Houghton Mifflin Harcourt Algebra 1, 2015
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Exercise 2 Page 377

Start by making a scatter plot.

Scatter Plot:

Correlation Coefficient: Close to 1
Prediction: 85.3

The table shown below represents the relation between the number of books read and the students' grades.

Books 0 0 0 0 1 1 1 2 2 3 5 8 10 14 20
Grade 64 68 69 72 71 74 76 75 79 85 86 91 94 99 98

In order to find an equation for the line of fit, we will make a scatter plot. Let x represent the number of books and y represent the grade.

One possible line of fit can be drawn as shown above. We will write the equation in the slope-intercept form. y= mx+ b In this form, m is the slope and b is the y-intercept. First, let's find the slope by using the Slope Formula. m=y_2-y_1/x_2-x_1 Here, (x_1,y_1) and (x_2,y_2) represent the points that are on the line. Since the points (0,72) and (14,99) appear to be on the line, we will substitute them into the formula.

m=y_2-y_1/x_2-x_1
m=99- 72/14- 0
m=27/14
m=1.92857
m=1.9

Thus, the slope for the line o fit is 1.9. y=1.9x+ b Next, we will determine the y-intercept. Since the line passes through the point (0,72), we can immediately determine the y-intercept as 72. Thus, we can write the equation as the following. y=1.9x+72 When we examine the scatter plot, we notice that points lie closely to the line and we also know that the slope of the line 1.9 is positive. Thus, there is a strong positive correlation and the correlation coefficient is close to 1. Now, let's predict the grade of a student who read 7 books.

y=1.9x+72
y=1.9( 7)+72
y=13.3+72
y=85.3

Our prediction of the grade of the student is 85.3.