Sabino took the ACT in 2003 and received a composite score of 22. If the mean and standard deviation for that year were, respectively, 20.8 and 4.8, what percent of all of the students who took the exams scored better than him?
To find the percentage of students who scored better than Sabino, we need to find the z-score for Sabino's composite score and then use a standard normal distribution table to find the percentage.
First, we need to calculate the z-score using the formula:
z = (x - μ) / σ
Where:
x = Sabino's composite score = 22
μ = mean composite score = 20.8
σ = standard deviation for composite scores = 4.8
z = (22 - 20.8) / 4.8
z = 1.2 / 4.8
z = 0.25
Now that we have the z-score, we can use a standard normal distribution table to find the percentage of students who scored better than Sabino.
Looking up the z-score of 0.25 in the standard normal distribution table, we find that the corresponding percentage is 59.87%.
Therefore, approximately 59.87% of the students who took the exams scored better than Sabino.