The grade average point for a student with an entrance exam score of 95 is:

  $\begin{array}{l}y=0.079x-3.451\\ y=0.079\cdot 95-3.451\\ y=7.505-3.451\\ y=4.05\end{array}$

Thus, the grade average point for a student with an entrance exam score of 95 is:4.05