The best predicted grade for a student with 11 absences is approximately 61.01.
To predict the grade for a student with 11 absences, assuming a significant correlation between the number of absences (x) and final grades (y) based on the provided data, you can use linear regression to find the best predicted grade.
Let's assume the data represents a linear relationship between the number of absences and final grades.
I'll demonstrate how to find the equation of the line using linear regression and then use that equation to predict the grade for a student with 11 absences.
Suppose the data looks something like this:
Absences (x) | Final Grades (y)
--------------------------------
5 80
7 75
10 70
12 65
15 60
18 55
20 50
22 45
25 40
Let's calculate the equation of the line using linear regression:
Calculate the mean of x and y:
Mean of x (absences) = (5 + 7 + 10 + 12 + 15 + 18 + 20 + 22 + 25) / 9 = 14
Mean of y (grades) = (80 + 75 + 70 + 65 + 60 + 55 + 50 + 45 + 40) / 9 = 60
Calculate the slope (m):
m = Σ((x - mean of x) * (y - mean of y)) / Σ((x - mean of x)^2)
Where Σ denotes the sum of values.
Calculate the slope:
Σ((x - mean of x) * (y - mean of y)) = (5-14)*(80-60) + (7-14)*(75-60) + ... + (25-14)*(40-60) = -180
Σ((x - mean of x)^2) = (5-14)^2 + (7-14)^2 + ... + (25-14)^2 = 270
Now, plug these values into the formula for the slope:
m = -180 / 270 = -0.6667 (approximately -0.67, rounded to two decimal places)
Calculate the y-intercept (b):
b = mean of y - m * mean of x
b = 60 - (-0.67) * 14
b = 60 + 9.38
b = 69.38
Now that we have the equation of the line (y = mx + b), which is:
y = -0.67x + 69.38
We can predict the grade for a student with 11 absences:
y = -0.67 * 11 + 69.38
y ≈ 61.01
Therefore, the best predicted grade for a student with 11 absences is approximately 61.01.